# ASF Puzzles & Conundrums Thread



## Kauri (22 November 2005)

*Re: ASF joke thread*

Which one is bigger....A or B..?


----------



## wayneL (22 November 2005)

*Re: ASF joke thread*



			
				Kauri said:
			
		

> Which one is bigger....A or B..?




Same size?


----------



## Kauri (23 November 2005)

*Re: ASF joke thread*



			
				wayneL said:
			
		

> Same size?




  Identical..


----------



## RichKid (5 December 2005)

Inspired by Joe's jokes thread, this is for riddles, puzzles, conundrums and that type of thing.


----------



## bellenuit (20 April 2015)

The above problem has gone viral around the internet over the past few weeks, so many of you may have seen it. It just requires some straightforward logic, no math.

Based on some of the responses I have seen, I would add the following assumptions to stop people being wrong tracked.

1. Albert knows that Bernard has been told the day of the birthday and Bernard knows that Albert has been told the month.
2. Each hears the statement the other makes. 
3. Both Albert and Bernard have sufficient information to make the assertions they made at the time they were made and each assertion is correct. 

Can you work out Cheryl's birthday?


----------



## luutzu (20 April 2015)

bellenuit said:


> View attachment 62328
> 
> 
> The above problem has gone viral around the internet over the past few weeks, so many of you may have seen it. It just requires some straightforward logic, no math.
> ...





June 17.

Because Cheryl was born under Gemini, and according to that star sign... Key Characteristics	Multi-talented, thinker, talkative, social, scattered, diverse interests, loves variety and excitement, haha


----------



## luutzu (21 April 2015)

bellenuit said:


> View attachment 62328
> 
> 
> The above problem has gone viral around the internet over the past few weeks, so many of you may have seen it. It just requires some straightforward logic, no math.
> ...




OK, not June 17.

Don't think their answer is right either.

I think it's August 17th.

I read their answer and it doesn't gel with me. It might be reasonable if they both do not hear each other, but they seem to be hearing whether the other answer or not.

---

My take, after reading theirs 

Can't be 18th or 19th else Bernard would know right away as there's only one of each those dates.

Can't be June because Albert know about 18th or 19th not being it, so if it's June he will know it's June 17th. But Albert doesn't know.

Bernard reason this thru, know what we're saying, so say since it can't be June, it must be August 17th because she told me it's 17th. If it's any other date, Bernard cannot now know because others have double.

Albert then heard this so he then knows, reason Bernard wouldn't have known for sure if the date repeats (have two or more of), so it must be 17th, so he now also know it's 17th August.

-----------

Their reasoning:

1. can't be 18th or 19th 
      - true
2. Can't be May or June because Albert, knowing the month, could have guessed 19th or 18th.  
      - can't be June yes; but she could have told Albert it's May and there's two option left out of 19th to guess from. So even if it's May, Albert can't know for sure anyway.


Anyway... this was for 3rd graders?


----------



## bellenuit (21 April 2015)

luutzu said:


> Anyway... this was for 3rd graders?




Apparently Singaporean 14 year old students, but it isn't that difficult if you are in to logic puzzles.


----------



## cynic (22 April 2015)

My logic tells me July 16. It seems to be the only date that enables all statements to be true.


----------



## trainspotter (22 April 2015)

Forget about the math ... what about the grammar? How do you just "become" friends with Cheryl? I want her phone number and forget about her birthday.

The answer is of course July 16th


----------



## luutzu (22 April 2015)

trainspotter said:


> Forget about the math ... what about the grammar? How do you just "become" friends with Cheryl? I want her phone number and forget about her birthday.
> 
> The answer is of course July 16th




How did you two get that?

I read their reasoning but couldn't work it out to be July 16th.

Their reason that it couldn't be May or June - because Albert would have guessed May 19th or June 18th - that doesn't make sense.

I get that it couldn't be June because Albert just exclude the June 18th and get the June 17th if he know it's June; but if he know it's May, there's two other options left in May beside May 19th.

To say that it can't be May or June for above reason... it must follow that it cannot be any other number either - because if it's, say 15th, then Bernard could figure out it's either 15th May or Aug etc. But he clearly couldn't knowing it's on the 16th.

-----
Say I'm Albert and I am told it's May. It could be 15th or 16th May... since I can't be sure, I said I don't know.

So the fact that Albert said he doesn't know first doesn't mean that it cannot be in May.

Now to Bernard... He heard that Albert doesn't know which mean it could not be June since June now only have one option left. So if it's June, Albert would have said he knew.

Since 14th, 15th, 16th all have two options in each month... How could Bernard know? He could only know, since he was told the date, if that date only have one option - hence the 17th of August.


--------

Anyway, got news for you kids... thinking this hard is not what take you far in the world. It's who you know, how well you can brown-nosed, and your ability to take orders without question.  [and no, I haven't ruin my children, yet]


----------



## cynic (22 April 2015)

1. Albert's first statement tells us that he is able to determine (from the month given) that Bernard cannot have been told either 18 or 19 (they appear once only and would therefore be sufficient to identify the date). Hence the months in which they appear (i.e.May and June) can now be ruled out.

2. Bernard's statement tells us that, the day number (provided by Cheryl), combined with information deduced from Albert's first statement, is sufficient to isolate the month thereby knowing the date. Given that 14 appears against both July and August those dates can now be conclusively ruled out. At this point the only remaining candidates are July 16, August 15 and 17 .

3. Albert's final statement seals the deal! 
The only way Albert can certainly know the date after Bernard's statement is if there is only one remaining date in the month provided to Albert by Cheryl. 
August still has two remaining options (15 and 17) so the month of August can now be ruled out leaving July 16 as the only possible answer!


----------



## cynic (22 April 2015)

trainspotter said:


> Forget about the math ... what about the grammar? How do you just "become" friends with Cheryl? I want her phone number and forget about her birthday.
> 
> The answer is of course July 16th




Are you sure that's wise? 

After all Cheryl didn't provide the year of her birth! 

For all we know she could be older than Methuselah! 

And we all know how coy some women can be about their age, particularly when getting a bit long in the tooth!


----------



## luutzu (22 April 2015)

cynic said:


> 1. Albert's first statement tells us that he is able to determine (from the month given) that Bernard cannot have been told either 18 or 19 (they appear once only and would therefore be sufficient to identify the date). Hence the months in which they appear (i.e.May and June) can now be ruled out.
> 
> 2. Bernard's statement tells us that, the day number (provided by Cheryl), combined with information deduced from Albert's first statement, is sufficient to isolate the month thereby knowing the date. Given that 14 appears against both July and August those dates can now be conclusively ruled out. At this point the only remaining candidates are July 16, August 15 and 17 .
> 
> ...




point 2 n 3 makes sense, but only if 1 is true.

1 can't be true because while you can rule out June, you can't rule out the possibility of Cheryl's month being May.


So we have it that Albert speak first, and he said he doesn't know. Him knowing the month, but he cannot know the day and month... why? Because after ruling out 19, 18... if it's in June Albert will know for sure. But if it's in May, it could be May 15, 16... other two months remaining all have >1 option... so knowing the month cannot lead Albert to know.

Bernard likely rule out 18, 19... He's the second being ask... He infer from Albert that it's not June17... so of the remaining 3 months, each month have >1 option... How could Bernard be certain it's on the 16th when there are two 16th?

So it cannot be 16 or 15, or 14... it must be the only one date that's left - 17th August.

----

For the reasoning to be 16th July... it would equally be plausible that it's the 15th Aug. Because if 17th June is ruled out, as 19 and 18 is ruled out then May and June must be ruled out, so if 17th June is out then 17th Aug must also be out else Bernard will know (based on the same reasoning) etc.


----------



## Ferret (22 April 2015)

cynic said:


> 1. Albert's first statement tells us that he is able to determine (from the month given) that Bernard cannot have been told either 18 or 19 (they appear once only and would therefore be sufficient to identify the date). Hence the months in which they appear (i.e.May and June) can now be ruled out.
> 
> 2. Bernard's statement tells us that, the day number (provided by Cheryl), combined with information deduced from Albert's first statement, is sufficient to isolate the month thereby knowing the date. Given that 14 appears against both July and August those dates can now be conclusively ruled out. At this point the only remaining candidates are July 16, August 15 and 17 .
> 
> ...




I worked out Aug 17.  Re the underlined above, why do you think both May 15 and May 16 must be ruled out at this stage?


----------



## skc (22 April 2015)

Ferret said:


> I worked out Aug 17.  Re the underlined above, why do you think both May 15 and May 16 must be ruled out at this stage?




If Albert was given May, he couldn't say that "I know that Bernard does not know" definitively. Because Albert doesn't know if Bernard was given the date "19". If Bernard was given 19 by Cheryl, it would invalidate what Albert said. If Albert was given May, he could only say "Bernard may or may not know".

Therefore, Albert was NOT given May.

BTW, the correct answer is 

Unfriend Cheryl unless she's really hot...


----------



## Habakkuk (22 April 2015)

luutzu said:


> Bernard likely rule out 18, 19... He's the second being ask...





This is a pure logic puzzle; no probability involved at all. It was set for 14-year olds.
You're making it complicated with all kinds of logic errors. If you wish, I can explain it clearly, but it's been done almost to death on other sites.
The best way to look at it is to consider the three (3) points-of-view: Albert's, Bernard's and yours. What does each person know after each statement?


----------



## cynic (22 April 2015)

luutzu said:


> point 2 n 3 makes sense, but only if 1 is true.
> 
> 1 can't be true because while you can rule out June, you can't rule out the possibility of Cheryl's month being May.
> ...




18 and 19 each occur once only. 

If either of these two numbers were given to Bernard by Cheryl, he would automatically be able to identify the  associated month. 

However, Albert knows from the month (given by Cheryl) that it is not possible for Bernard to solely determine the birthdate (from the day number Cheryl provided). 

Hence any month featuring a day that would otherwise enable Bernard to identify the month, would render Albert's first statement invalid. 

Consequently, May and June can be excluded.

Edit: Just noticed SKC beat me to it!


----------



## luutzu (22 April 2015)

skc said:


> If Albert was given May, he couldn't say that "I know that Bernard does not know" definitively. Because Albert doesn't know if Bernard was given the date "19". If Bernard was given 19 by Cheryl, it would invalidate what Albert said. If Albert was given May, he could only say "Bernard may or may not know".
> 
> Therefore, Albert was NOT given May.
> 
> ...




yea... makes perfect sense, now.... but what if Albert really likes Cheryl and live in denial so just blurt out that "Bernard is a fool, there is no way he would know." 

Yea i'd unfriend Cheryl - if it's this hard to get her birth date, forget about the bases. But then maybe she really like them both and so were testing... now that the test has been met, there is a definite chance.


----------



## Value Collector (22 April 2015)

If a baseball and a bat cost $1.10 together, and the bat costs $1.00 more than the ball, how much does the ball cost?


----------



## pixel (22 April 2015)

Value Collector said:


> If a baseball and a bat cost $1.10 together, and the bat costs $1.00 more than the ball, how much does the ball cost?




5c.
why?
*And where do you get a bat for $1.05?*


----------



## Value Collector (22 April 2015)

pixel said:


> 5c.
> why?




The most common answer is 10cents, which as you know is wrong.

Apparently people who answer 10cents instinctively are more likely to be religious, because they rely on intuitive thought rather than thinking analytically.


----------



## craft (22 April 2015)

Value Collector said:


> Apparently people who answer 10cents instinctively are more likely to be religious, because they rely on intuitive thought rather than thinking analytically.




Show me the evidence for that statement - or is it just an intuitive opinion.


----------



## pixel (22 April 2015)

Value Collector said:


> The most common answer is 10cents, which as you know is wrong.
> 
> Apparently people who answer 10cents instinctively are more likely to be religious, because they rely on intuitive thought rather than thinking analytically.




ahh! so that's why our Jesuit PM gets his foot so often into his mouth:
He answers first and thinks about it later; 
and when his first answer gets him into more strife, rinse and repeat.


----------



## trainspotter (22 April 2015)

cynic said:


> Are you sure that's wise?
> 
> After all Cheryl didn't provide the year of her birth!
> 
> ...




I would have thought they became friends ? How can you become something unless you are a shape shifter? Could be totally skewiff on this one but it does not sound right at all. I have become friends with Cheryl. Cheryl could be a Troglodyte for all I care ... she is certainly taking the P!SS out of poor old Bernard and Albert. Talking in riddles like that to her newest besties. I became friends with Cheryl. Sounds better don't it?


----------



## trainspotter (22 April 2015)

pixel said:


> 5c.
> why?
> *And where do you get a bat for $1.05?*




Does not say what kind of bat now does it? It only states the ball is of the baseball variety. 

So where do you get a baseball for 5 cents is the real riddle to this fiscal conundrum.


----------



## Value Collector (22 April 2015)

craft said:


> Show me the evidence for that statement - or is it just an intuitive opinion.




It's just what was claimed on the website I copied and pasted that from, I am not claiming to know it's a fact, hence why I used the word "apparently" at the start of the sentence.


----------



## bellenuit (4 May 2015)

I came across this one in a British newspaper today (hence the references to the British political parties). I haven't attempted it yet myself.

_There are five houses with the outside walls painted in five different ways. David, Ed, Nick, Nicola and Nigel each live in one of the houses. They each drink a certain type of coffee, have a preferred mode of transport and keep a certain pet. No owners have the same pet, the same preferred mode of transport or drink the same type of coffee.

WHO OWNS THE FISH?

Nicola lives in the tartan house.
Ed has a guinea pig
David drinks mochaccino
The paisley house is on the left of the gingham house
The owner of the paisley house drinks flat whites.
The person who drives by car has a squirrel.
The owner of the striped house travels by bike.
The person living in the centre house drinks double espresso.
Nick lives in the first house.
The person who travels by train lives next to the one who has a pitbull.
The person who has a badger lives next to the person who travels by bike.
The person who travels by plane drinks chai latte.
Nigel goes everywhere by foot.
Nick lives next to the polka dot house.
The person who travels by train has a neighbour who drinks decaf.

For the sake of clarity: the houses are adjacent to each other in a line and the directions are from the perspective of someone looking at the houses. And any resemblance of David, Ed, Nick, Nicola and Nigel to the party leaders is purely coincidental. 
_


----------



## cynic (4 May 2015)

That'd be Nigel.


----------



## SirRumpole (4 May 2015)

cynic said:


> That'd be Nigel.




Why?


----------



## cynic (4 May 2015)

SirRumpole said:


> Why?




Why not?

If you take the time to work through the puzzle, you'll see that all other pets are assigned.

Edit: I can step you through the process if you like, but it'll take me a few minutes to remember the sequence as there were quite a few steps with this one.


----------



## SirRumpole (4 May 2015)

cynic said:


> Why not?
> 
> If you take the time to work through the puzzle, you'll see that all other pets are assigned.
> 
> Edit: I can step you through the process if you like, but it'll take me a few minutes to remember the sequence as there were quite a few steps with this one.




At your convenience


----------



## cynic (4 May 2015)

SirRumpole said:


> At your convenience




(i) Nick lives in the first house.

(ii) Nick lives next to the polka dot house.

We now know that the second house is polka dot.

(iii) The person living in the centre house drinks double espresso.

We now know that the third house is occupied by the double espresso drinker.

(iv) The paisley house is on the left of the gingham house.

This cannot be the second house (as we know it to be polka dot) nor can it be the first house as it is on the left of the polka dot house. 

(v) The owner of the paisley house drinks flat whites.

This cannot be the third house as we know the occupant drinks double espresso, and, given that the fifth house isn't on anything's left it can only be the fourth house.

We now know that the fourth house is paisley and that the occupant drinks flat whites.

We also now know that the fifth house is gingham.

(vi) Nicola lives in the tartan house.

We now know that the only houses with unaccounted colours are the first and third, but we know that Nick is in the first so Nicola must live in the third house which we now also know to be tartan.

(vii) The owner of the striped house travels by bike.

There is only one remaining house with unaccounted colour, namely the first, in which Nick resides. Ergo Nick lives in the striped first house and travels by bike.

(viii) The person who has a badger lives next to the person who travels by bike.

We now know that the badger belongs to the second house occupant.

(xi) The person who travels by plane drinks chai latte.

(x) David drinks mochaccino

We now know that Nick (travels by bike and is certainly not named David) doesn't drink either  mochaccino or chai latte, hence those drinkers belong in either the second or fifth house as the third and fourth are already accounted by other beverage drinkers.

Therefore Nick must be the decaf drinker.

(xi) The person who travels by train has a neighbour who drinks decaf.

Therefore the Train traveller lives in the second house, is named David , drinks mochaccino and owns a badger.

We also now know that the chai latte drinker lives in the fifth house and travels by plane.

(xii) The person who travels by train lives next to the one who has a pitbull.

We now know that Nick has the pitbull.

(xiii) Nigel goes everywhere by foot.

Nigel must now be in the fourth house drinking flat whites (we already know the fifth to be occupied by the mochaccino drinking plane traveler). 

(xiv) Ed has a guinea pig

(xv) The person who drives by car has a squirrel.

We now know that Nicola owns the squirrel and drives a car and lives in the tartan third house drinking double espresso.

We also now know that Ed travels by plane, drinks chai latte and lives with the guinea pig in the gingham fifth house.

This means that Nigel is the only remaining resident that could possibly own the fish.


----------



## SirRumpole (4 May 2015)

Thanks for that.

Amazing how logical it is when you know how.


----------



## cynic (4 May 2015)

There was a small typo in my previous post:



cynic said:


> ...
> 
> (xi) The person who travels by plane drinks chai latte.
> ...



The aforequoted line was supposed to read : "(*ix*) The person who travels by plane drinks chai latte."


----------



## bellenuit (5 May 2015)

cynic said:


> That'd be Nigel.




I agree and my logic follows similar reasoning to yours although a slightly different sequence.

I made a table with House Number (1 - 5) across the top and the various attributes down the side.

I then listed the the various statements and like you changed the sequence to suit the solution. I attached a letter to each statement and as I filled in the table I put the respective letter(s) that determined each element next to that element.

This is my solution. The table elements should be entered in the alphabetic sequence of the letters attached to them (there is no l).


----------



## bellenuit (5 May 2015)

bellenuit said:


> I came across this one in a British newspaper today (hence the references to the British political parties). I haven't attempted it yet myself.
> 
> _There are five houses with the outside walls painted in five different ways. David, Ed, Nick, Nicola and Nigel each live in one of the houses. They each drink a certain type of coffee, have a preferred mode of transport and keep a certain pet. No owners have the same pet, the same preferred mode of transport or drink the same type of coffee.
> 
> ...




BTW, this is referred to as Einstein's Riddle and I read about it here. I would be surprised if only 2% of the population can solve it as I didn't think it very difficult. It involved a lot of steps, but the logic was fairly straightforward.

*Einstein's Election Riddle: are you in the two per cent that can solve it?*

http://www.theguardian.com/science/...are-you-in-the-two-per-cent-that-can-solve-it


----------



## bellenuit (5 May 2015)

The Guardian have just published their solution on Twitter. Nigel is correct.

http://www.theguardian.com/science/...-election-riddle-are-you-in-that-two-per-cent


----------



## cynic (5 May 2015)

bellenuit said:


> BTW, this is referred to as Einstein's Riddle and I read about it here. I would be surprised if only 2% of the population can solve it as I didn't think it very difficult. It involved a lot of steps, but the logic was fairly straightforward.
> 
> *Einstein's Election Riddle: are you in the two per cent that can solve it?*
> 
> http://www.theguardian.com/science/...are-you-in-the-two-per-cent-that-can-solve-it




Whilst I did find the underlying logic reasonably straightforward, there was still the need to employ a level of creativity whilst formulating a strategic approach. 
Given that I seldom encounter people sufficiently gifted in the balanced application of logic and creativity, the quoted 2% figure is probably higher than I might typically have expected.



bellenuit said:


> The Guardian have just published their solution on Twitter. Nigel is correct.
> 
> http://www.theguardian.com/science/...-election-riddle-are-you-in-that-two-per-cent



Did we just miss an opportunity to rain on the parade?


----------



## Tisme (5 May 2015)

I remember my pursuit of lateral thinking started in earnest when my grade six teacher posed this question:



A man form Kenya, with black hair and black gloves, otherwise dressed completely in black, wearing a black ski mask, stands at an intersection looking north in a totally black-painted town that is facing west. All of the street lamps in town are broken. There is no moon. A matt black car heading south, with no reflectors nor lamps drives straight toward him, but swerves in time to miss him. How did the driver know to swerve?


----------



## craft (5 May 2015)

bellenuit said:


> BTW, this is referred to as Einstein's Riddle and I read about it here. I would be surprised if only 2% of the population can solve it as I didn't think it very difficult. It involved a lot of steps, but the logic was fairly straightforward.
> 
> *Einstein's Election Riddle: are you in the two per cent that can solve it?*
> 
> http://www.theguardian.com/science/...are-you-in-the-two-per-cent-that-can-solve-it




I always knew the Einstein puzzle as a challenge to assumptions. But it was worded slightly differently. “Who keeps fish”?

In that case the answer is - You can’t know.

You can know who doesn’t keep fish but you can’t know that (Nigel in this scenario) keeps fish – he could have any pet other then the four mentioned.

By wording it who owns the fish implies there is a fish and changes it to a straight logic test rather than a challenge to assumptions – perhaps the 2% related to the original wording of the riddle and refers to the % who challenge assumption rather than the % who can do logic puzzles. Who knows the basis or robustness of the 2% claim?


----------



## craft (5 May 2015)

Tisme said:


> I remember my pursuit of lateral thinking started in earnest when my grade six teacher posed this question:
> 
> 
> 
> A man form Kenya, with black hair and black gloves, otherwise dressed completely in black, wearing a black ski mask, stands at an intersection looking north in a totally black-painted town that is facing west. All of the street lamps in town are broken. There is no moon. A matt black car heading south, with no reflectors nor lamps drives straight toward him, but swerves in time to miss him. How did the driver know to swerve?




This one also seems to be questioning assumptions as there is nothing to stop it being full sun middle of the day.


----------



## galumay (5 May 2015)

craft said:


> This one also seems to be questioning assumptions as there is nothing to stop it being full sun middle of the day.




That was the conclusion I came to, - that it was daytime.


----------



## skc (5 May 2015)

Tisme said:


> I remember my pursuit of lateral thinking started in earnest when my grade six teacher posed this question:
> 
> 
> 
> A man form Kenya, with black hair and black gloves, otherwise dressed completely in black, wearing a black ski mask, stands at an intersection looking north in a totally black-painted town that is facing west. All of the street lamps in town are broken. There is no moon. A matt black car heading south, with no reflectors nor lamps drives straight toward him, but swerves in time to miss him. How did the driver know to swerve?




Lol. A great question for when we were kids. These days, children will probably say things like "automatic avoidance system" was activated.


----------



## galumay (5 May 2015)

skc said:


> Lol. A great question for when we were kids. These days, children will probably say things like "automatic avoidance system" was activated.




I am pleased to say my 11 year old worked it out!


----------



## Craton (5 May 2015)

galumay said:


> That was the conclusion I came to, - that it was daytime.




At the risk of being labelled, my way of thinking was in the whites of the eyes.


----------



## bellenuit (11 May 2015)

This is a subset of a puzzle that has proved controversial over the years.

A king has dominion over several islands each filled with slaves. The slaves have either blue or brown eyes. Some islands have just blue eyed slaves, some have just brown eyed slaves and some have a mixture of both.

The slaves on these islands are very smart and logical. Each has a burning ambition to be made free and travel to the mainland where the king and all the free people live.

The only way the slaves can become free is by deducing the colour of their own eyes and having done so they must depart the island that very day for the mainland on one of the many boats that go back and forth each day. This is not so easy as although the slaves can see the colour of everybody else's eyes, there are no reflective surfaces that allow them to see the colour of their own eyes. Additionally, slaves are never allowed discuss other slaves' eye colours so cannot tell or indicate in any way to another slave what eye colour that other slave has.

On one island there are just 3 slaves, all with blue eyes. One Monday morning, a king's official is visiting the island to check on things. Later that morning, as he is about to get on one of the many ferries back to the mainland, he is asked by the captain: "Does everyone on this island have brown eyes?". He answers: "No". Both the question and the response are overheard by the 3 slaves.

The question is, what (apart from the normal mundane daily drudgery of slaves) happens that week on the island once the slaves hear this seemingly redundant information?


----------



## cynic (11 May 2015)

I'll take a stab at it.



All three slaves individually arrive at the erroneous conclusion that they are the only slave on the island with brown eyes.

Upon observing that all three have boarded the next ferry off the island, all three slaves then realise their unanimous error, and now know that they must all have blue eyes.


----------



## skc (11 May 2015)

bellenuit said:


> This is a subset of a puzzle that has proved controversial over the years.
> 
> A king has dominion over several islands each filled with slaves. The slaves have either blue or brown eyes. Some islands have just blue eyed slaves, some have just brown eyed slaves and some have a mixture of both.
> 
> ...




There is no mention of what happens when you guess wrong? Given that it's a 50/50 guess... a slave should simply guess blue on day 1 and brown on day 2. It's a wonder that there are any slaves left at all.


----------



## bellenuit (11 May 2015)

skc said:


> There is no mention of what happens when you guess wrong?




I didn't think I needed to be explicit about that as the text clearly said "_the only way the slaves can become free is by *deducing* the colour of their own eyes_".

Deducing and guessing are not the same.  

I will add the following clarification to prevent guessing. 

_Slaves must at all times strive to logically determine what their eye colour is based on everything they know and having made the determination that they know without doubt what their eye colour is must leave for the mainland that same day otherwise they will be put to death as it is not allowed to remain on the island if you know your eye colour. If a slave boards the ferry to leave, he must explain to the captain his reasoning for assuming he now knows his eye colour. If the logic is incorrect or if the slave simply made a guess and cannot demonstrate any logic to his determination, then the slave will immediately be put to death._


----------



## cynic (11 May 2015)

On further consideration. The slaves realise that the ferry captain was colour blind and seeing blue as brown. The slaves now know that they are all blue eyed.


----------



## trainspotter (11 May 2015)

bellenuit said:


> On one island there are just 3 slaves, all with blue eyes. One Monday morning, a king's official is visiting the island to check on things. Later that morning, as he is about to get on one of the many ferries back to the mainland, he is asked by the captain: "Does everyone on this island have brown eyes?". He answers: "No". Both the question and the response are overheard by the 3 slaves.
> 
> The question is, what (apart from the normal mundane daily drudgery of slaves) happens that week on the island once the slaves hear this seemingly redundant information?




On the 3rd day they all leave as they have deduced that they all have blue eyes. Once the kings official announces that the islanders have blue eyes and confirming with the captain it becomes common knowledge. The first slave can see that the other 2 have blue eyes. The second slave can see that the other 2 have blue eyes. The third slave can see the other 2 have blue eyes. Then Robinson Crusoe gets all the work done by Friday


----------



## bellenuit (11 May 2015)

trainspotter said:


> Once the kings official announces that the islanders have blue eyes and confirming with the captain it becomes common knowledge. The first slave can see that the other 2 have blue eyes. The second slave can see that the other 2 have blue eyes. The third slave can see the other 2 have blue eyes. Then Robinson Crusoe gets all the work done by Friday




I don't see your logic. The official didn't say that everyone has blue eyes. He said NO to the question: "Does everyone have brown eyes". Wasn't that already common knowledge before the official arrived?


----------



## trainspotter (11 May 2015)

bellenuit said:


> I don't see your logic. The official didn't say that everyone has blue eyes. He said NO to the question: "Does everyone have brown eyes". Wasn't that already common knowledge before the official arrived?




Nope because the slaves are not allowed to talk to each other about their eye colour. By the official stating the question and the captain answering in the affirmative that they do not ALL have brown eyes. There is the possibility that the slaves think that one of them has brown eyes but by the third day they would have deduced they all have blue eyes because one slave can see the other 2 have blue eyes. The second slave can see the other 2 have blue eyes. The third slave can see the other 2 have blue eyes so with CERTAINTY they can say that they ALL have blue eyes. Follow my logic?


----------



## bellenuit (11 May 2015)

trainspotter said:


> Nope because the slaves are not allowed to talk to each other about their eye colour. By the official stating the question and the captain answering in the affirmative that they do not ALL have brown eyes. There is the possibility that the slaves think that one of them has brown eyes but by the third day they would have deduced they all have blue eyes because one slave can see the other 2 have blue eyes. The second slave can see the other 2 have blue eyes. The third slave can see the other 2 have blue eyes so with CERTAINTY they can say that they ALL have blue eyes. Follow my logic?




Nope, I don't follow your logic. Before the official arrived, each slave could see that the other two had blue eyes. They also knew that all 3 of them didn't have brown eyes because they could see two others with blue eyes. So why did they on hearing the official state what they already knew make them all leave on the Wednesday?

I get the feeling you have come across the problem before, but have forgotten the deductive steps that they follow to arrive at knowing their own eye colour. Look at it from the perspective of the slave arriving at the boat and explaining to the captain why he now knows he has blue eyes. As far as I can see from your statement, the slave gives this as a reason:

_It is the 3rd day and I know all 3 of us do not have brown eyes. I can see the other two have blue eyes, therefore I have blue eyes._

I fail to see why the captain would accept this as a logically deduced conclusion and not kill him.

You are on the right track, but I don't agree that you have given sufficient grounds for them leaving on the Wednesday.


----------



## Tisme (11 May 2015)

I'm guessing the prospect of freedom would drive them to kill the other two in hope they both have brown eyes?


----------



## bellenuit (11 May 2015)

Tisme said:


> I'm guessing the prospect of freedom would drive them to kill the other two in hope they both have brown eyes?




? But each slave can see that the other two have blue eyes.


----------



## Tisme (11 May 2015)

bellenuit said:


> ? But each slave can see that the other two have blue eyes.





Mere details  And you just found the answer then


----------



## bellenuit (11 May 2015)

Tisme said:


> Mere details  And you just found the answer then




I have no idea what you mean.


----------



## cynic (12 May 2015)

I think i've  finally gotten it!

Each slave realises that if they have brown eyes then the other two slaves will see one blue eyed and one brown eyed slave. Each slave further realises that if one were to see two brown eyed slaves then that slave would know his/her eye colour must be blue and would leave the island that day. On the Tuesday when the slaves find that nobody left , they all recognise that this can only mean that at most there can only be one brown eyed slave.

However, on Wednesday morning when all slaves are still present, they are then able to deduce that there are no brown eyed slaves and safely depart the island.


P.S. I think this might be similar to T.S. answer.


----------



## McLovin (12 May 2015)

So the captain the question/answer asked by the captain/servant could be rephrased as "There is at least one blue eyed slave". 

If there is one blue eyed slave, he will leave on the first day because he can only see two brown eyed slaves. 

If there are two blue eyed slaves then they will not leave on the first day. From the point of view of a blue eyed slave he sees one brown eyed slave and one blue eyed slave but does not know his own eye colour. On the second day, because the other blue eyed slave has not left indicates to him that he must be blue eyed also; if the other blue eyed slave had seen two sets of brown eyes he would have left the day before.

If all three are blue eyed, then on the first and second day none of them would leave. On the third day, from the point of view of a blue eyed slave, he can see two other blue eyed slave. They have not left on the second day which must mean he also has blue eyes.

So they all leave on the third day knowing they have blue eyes.


----------



## trainspotter (12 May 2015)

What McLovin said ... when did Wednesday come into it?


----------



## cynic (12 May 2015)

trainspotter said:


> What McLovin said ...



+1 
McLovin did explain it better.

However, it seems pretty clear to me that you arrived at the solution first.



trainspotter said:


> ... when did Wednesday come into it?




I think it arrived just after the Tuesday following the preceding Monday!


----------



## McLovin (12 May 2015)

trainspotter said:


> View attachment 62551




Of course! Why didn't I think of that! Just take your eyeballs out and look at them.


----------



## bellenuit (12 May 2015)

McLovin said:


> So the captain the question/answer asked by the captain/servant could be rephrased as "There is at least one blue eyed slave".
> 
> If there is one blue eyed slave, he will leave on the first day because he can only see two brown eyed slaves.
> 
> ...




Yes, McLovin and cynic, you are correct. This is perhaps the most concise explanation I have come across and thanks for saving me having to write up a solution.

This is a subset of very controversial puzzle and many do not agree with the answer. The original has more than just three islanders. However, if you don't accept the above answer for just 3 slaves, you will not accept it for more than 3. Equally, if you accept it for 3, you will see how it can logically be expanded to more than 3.

The original has a large number of slaves and they can be brown eyed and (at least one) blue eyed. The solution is basically the same, except you continue the logic beyond 3 to the total number of blue eyed slaves. The generalised result is that if there are x blue eyed people on the island, irrespective of the number of brown eyed people, then they will all leave on day x.

One person who disagrees with the result has compared the problem to this other one.

_It's a Sunday. A very smart and logical prisoner is to be executed at midday on one of the 5 weekdays that follow and is told by the executioner that he (the prisoner) will not be able to determine beforehand what day this will be. The prisoner is very pleased on hearing this and says that means you can't execute me at all. When asked why is that, he replies: "You cannot execute me on Friday, since that is the last execution day, so I would know beforehand on Friday morning that Friday is going to be the day. If you cannot execute me on Friday, then Thursday is the last day available, but then I would know Thursday morning that Thursday is to be the day, so it can't be Thursday either." He continues using the same logic to rule out the possibility of being executed Wednesday, Tuesday or Monday._

_*To his surprise he is taken out Wednesday and executed.*_

So like the blue eyed problem, the solution appears logical and consistent with the problem parameters, but we feel there is a gotcha somewhere there.


----------



## trainspotter (12 May 2015)

TRUE OR FALSE

Are all Caucasian babies born with blue eyes?




I can't believe my logic did not make sense bellenuit?


----------



## skc (12 May 2015)

bellenuit said:


> Yes, McLovin and cynic, you are correct. This is perhaps the most concise explanation I have come across and thanks for saving me having to write up a solution.
> 
> This is a subset of very controversial puzzle and many do not agree with the answer. The original has more than just three islanders. However, if you don't accept the above answer for just 3 slaves, you will not accept it for more than 3. Equally, if you accept it for 3, you will see how it can logically be expanded to more than 3.
> 
> ...




Thanks Bellenuit... your questions always managed to lose me a few hours of precious sleep time!

I am still trying to get my head around the answer. Yes it seems to make sense but here's what I can't figure out.

The only useful information from the overheard conversation is that there is at least one blue eye slave on the island, and that each slave knows that. However, this is not at all different to the situation BEFORE the conversation took place. Each slave sees two pairs of blue eyes... therefore each know themselves that there is at least one pair of blue eye. At the same time, each is certain that the other 2 slaves see at least one pair of blue eye. So... the captain's conversation has added no new information.

If I was to follow this logical... how can no new information suddenly trigger a sequence of inductions? I am not questioning the answer... but I wonder if someone can offer an explanation to this?


----------



## cynic (12 May 2015)

skc said:


> ...
> If I was to follow this logical... how can no new information suddenly trigger a sequence of inductions? I am not questioning the answer... but I wonder if someone can offer an explanation to this?



It does provide new information!

Although we know that each slave is able to recognise the presence of at least one blue eyed slave, the slaves don't know that their fellows have this same awareness until after all heard the question and answer regarding all slaves eye colour not being brown!


----------



## bellenuit (12 May 2015)

trainspotter said:


> I can't believe my logic did not make sense bellenuit?




Tr,

You obviously were correct to say the third day, but I just couldn't see how your answer arrived at that conclusion. 

_There is the possibility that the slaves think that one of them has brown eyes but by the third day they would have deduced they all have blue eyes because one slave can see the other 2 have blue eyes. The second slave can see the other 2 have blue eyes. The third slave can see the other 2 have blue eyes so with CERTAINTY they can say that they ALL have blue eyes._

I'm sorry if I missed something in your answer that I should see.

Although McLovin set the answer out in a concise manner, the actual decision process (assuming the slaves are A, B and C) that arrives at the answer involves a nested set of hypotheses.

A hypothesises that he (A) has brown eyes and then puts himself in B's shoes assuming his hypothesis is correct. He then is thinking since B is evaluating all possibilities to deduce his (B's) eye colour, then B would, as part of his (B's) deductive process, also hypothesise that he (B) has brown eyes. So if both hypotheses are correct, C must know that his eye colour has to be blue as both A and B are brown and there is one blue eyed slave. So C would have left the island on Monday. 

Since C didn't leave the island on Monday, B must know his hypothesis that he (B) has brown eyes is incorrect, so he would have left the island on Tuesday (he obviously had to wait until Monday was by to be sure C didn't leave).

But B leaving the island on Tuesday is predicated on A's hypothesis that he (A) has brown eyes. So when B didn't leave the island on Tuesday, A now knows that his hypothesis about the colour of his own eyes is incorrect, so A must have blue eyes. He must wait until Tuesday is by to be sure B has not left, so he leaves on Wednesday. Since each of the three slaves will have used the same deduction process they will each conclude on Wednesday that each have blue eyes and will thus all leave on Wednesday.

While I could see the above reasoning being implicit in Cynics and McLovin's responses, though not stated that way, I could not see that being implicit in your answer. Sorry if that wasn't the case.


----------



## cynic (12 May 2015)

cynic said:


> It does provide new information!
> 
> Although we know that each slave is able to recognise the presence of at least one blue eyed slave, the slaves don't know that their fellows have this same awareness until after all heard the question and answer regarding all slaves eye colour not being brown!




Just to further elaborate, it's not just about knowing that there are blue eyed slaves present, it's about knowing that all slaves know this and that all slaves know that all slaves know!!


----------



## trainspotter (12 May 2015)

cynic said:


> It does provide new information!
> 
> Although we know that each slave is able to recognise the presence of at least one blue eyed slave, the slaves don't know that their fellows have this same awareness until after all heard the question and answer regarding all slaves eye colour not being brown!




It's called "common knowledge". Prior to this they were not allowed to discuss their eye colour and had no way of knowing their own as their were no reflective materials to look into. Once the official asked the question and the captain answered in the affirmative then it became "common knowledge". Therefore they could deduce that if by the third day no one had left the island then with CERTAINTY they knew they all had blue eyes as they could see each others eye colour. I think that's how it goes?

Or so Wiki reckons:-



> What's most interesting about this scenario is that, for k > 1, the outsider is only telling the island citizens what they already know: that there are blue-eyed people among them. However, before this fact is announced, the fact is not common knowledge.
> 
> For k = 2, it is merely "first-order" knowledge. Each blue-eyed person knows that there is someone with blue eyes, but each blue eyed person does not know that the other blue-eyed person has this same knowledge.
> 
> ...




http://en.wikipedia.org/wiki/Common_knowledge_(logic)


----------



## skc (12 May 2015)

cynic said:


> Just to further elaborate, it's not just about knowing that there are blue eyed slaves present, it's about knowing that all slaves know this and that all slaves know that all slaves know!!




But they do know beforehand. A sees B and C having blue eyes. A also knows that B sees C with blue eyes. A also knows that C sees B with blue eyes. Therefore, A knows that all slaves know there is at least one blue eye slave on the island. It is already common knowledge before the captain has spoken...

What am I missing still..


----------



## bellenuit (12 May 2015)

skc said:


> I am still trying to get my head around the answer. Yes it seems to make sense but here's what I can't figure out.
> 
> The only useful information from the overheard conversation is that there is at least one blue eye slave on the island, and that each slave knows that. However, this is not at all different to the situation BEFORE the conversation took place. Each slave sees two pairs of blue eyes... therefore each know themselves that there is at least one pair of blue eye. At the same time, each is certain that the other 2 slaves see at least one pair of blue eye. So... the captain's conversation has added no new information.




skc, it took me a while to get my head around that too, but then I realised it is providing new information. Some of the others have explained why it provides new information. This is my way of putting it.  If you look at McLovin's answer:

_If there is one blue eyed slave, he will leave on the first day because he can only see two brown eyed slaves. 

If there are two blue eyed slaves then they will not leave on the first day. From the point of view of a blue eyed slave he sees one brown eyed slave and one blue eyed slave but does not know his own eye colour. On the second day, because the other blue eyed slave has not left indicates to him that he must be blue eyed also; if the other blue eyed slave had seen two sets of brown eyes he would have left the day before.

If all three are blue eyed, then on the first and second day none of them would leave. On the third day, from the point of view of a blue eyed slave, he can see two other blue eyed slave. They have not left on the second day which must mean he also has blue eyes._

Each of these paragraphs are predicated on the one before it being true. For the 3 blue eyed slaves scenario, they leave on the third day because if there were only two they would have left on the second day and they didn't. But for the 2 blue eyed slaves scenario, they leave on the second day because if there were only one he would have left on the first day, but he didn't. For the one blue eyed slave only, he needs that external information to arrive at a deduction that he has blue eyes. In other words, if there was just one blue eyed person on the island, without hearing what the official said, he would not be able to deduce what colour eyes he has. It is on hearing that piece of information from the official that he could determine his eye colour, which then allowed the two blue eyed scenario to be resolved which in turn allowed the 3 blue eyed scenario to be resolved.

I mentioned that this is a controversial problem and many don't accept the conclusion. IMO they are wrong. The conclusion is correct.


----------



## cynic (12 May 2015)

skc said:


> But they do know beforehand. A sees B and C having blue eyes. A also knows that B sees C with blue eyes. A also knows that C sees B with blue eyes. Therefore, A knows that all slaves know there is at least one blue eye slave on the island. It is already common knowledge before the captain has spoken...
> 
> What am I missing still..




Take it a step or two further.

A,B and C already know that each of their counterparts know that each of them can see at least one blue eyed slave. 

What each of them doesn't know beforehand is that all slaves know that all slaves know that each will see at least one blue eyed slave!


----------



## keithj (13 May 2015)

skc said:


> But they do know beforehand. A sees B and C having blue eyes. A also knows that B sees C with blue eyes. A also knows that C sees B with blue eyes. Therefore, A knows that all slaves know there is at least one blue eye slave on the island. It is already common knowledge before the captain has spoken...
> 
> What am I missing still..



Try thinking about it with only 2 blue eyed slaves....
There's 4 possible scenarios -
Brown Brown
Blue Brown
Brown Blue
Blue Blue

They both can eliminate 1st scenario.  And each can eliminate either scenario 2 or 3.  So each has to consider 2 possible (but different) scenarios.
Then the captains makes a seemingly redundant comment that there is at least one blue eyed....
If the 2nd or 3rd scenario is correct then one slave will know the exact scenario and would leave (because they can see only a brown eyed slave & therefore they must be the blue eyed one).

As no-one leaves each slave knows the other has eliminated a scenario on their behalf.

Add a 3rd slave and an extra day.....


----------



## skc (13 May 2015)

keithj said:


> Try thinking about it with only 2 blue eyed slaves....
> There's 4 possible scenarios -
> Brown Brown
> Blue Brown
> ...




OK... I think the above explained it.

What they used to know was that, they know other knows that there is at least one blue eye.

What they didn't used to know, was whether the blue-eye slave that other knows was the same slave.

Tricky.


----------



## bellenuit (14 May 2015)

This is a lot easier than the blue-eyed island puzzle.

*The Special Room*

A special room is on the 4th floor of a big house. The room has been locked for the last 50 years (a long story, that's what makes it special) and contains 3 light fittings at head height. There is one on each wall except the wall where the the door is. Each light fitting is controlled by a separate switch (one for each light), but due to an architectural oddity the switch panel is on the ground floor by the entrance to the house.

You have just bought the house and come to collect the keys from the agent late one afternoon. The agent explains about the special room. She checked it that morning and being the first to enter it in 50 years found it obviously very musty but otherwise OK. She explains about the light fittings and said she checked them out earlier that day and all the globes were working. On the way out she shows you the panel by the front door that has the three switches. Then she leaves. You check out the house and when you get to the special room, you realise that you didn't turn on the light switches. It is almost dark, but you can make out the 3 light fittings.

You head down to turn the lights on for the special room and you realise the agent forgot to tell you which switch controls which light. Not having your phone with you and being lazy and logical, you decide to work it out for your self. 

*The problem is: with all switches initially off and knowing all the light globes are working, what is the least number of trips you need to make from the switch panel to the special room on the 4th floor to determine which switch controls which fitting. Explain your answer.*

You can assume that no special equipment is to be used, you are alone without help and that you cannot determine from outside the house or anywhere else in the house the status of a light in the special room other that going to the special room. There are no tricks to the question.


----------



## Boggo (14 May 2015)

bellenuit said:


> You head down to turn the lights on for the special room and you realise the agent forgot to tell you which switch controls which light. Not having your phone with you and being lazy and logical, you decide to work it out for your self.
> 
> *The problem is: with all switches initially off and knowing all the light globes are working, what is the least number of trips you need to make from the switch panel to the special room on the 4th floor to determine which switch controls which fitting. Explain your answer.*




This is assuming that you are very fit you could do it in one trip.
You could turn on two of the switches and allow the lights to warm up, then switch one off and run up the stairs like buggery before the one you had on cooled down.
You should now have one on, one hot and the one that hasn't been switched on would still be cold ???????


----------



## pixel (14 May 2015)

Boggo said:


> This is assuming that you are very fit you could do it in one trip.
> You could turn on two of the switches and allow the lights to warm up, then switch one off and run up the stairs like buggery before the one you had on cooled down.
> You should now have one on, one hot and the one that hasn't been switched on would still be cold ???????




Smart thinking, 99.
Methinks you got it in one


----------



## Boggo (14 May 2015)

pixel said:


> Smart thinking, 99.
> Methinks you got it in one




Not sure yet, that 4th floor, big house bit could be a factor


----------



## cynic (14 May 2015)

It can be done in two complete trips. The first trip requires two switches flipped on and the lights noted. The light that isn't on belongs to the unflipped switch. 

The second trip requires that one of the two flipped switches be unflipped.  

The remaining light belongs to the flipped  switch and the other light (that was previously on) belongs to the recently unflipped switch.


----------



## pixel (15 May 2015)

cynic said:


> It can be done in two complete trips. The first trip requires two switches flipped on and the lights noted. The light that isn't on belongs to the unflipped switch.
> 
> The second trip requires that one of the two flipped switches be unflipped.
> 
> The remaining light belongs to the flipped  switch and the other light (that was previously on) belongs to the recently unflipped switch.




If you want to allow two trips, you can simply flick switch One, climb the stairs at your leisure, mark the light that is burning with a "1". Then go back down, repeat the same with switch Two and lamp "2". No need to test switch Three.

I still prefer Boggo's solution


----------



## cynic (15 May 2015)

pixel said:


> If you want to allow two trips, you can simply flick switch One, climb the stairs at your leisure, mark the light that is burning with a "1". Then go back down, repeat the same with switch Two and lamp "2". No need to test switch Three.
> ...




Yes - that'd work also. 

It'll be interesting to find out if the one trip approach works for those high efficiency globes that barely heat.


----------



## bellenuit (15 May 2015)

Boggo said:


> You could turn on two of the switches and allow the lights to warm up, then switch one off and run up the stairs like buggery before the one you had on cooled down.
> You should now have one on, one hot and the one that hasn't been switched on would still be cold ???????




Yes, Boggo. That is the answer



Boggo said:


> This is assuming that you are very fit you could do it in one trip.






			
				cynic said:
			
		

> It'll be interesting to find out if the one trip approach works for those high efficiency globes that barely heat.




Cynic/Boggo When I first heard the problem, which was phrased differently, I came up with the same answer as Boggo, but thought it had the same potential weakness in that it mightn't work for new style globes. I thought about specifying that they were incandescent globes, but realised that the word "incandescent" would be too big a clue and make the answer obvious. So the whole spiel about the room being locked for 50 years was simply to ensure no one could claim the answer was wrong because modern day globes leave off little heat. The globes are obviously from the 1960s at the latest and I am sure most bedrooms would only have incandescent globes available at that time. Globes from that era hold their heat for at least 10 minutes, but as only a few minutes are needed to climb the stairs, they would still be discernibly warm to the touch. Although fluorescent tubes were around then and are cooler, they still were not as efficient as today's variety and also would be discernibly warm after being off for 2 minutes say. In any case, I specified globes, not tubes, so I think Boggo's answer is without doubt correct.


----------



## cynic (15 May 2015)

bellenuit said:


> Yes, Boggo. That is the answer




Darn it! Foiled again!!

P.S. Well done Boggo!


----------



## bellenuit (15 May 2015)

A really hard one this time.

This is a problem I came across years ago. I didn't work it out, but was staggered by the answer (I doubt if beforehand anyone would even guess close to the answer, but it it is purely mathematical and the answer is beyond question). It does require a lot of math, particularly probability theory. I have never been able to remember what the full answer is, but have been able to get a close approximation that is not far off the correct answer. The close approximation is still unbelievable by most people, but when they see the math they cannot dispute it. I will pose it here in the hope that some one can provide the exact solution. It is a simply described problem, but one must pay close attention to the actual wording. There is no trickery involved.

Random numbers between 0 and +infinity are generated by a computer and each is printed on a separate sheet of paper. This is done until several thousand numbers are generated. Each sheet of paper is placed face down  on a table (I know, it must be a very big table) in no particular order. The computer also prints out on a separate sheet the highest number it generated. This is for the use of the administrator to check the result.

A brilliant logician is invited into the room and offered this challenge (he is told the number of sheets of paper on the table, which he can count anyway if he wanted to, and that each piece contains a random number between 0 and + infinity): "You are to turn up sheets of paper one at a time and after turning up a sheet and seeing its number you are to declare whether that number is the highest number among all the sheets of paper that are on the table. If you say it isn't you continue on turning. If you say it is, you stop". *The question is: what is the probability that the logician will stop at the correct number that is the highest (e.g. corresponds to the number that the administrator was given by the computer).*

You can assume that being a brilliant logician he will use a method that maximises his chances of being correct.

As most people often get this part wrong, I will reiterate. He can only declare the number he has just turned to be the highest and stop. He cannot say, for instance, that a piece of paper he turned earlier is the highest. If he thinks an earlier number is the highest, then it is too late and that attempt is seen as a failure.


----------



## trainspotter (15 May 2015)

Wan't this one done in November 2009? I have a very faint memory of wolframalpha being involved??


----------



## keithj (15 May 2015)

bellenuit said:


> You can assume that being a brilliant logician he will use a method that maximises his chances of being correct.



The method a logician would use to find the highest number is 

work out the probability of the location of the 2nd highest number
turn over that many papers & remember the highest number found so far
finally turn over more papers until he finds one higher & declares it the highest.

I didn't use any maths or probability theory to arrive at this figures.  I generated 4000 bits of paper with a random number on it, and asked my PC to try the method 1,000,000 times.

My PC seemed to think that on average the 2nd highest number would be found somewhere in the first 18.7% of the bits of paper.

It doesn't matter what the highest number is (infinity isn't an issue) - the only thing that matters is that after around 18.7% of the bits of paper have been sighted he can expect to have seen the 2nd highest.

And therefore if any subsequent number seen that is higher is likely to be the highest.



And the 2nd part of the question is what is the probability he is right.....

My PC checked 1,000,000 times to see how many times the highest number of all the bits of paper was not in the 1st 18.7%.

If he uses this method, my PC reckons he has a probability of around 72.5% chance of getting it right.


----------



## bellenuit (15 May 2015)

trainspotter said:


> Wan't this one done in November 2009? I have a very faint memory of wolframalpha being involved??




Yes, I remember posting it before, but I don't recall where or if it was on ASF. I have yet to find the full solution other than approximations, so I was hoping some new minds working on it might prove fruitful.


----------



## bellenuit (15 May 2015)

keithj said:


> The method a logician would use to find the highest number is
> 
> work out the probability of the location of the 2nd highest number
> turn over that many papers & remember the highest number found so far
> ...




You are on the right track, but I see some errors already. 

_My PC seemed to think that on average the 2nd highest number would be found somewhere in the first 18.7% of the bits of paper._

Try a Mac instead. Surely basic probability would indicate that the on average the 2nd highest number would be found somewhere in the first 50%?


----------



## cynic (15 May 2015)

bellenuit said:


> You are on the right track, but I see some errors already.
> 
> _My PC seemed to think that on average the 2nd highest number would be found somewhere in the first 18.7% of the bits of paper._
> 
> Try a Mac instead. Surely basic probability would indicate that the on average the 2nd highest number would be found somewhere in the first 50%?




Also the highest number would surely have an equal chance of falling in the first 50%.


----------



## cynic (15 May 2015)

I haven't tested my approach with the maths yet, but at a glance I suspect that the presence of positive infinity will obfuscate the practical reality.

So for the moment I'll give my quick and dirty answer to how I'd approach the situation.

I'd sequentially turn over the first floor(N/2) papers. (I would only stop if I found positive infinity!)

Upon having examined the first half and noting the highest number, I'd then sequentially turn over the [N-floor(N/2)] papers until I found a new highest number and then stop.

There's a 50% chance that the highest number wouldn't already have been turned over (in the first half) which means there's a 50% chance that there are one or more larger numbers in the residual half.

My chance of being wrong would be 0.5 + sum[(probabilities > 1 higher numbers in second half)X(probability highest appearing later than first higher)].


----------



## cynic (15 May 2015)

It just occurred to me that, following on from my earlier logic, the second half would then be viewable in the same context as the first. (i.e. that there is a highest number that could just as easily fall within either the first or second half of the residual half.)

 The perpetuation of such logic might easily result in turning over every card in the expectation that the last will ultimately be the highest!


----------



## keithj (15 May 2015)

He should turn over a sample set of 37% of the papers & note the highest number.

There is then (coincidentally) a 37% chance that the next paper he turns over that has a higher number on it will be the highest in the whole set.

By using a sample size of 37% of the papers, this balances the probability of actual highest number NOT being in that sample & also that the sample contains a sufficiently high number (not necessarily the 2nd highest) to exclude all but the highest in the remainder

And additionally the probability of any numbers higher than the highest in the sample set, but not the actual highest that are (fortunately) NOT turned over from the remaining papers.



However, still using brute force & a PC....


----------



## bellenuit (15 May 2015)

keithj said:


> He should turn over a sample set of 37% of the papers & note the highest number.
> 
> There is then (coincidentally) a 37% chance that the next paper he turns over that has a higher number on it will be the highest in the whole set.
> 
> ...




Yes. Close enough to the optimum. He turns over the first 1/e pieces of paper, where e is the natural number e=2.718 approx, and notes the highest number among them. He then starts turning the remaining pieces of paper and stops at the first that is higher. The actual probability of that being the highest of all pieces of paper is also 1/e.  

1/e = 36.787%

I usually explain it using 1/3 rather than 1/e as it gets the principle across (most would assume that the chances are close to zero) and is close enough to the optimum result anyway.

Using 1/3 as the ratio representing the number that are turned up first (that is why the logician must know the total number of pieces of paper) you get something like this.

What is the probability that the 2nd highest number is in the first 1/3, but the highest is not (because if this happens his method will cause him to stop on the highest). This probability is:

1/3 * 2/3 = 2/9 or 22.2%

But that is just one way that method would yield a win. It would also yield a win if the 3rd highest number is in the first 1/3, the 1st and 2nd highest are not in the first third and he turns the highest before the 2nd highest. The probability of that happening is:

1/3 * 2/3 * 2/3 * 1/2 = 4/54 or 7.4% 

Since these are mutually exclusive events, and since both would yield a win, you can add the answers together and we now are at 29.6% probability.

You could still go on with additional mutually exclusive events that increase the final answer. The next one would be the probability that the 4th highest is in the first 1/3, the 1st, 2nd and 3rd highest are not in the first 1/3, and he turns the highest before he turns the 2nd or 3rd highest. The probability of that is:

1/3 * 2/3 * 2/3 * 2/3 * 1/3 =8/243 or 3.3%

The total is now at 32.9%.

So without too much math, it is easy enough to explain how rather than being an impossible task, the logician has about a 1 in 3 chance of being right. The interesting thing is once the optimum method is worked out, the actual process of stoping at the highest number is purely mechanical and doesn't take any thinking.

As I mentioned, the optimum ratio is 1/e and the math to work that out is beyond me. But I have discovered that this problem is on Wikipedia (called the Secretary Problem) and all is explained there for those interested.

http://en.wikipedia.org/wiki/Secretary_problem


----------



## cynic (16 May 2015)

bellenuit said:


> ...As I mentioned, the optimum ratio is 1/e and the math to work that out is beyond me. But I have discovered that this problem is on Wikipedia (called the Secretary Problem) and all is explained there for those interested.
> 
> http://en.wikipedia.org/wiki/Secretary_problem



After looking at that article, I'm pretty sure that 1/e is an approximation. The optimal ratio will approach 1/e as n (the total number of papers) approaches infinity.

The probability sum is essentially a real multiple of the difference between two harmonic numbers. This explains the relationship to e and the reason why the approximation can be expected to become increasingly accurate as n approaches infinity.


----------



## trainspotter (16 May 2015)

bellenuit said:


> Yes, I remember posting it before, but I don't recall where or if it was on ASF. I have yet to find the full solution other than approximations, so I was hoping some new minds working on it might prove fruitful.




16th November 2009 you wrote this in ASF (took me a while to find it )  

The Random Number Generator - One of the best Maths problems ever IMO



> People seem to be now on the right track, so I will give the solution as best as I can. brty was close to the mark when he/she posted "By waiting until the first 500 numbers have been drawn,and then putting your name to the first number that is higher than any in the first 500, should give a probability of 1 in 4 of being correct" That is the strategy, but the number to pass over is not the optimum. 2020 was closer when he/she said " my gut feel is that it is about the 33% mark, that you turn up 333 of them, then continue until (as others have said) you get a bigger number than all the numbers to date = the biggest in the upturned pack".
> 
> As I mentioned at the very beginning you can get an approximate answer with very basic probability theory, but to get the exact answer requires a better knowledge of maths than I have. In fact my maths has improved enormously over the last day or so when I started using the Wolfram Alpha computational engine, but more about that below.
> 
> ...




*whew* for a second there I thought my memory was going !!


----------



## bellenuit (16 May 2015)

trainspotter said:


> 16th November 2009 you wrote this in ASF (took me a while to find it )
> 
> The Random Number Generator - One of the best Maths problems ever IMO
> 
> ...




What thread was that? I thought I might have posted it on ASF before, but the only puzzle thread I could find was this one and it wasn't here.

Seeing the answer is fully explained in Wiki, it will be hard to post any of the classic puzzles without someone being able to look up the answer. You would certainly have to call them by a different name to what the original was called and change the narrative a bit.


----------



## cynic (18 May 2015)

bellenuit said:


> What thread was that? I thought I might have posted it on ASF before, but the only puzzle thread I could find was this one and it wasn't here.
> 
> Seeing the answer is fully explained in Wiki, it will be hard to post any of the classic puzzles without someone being able to look up the answer. You would certainly have to call them by a different name to what the original was called and change the narrative a bit.




Yes! What's happening seems analogous to people relying upon calculators instead of learning their times tables.

Your postings to this thread have attracted much interest from some of the logically and mathematically inclined members of this forum.

Perhaps some of us could invent puzzles of our own and post them here.


----------



## barney (18 May 2015)

cynic said:


> Your postings to this thread have attracted much interest from some of the logically and mathematically inclined members of this forum.





Yeah its a good fun thread ..... although I'm still not convinced about the brown and blue eyed Slaves.  My logic behind not being convinced is thus ....

If the Captain of the boat had lined the 3 Slaves up (facing away from him of course) and asked them all (both separately and/or collectively)  ..... Do all the Slaves on this Island have brown eyes,  they would have all said NO.  Even if they heard each others answer, what does that teach them that they didn't already know?   I don't see why a third party saying NO gives any added info, but I am quite prepared to admit my brain is simply not up to the "twist"

Here is a simple one to start the week.  I just got it off the net so hopefully a few haven't seen it.  PS. If it looks too easy, try it after 3 beers!

*At a recent downhill mountain bike race, four entrants entered the challenging slalom event.


Alan came first.

The entrant wearing number 2 wore red, whereas John didn't wear yellow.

The loser wore blue and Steve wore number 1.

Kev beat Steve and the person who came second wore number 3.

The entrant in yellow beat the entrant in green.

Only one of the entrants wore the same number as their final position.


 Can you determine who finished where, the number and colour they wore?*


----------



## bellenuit (18 May 2015)

barney said:


> Y
> If the Captain of the boat had lined the 3 Slaves up (facing away from him of course) and asked them all (both separately and/or collectively)  ..... Do all the Slaves on this Island have brown eyes,  they would have all said NO.  Even if they heard each others answer, what does that teach them that they didn't already know?   I don't see why a third party saying NO gives any added info, but I am quite prepared to admit my brain is simply not up to the "twist"[/B]




As cynic and others have pointed out, the crux of the problem (though not explicitly stated in McLovin's answer) is not what each slaves knows, but what each slave knows the other slaves know. In this case the information is not redundant. It is the reaction of the other slaves that allows the slaves to determine their own eye colour. Let's call the slaves A, B and C and look at it from the point of A trying to see what B might be thinking.

As part of his deductive reasoning, A sets up a hypothesis for testing that may or may not be correct. The hypothesis is that he, A, has brown eyes. He then tries to put himself in B's shoes. Although A knows C knows B has blue eyes, B doesn't know his own eye colour, so B doesn't know what C knows about B's eye colour. 

This might make it clearer:

1 A hypothesises that he, A, has brown eyes and then looks at it from B's viewpoint.

   2. So B, hypothesising that he, B, has brown eyes, must look at C and think C must have this info:
      (remember this is what A is thinking B is thinking in regards to C)

     3.1 C must assume that he, C, has brown eyes and thus everyone has brown eyes OR
     3.2 C must assume that he, C, has blue eyes and he is the only one with blue eyes .

This nested hypothesis is in A's mind and is quite valid, though we as outside observers know isn't the case.

So 3.1 and 3.2 is assuming that 2 is true and 2 is assuming 1 is true. At this point look at 3.1.  This is a valid hypothesis and it's assumption is that everyone has brown eyes. That is why the information is not redundant. That information about not everyone having brown eyes allows B to realise that with the "new"information, C must leave the island Monday (because if 1 and 2 were true, and  3.1 now ruled out with the "new" information, C would know his eye colour is blue and leave the island). When C doesn't leave the island, B (in A's mind) would know hypothesis 2 is incorrect and that he, B, must have blue eyes and then he leaves Tuesday. But when that doesn't happen, A must assume that hypothesis 1 is incorrect and thus he, A, has blue eyes, leaving Wednesday. Everything being symmetrical across a, B and C, they all leave Wednesday.

So the information isn't redundant because it allows a potential conclusion in the nested hypotheses to be ruled out which then sets off a chain reaction of conclusions. Without that information being provided, 3.1 and 3.2 would never be resolved and no conclusion could be drawn from C's lack of reaction, so 2 or 1 could also not be resolved.


----------



## cynic (18 May 2015)

barney said:


> Yeah its a good fun thread ..... although I'm still not convinced about the brown and blue eyed Slaves.  My logic behind not being convinced is thus ....
> 
> If the Captain of the boat had lined the 3 Slaves up (facing away from him of course) and asked them all (both separately and/or collectively)  ..... Do all the Slaves on this Island have brown eyes,  they would have all said NO.  Even if they heard each others answer, what does that teach them that they didn't already know?   I don't see why a third party saying NO gives any added info, but I am quite prepared to admit my brain is simply not up to the "twist"



After reading some of the responses to this thread, I am of the opinion that you are by no means alone in that regard. 

The important thing is to think about what the slaves suddenly know afterwards that they couldn't possibly know beforehand.

Whenever I try to articulate it, it looks like a typo: all slaves then know that all slaves know that all slaves know that there is at least one blue eyed slave.

Without that crucial piece of information the slaves simply cannot draw any conclusions from the reaction (or absence thereof) of their fellows on the island over the successive few days.


I'll try and formulate a series of questions, later today, that will hopefully make the distinction clearer. 


> Here is a simple one to start the week.  I just got it off the net so hopefully a few haven't seen it.  PS. If it looks too easy, try it after 3 beers!
> 
> *At a recent downhill mountain bike race, four entrants entered the challenging slalom event.
> 
> ...



Yes I can, but that's only because I've run out of beer!!

Edit: Just noticed that Bellenuit has beaten me to the punch on clarification of the slaves puzzle. Thanks Bellenuit.


----------



## barney (18 May 2015)

LOL  ..... Thanks for such a detailed response "Bell" ........ 

Leave it with me for a while  .....  My brain only runs in Stereo, not Quadraphonic!


----------



## barney (18 May 2015)

cynic said:


> After reading some of the responses to this thread, I am of the opinion that you are by no means alone in that regard.




Even after Bells detailed response I still have questions to be answered  ..... Its almost as hard to write a question to articulate the problem as it is to try and solve the problem (in my brain )



cynic said:


> Yes I can, but that's only because I've run out of beer!!




I should have posted it up after 5pm


----------



## Tisme (18 May 2015)

bellenuit said:


> This is a subset of a puzzle that has proved controversial over the years.
> 
> A king has dominion over several islands each filled with slaves. The slaves have either blue or brown eyes. Some islands have just blue eyed slaves, some have just brown eyed slaves and some have a mixture of both.
> 
> ...




I didn't want to think about this because I would presume that if someone asks if everyone has brown eyes, it would be based on observing someone with brown eyes, but disregarding that logic:

Each one would look at the other two and see blue eyes. Each would suppose by intuition the other two would take off the next day, but because neither do, surety rules apply hence and logically mean (if people = x and if days = y) y = x-1 days would pass to make sure all three could be sure they all blue eyes. Therefore it would be Wednesday before they pack up the esky and pink zinc and head for the ship.


----------



## trainspotter (18 May 2015)

barney said:


> Yeah its a good fun thread ..... although I'm still not convinced about the brown and blue eyed Slaves.  My logic behind not being convinced is thus ....
> 
> If the Captain of the boat had lined the 3 Slaves up (facing away from him of course) and asked them all (both separately and/or collectively)  ..... Do all the Slaves on this Island have brown eyes,  they would have all said NO.  Even if they heard each others answer, what does that teach them that they didn't already know?   I don't see why a third party saying NO gives any added info, but I am quite prepared to admit my brain is simply not up to the "twist"
> 
> ...





It should be simple enough to create a Matrix with names, colours, numbers and finishing positions? 

1) Alan number 2, red
2) Kev number 3, yellow
3) Steve number 1, green
4) John number 4, blue


----------



## cynic (18 May 2015)

trainspotter said:


> It should be simple enough to create a Matrix with names, colours, numbers and finishing positions?
> 
> 1) Alan number 2, red
> 2) Kev number 3, yellow
> ...




Yes! That's what I got. But I cheated by not drinking beer beforehand.


----------



## bellenuit (18 May 2015)

trainspotter said:


> It should be simple enough to create a Matrix with names, colours, numbers and finishing positions?
> 
> 1) Alan number 2, red
> 2) Kev number 3, yellow
> ...




Same with me


----------



## barney (18 May 2015)

cynic said:


> Yes! That's what I got. But I cheated by not drinking beer beforehand.




LOL   You are a funny man Mr Cynic.

Indeed all are correct  ...... I knew that would be easy for you guys


----------



## bellenuit (18 May 2015)

A not too difficult one.

*You are being interviewed for a job in a factory that makes fuses for explosives (the type that you light and will burn from end to end for a specific length of time). As an intelligence test, the interviewer gives you this problem. 

He shows you two of the fuses they make that are identical. Both are yellow at one end and green at the other. They are flexible and can be twisted any way you want. The fuses can be lit from either end and burn for exactly two minutes. They do not burn at a constant rate and the burn rate (cms/sec) will vary along their lengths, sometimes going faster and sometimes going slower. However, because they are manufactured identically, the two fuses, if lit from the same coloured end, will burn exactly at the same rate as each other, both increasing or decreasing in speed at the same time. As they burn they completely vaporise and leave no residue trail of any sort. Fuses can ignite each other if the burning point of one comes into contact with the other.

The interviewer hands you the two fuses and shows you the heat resistant table to use for your test. This table also won't stain or show a trail of a burnt fuse. He gives you a box of matches with just two matches, each which only burns for a couple of seconds. He says: "Choose your begin time and then tell me when 90 seconds has elapsed after that". You are not allowed use any device other than the fuses and matches.

You can assume that the fuse will ignite the instant it is touched by the flame.

Extra points for being able to gauge with the most accuracy when 90 seconds is up.*


----------



## cynic (18 May 2015)

I would lay the two fuses on the table so that they intersect each other 3/4 of the way along with the green end of one fuse and yellow end of other a 3/4 length away from the intersection.

I'd then light each fuse simultaneously from the end that is farthest from the intersection (i.e. the green end on one fuse and yellow end on the other) and declare the 90 seconds to have begun.

When the flame of one fuse first meets the intersection, I'd observe the now burning residual portions of the 1/4 lengths. As soon as they appeared to be approximately equal length I'd declare the 90 seconds over.

Edit: It just occurred to me that the 90 seconds could be declared over when the residual portion of the slower 3/4 length burnt out.


----------



## bellenuit (18 May 2015)

cynic said:


> I would lay the two fuses on the table so that they intersect each other 3/4 of the way along with the green end of one fuse and yellow end of other a 3/4 length away from the intersection.
> 
> I'd then light each fuse simultaneously from the end that is farthest from the intersection (i.e. the green end on one fuse and yellow end on the other) and declare the 90 seconds to have begun.
> 
> ...




_When the flame of one fuse first meets the intersection, I'd observe the now burning residual portions of the 1/4 lengths. As soon as they appeared to be approximately equal length I'd declare the 90 seconds over._

That wouldn't work, if I am understanding you correctly. Assume as an example, that A burns really fast for the first 3/4 of its length and really slow for the rest. The other, B, since it is in reverse, will be slow for the 1st quarter and then fast for the remaining 3/4. Then A will obviously burn to the intersection first and ignite B. But since A now drops to its slow speed and B is at its top speed, B will obviously burn its remaining 1/4 at a faster rate than A, so the residual portions will never be equal length.

_Edit: It just occurred to me that the 90 seconds could be declared over when the residual portion of the slower 3/4 length burnt out._

Again I think this is wrong. If we assume A burns 120 times faster in the 1st 3/4 than the last 1/4, then it will reach the intersection in under a second, where it ignites B. But B is now in the fast portion of its burn too, so will complete the full burn in less than a  second as well. All up under 2 seconds.


----------



## Tisme (18 May 2015)

bellenuit said:


> A not too difficult one.
> 
> *You are being interviewed for a job in a factory that makes fuses for explosives (the type that you light and will burn from end to end for a specific length of time). As an intelligence test, the interviewer gives you this problem.
> 
> ...




Light one fuse at both ends and the 60 second mark will be measurable when it burns out. Immediately light the second fuse and allow to burn to the same length that the first fuse burnt itself out.


----------



## bellenuit (18 May 2015)

Tisme said:


> Light one fuse at both ends and the 60 second mark will be measurable when it burns out. Immediately light the second fuse and allow to burn to the same length that the first fuse burnt itself out.




Your getting hot. One logistical problem though. The first leaves no residual marks, so how will you determine the same length as the first with what you have available. Also, since you have just matches, how do you light the first at both ends? Will there be a lag in moving the lit match from one end to the other or do you use both matches to do this?


----------



## cynic (18 May 2015)

bellenuit said:


> _When the flame of one fuse first meets the intersection, I'd observe the now burning residual portions of the 1/4 lengths. As soon as they appeared to be approximately equal length I'd declare the 90 seconds over._
> 
> The wouldn't work, if I am understanding you correctly. Assume as an example, that A burns really fast for the first 3/4 of its length and really slow for the rest. The other, B, since it is in reverse, will be slow for the 1st quarter and then fast for the remaining 3/4. Then A will obviously burn to the intersection first and ignite B. But since A now drops to its slow speed and B is at its top speed, B will obviously burn its remaining 1/4 at a faster rate than A, so the residual portions will never be equal length.




Yes I do see your point. I didn't think that one through with sufficient care.


> _Edit: It just occurred to me that the 90 seconds could be declared over when the residual portion of the slower 3/4 length burnt out._
> 
> Again I think this is wrong. If we assume A burns 120 times faster in the 1st 3/4 than the last 1/4, then it will reach the intersection in under a second, where it ignites B. But B is now in the fast portion of its burn too, so will complete the full burn in less than a  second as well. All up under 2 seconds.



Again I see your point.

How about intersecting them as previously described but with the intersection exactly halfway this time and declaring the 90 seconds up when the north and west pointing 1/2 portions (presuming that the fuses were lit at the ends pointing east and south) appear to have combined residual lengths equal to 1/2.


----------



## cynic (18 May 2015)

Following on from Tisme's logic I think I might know an answer (thanks Tisme).

Curl the first fuse into a loop so that both ends can be lit by the first match at once.

Curl the second fuse into a loop and then twist it into a double loop (again so that both ends can be lit by a single match).

Light the first looped fuse and a minute later when it has burnt out light the double looped fuse. A further 30 seconds will have elapsed when the second fuse has burnt out.

Adjust for the seconds it took to light each of the fuses in the 90 seconds estimate.


----------



## bellenuit (18 May 2015)

cynic said:


> Yes I do see your point. I didn't think that one through with sufficient care.
> 
> Again I see your point.
> 
> How about intersecting them as previously described but with the intersection exactly halfway this time and declaring the 90 seconds up when the north and west pointing 1/2 portions (presuming that the fuses were lit at the ends pointing east and south) appear to have combined residual lengths equal to 1/2.




I think the math of the combined residual lengths equal 1/2 is probably correct, but the logistics of doing it are horrendous. Firstly, since there is no residue left from what has already being burnt, gauging not just what 1/2 of the full length is would be very difficult, but trying to visually combine two differently orientated non-touching and decreasing lengths to assess whether they are the same length as 1/2 something that has already vanished would be a bit much. You could possibly place the matches at the intersection and one end of the fuse to gauge what 1/2 the full length is, but I still think visually comparing the sum of the lengths of the two decreasing residual fuses to that length would not be easy and inaccurate in comparison to other solutions.

Doable, but not the best solution.


----------



## bellenuit (18 May 2015)

cynic said:


> Following on from Tisme's logic I think I might know an answer (thanks Tisme).
> 
> Curl the first fuse into a loop so that both ends can be lit by the first match at once.
> 
> ...




I have issues with this part.

_Light the first looped fuse and a minute later when it has burnt out light the double looped fuse. A further 30 seconds will have elapsed when the second fuse has burnt out._

When you say a double loop, do you mean a figure 8 or two zeros on top of each other. I don't believe either would work.

Again thinking about a situation where the burn rate for the first 3/4's is 120 times (or some big number) faster than the last quarter. For the stacked zero configuration, it would zip around in under a second (as one of the loops is entirely at the top speed and that will ignite the other). For the figure 8 config, the intersection is where the 1/4 and 3/4 points cross. The fast end will reach that intersection and continue around the outside loop back to that intersection in under a second. The other quarter will have been lit at both ends, firstly at the start (when the joined slow and fast ends were lit) and secondly when the fast end first crossed the intersection. But that 1/4 segment is really really slow, even though lit from both ends. It takes about 119 seconds (approx) to burn from 1 end, so will take almost a minute to burn if lighting at both ends.


----------



## McLovin (18 May 2015)

Lay the two fuses next to each other with the same coloured ends. Light one of the fuses at both ends and one at one end, with the first match. When the first fuse extinguishes, light the remaining unlit end with the second match.

So the first fuse will go for 1 minute and then once the unlit end of the second fuse is lit it will run for a further 30 seconds. Giving 90 seconds all up.

This won't work if the matches go out very quickly (ie you're not supposed to be able to light both ends with the same match).

Thanks for posting these bellenuit, it really gets the grey matter firing!


----------



## skc (18 May 2015)

bellenuit said:


> A not too difficult one.
> 
> *You are being interviewed for a job in a factory that makes fuses for explosives (the type that you light and will burn from end to end for a specific length of time). As an intelligence test, the interviewer gives you this problem.
> 
> ...




1. Place fuse 1 with both ends at one edge of the table. The two halves of the fuse shall be parallel and as close to each other without touching.

2. Ignite both ends of the fuse at the same time.

3. Hold the second fuse at the ready. Note where fuse 1 is about to burn out and ignite the correct end of fuse 2 using the very last spark (or as much as possible) from fuse 1. The correct end is the end closest to the burnt out location for fuse 1. At the same time, place remainder of fuse 2 towards the edge of the table where the ends of fuse 1 were placed.

4. Declare 90 second count begins.

5. When fuse 2 burns to the edge of the table, 60 seconds would have passed. At this time, ignite the remaining end of fuse 2. Another 30 seconds shall pass when fuse 2 is burnt out. Making a total of 90 seconds.


----------



## cynic (19 May 2015)

It looks to me like you might've gotten that job, SKC.


----------



## bellenuit (19 May 2015)

skc said:


> 1. Place fuse 1 with both ends at one edge of the table. The two halves of the fuse shall be parallel and as close to each other without touching.
> 
> 2. Ignite both ends of the fuse at the same time.
> 
> ...




also,



			
				McLovin said:
			
		

> Lay the two fuses next to each other with the same coloured ends. Light one of the fuses at both ends and one at one end, with the first match. When the first fuse extinguishes, light the remaining unlit end with the second match.
> 
> So the first fuse will go for 1 minute and then once the unlit end of the second fuse is lit it will run for a further 30 seconds. Giving 90 seconds all up.
> 
> This won't work if the matches go out very quickly (ie you're not supposed to be able to light both ends with the same match).




Both of these solutions are right and will give an approx 90 second duration. As Mc Lovin noticed with his solution, there may be an issue with lighting the 2nd end of the fuse that is lit at both ends. Will the match stay lit. Assuming it does, then there will be some small inaccuracy with the time lag in moving the match from one end to the other (the lengths weren't specified so we are not sure how big this lag is). SKCs solution doesn't seem to have this problem, but could be logistical difficult trying to place the now burning 2nd fuse over the same exact path as that section of fuse 1 without some sort of fumbling (for instance, the short portion of fuse 2 is on the table and the long portion will overhang: so how do you stop it falling off?). It also assumes that when lit, the fuses stay perfectly still and don't move or curl in anyway, because if they do, then the end point where fuse 1 burns out may not be where that exact point on fuse 1 originally started, meaning fuse 2 placed on that burn out point may not take exactly 60 seconds to reach the table edge. 

I would think SKCs solution would give the most accurate result if those factors don't come into play, though McLovin's has the element of simplicity.

Both realised a key element in the solution in that if one of the fuses is lit at both ends, irrespective of the burn rate at different parts of the fuse, the burns will always meet at the 60 second point (which may be quite different to the half length point).

The solution that I like has elements of both of the above and also an idea that cynic suggested but didn't pursue.

Make a circle with one fuse so that both ends of it meet at one point. Place the 2nd fuse in a line with one end (it doesn't matter which end) starting from this same point but moving outwards from the circle. Light the three ends that meet with one match and start your count. When the circle fuse burns out (which will be at 60 seconds) light the other end of the 2nd fuse. At this time, the 2nd fuse will already have been burning for 60 seconds from the circle end, so this new burn will meet up at the old burn in 30 seconds, which gives your count.

It is very similar to McLovin's and has the same simplicity, but avoids the lag between having to light both ends of one fuse when the ends are not together. It also doesn't require any fuse placement action be taken once the fuses are initially laid out, which make it simpler that SKCs.


----------



## Tisme (19 May 2015)

bellenuit said:


> Your getting hot. One logistical problem though. The first leaves no residual marks, so how will you determine the same length as the first with what you have available. Also, since you have just matches, how do you light the first at both ends? Will there be a lag in moving the lit match from one end to the other or do you use both matches to do this?




I thought that would be self evident:
bring both ends together and light first fuse
lay lit fuse next to (parallel) unlit fuse
place burnt match on burnout point
have second match lit in readiness


----------



## skc (19 May 2015)

bellenuit said:


> It is very similar to McLovin's and has the same simplicity, but avoids the lag between having to light both ends of one fuse when the ends are not together. It also doesn't require any fuse placement action be taken once the fuses are initially laid out, which make it simpler that SKCs.




McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out.


----------



## McLovin (19 May 2015)

skc said:


> McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out.




Yeah good idea. That gets around the lit match problem.


----------



## Tisme (19 May 2015)

skc said:


> McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out.




Not sure that will work due to the non linear fashion of the burn, but worth doing the maths I suppose....later on.


----------



## cynic (19 May 2015)

Tisme said:


> Not sure that will work due to the non linear fashion of the burn, but worth doing the maths I suppose....later on.




Good point. 

If the slowest burning minute length of the fuse is sufficiently long, it would be possible to light the remaining unlit end by curling it to touch its burning end. This would leave one spare match and show the interviewer that one has an appreciation for economy as well as accuracy.


----------



## luutzu (19 May 2015)

cynic said:


> Good point.
> 
> If the slowest burning minute length of the fuse is sufficiently long, it would be possible to light the remaining unlit end by curling it to touch its burning end. This would leave one spare match and show the interviewer that one has an appreciation for economy as well as accuracy.




Place two fuse next to each other, same coloured ends together.

Use match 1 to light both fuse at, say Green ends and at the same time use match 2 to lit the other end - but only lit 1 fuse.

So you have 1 fuse (F1) burning at both ends (started at the same time with 2 matches); and the other fuse also burning at but only one end (was lit when you burn both fuses).

When the two ends of F1 meet, that's 60sec; at this point it's also been 60sec on F2 - with remaining 60sec to burn through for F2.

At point where F1 is about to burnt out, you hold F2 and place unburnt end there to be ignited by F1's 2 ends.

with F1 now gone and F2 burning at both ends... knowing there's only 60sec left when F2 was lit by F1.. .when F2 burnt out that's 30 sec. Hence 90.


Or you lit both and count 1 mississippi, 2 mississippi to 90.


----------



## skc (19 May 2015)

luutzu said:


> Or you lit both and count 1 mississippi, 2 mississippi to 90.




Nailed it.


----------



## bellenuit (19 May 2015)

bellenuit said:


> Make a circle with one fuse so that both ends of it meet at one point. Place the 2nd fuse in a line with one end (it doesn't matter which end) starting from this same point but moving outwards from the circle. Light the three ends that meet with one match and start your count. When the circle fuse burns out (which will be at 60 seconds) light the other end of the 2nd fuse. At this time, the 2nd fuse will already have been burning for 60 seconds from the circle end, so this new burn will meet up at the old burn in 30 seconds, which gives your count.




I still think the solution I offered last night is still the simplest and most accurate of everything I have read so far.

SKCs amendment to McLovin's solution (_McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out_) is just this restated. 

Tisme's potential issue (_Not sure that will work due to the non linear fashion of the burn, but worth doing the maths I suppose....later on_) is not a problem. The math is fine. A fuse lit at both ends will always burn out in half the time that the same fuse lit at just one end will. This applies to the first fuse burnt in full (yielding 60 seconds) and the second fuse that is already underway for 60 seconds and then has the other end lit. That remaining segment of fuse will burn out 30 seconds after the other end is lit.


----------



## bellenuit (19 May 2015)

Two easy ones.

#1

*Swiss Rail run a non stop goods train on a single line track between two towns separated by a large mountain range (part of the Alps). The train leaves every morning at 9:00 am and reaches its destination at 5:00 pm. Because of the danger of fallen rocks and snow drifts, the train only travels during the day, so it makes the return journey the following morning at 9:00 am reaching its destination at 5:00 pm. Because of the nature of the path, it doesn't travel at a constant speed, but will be slow going up hill and a lot faster going down hill and of course these same hills will assist or resist in the opposite way on the return journey.

Both towns are in flat valleys that give the driver sufficient time to moderate the speed upwards or downwards so that the train always arrives exactly on time, should the train be behind or ahead of schedule. Being exact to the scheduled arrival time is more important to Swiss Rail than being early

The question. One Monday morning the train makes its outbound journey and the following day it returns. Without the driver intentionally trying to do so, will the train (using its exact middle point as reference) ever be at the same point on the tracks at the same clock time on that Tuesday's return journey as it was on Monday's outward journey?*

and

#2.

*A blind person is given a standard (52) pack of playing cards that have 10 cards facing up and the remaining facing down. The cards have been shuffled, so the upward facing cards are distributed randomly throughout the deck. The pack is brand new and there is no way to tell if a card is facing up or down by feel. 

The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards.

Without the assistance of anyone else or a device of any sort, can he do it?*


----------



## barney (19 May 2015)

skc said:


> Nailed it.





LOL  ..... great thread!!     Humour and brain power


----------



## cynic (19 May 2015)

bellenuit said:


> Two easy ones.
> 
> #1
> 
> ...




Are both towns at the same level?


> and
> 
> #2.
> 
> ...



Do both stacks have to add up to the entire deck?


----------



## Tisme (19 May 2015)

skc said:


> Nailed it.




yeah both of you


----------



## keithj (19 May 2015)

bellenuit said:


> *A blind person is given a standard (52) pack of playing cards that have 10 cards facing up and the remaining facing down. The cards have been shuffled, so the upward facing cards are distributed randomly throughout the deck. The pack is brand new and there is no way to tell if a card is facing up or down by feel.
> 
> The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards.
> 
> Without the assistance of anyone else or a device of any sort, can he do it?*



Deal 6 equal stacks each containing 4 cards (& chuck the rest in the bin)

There is guaranteed to be at least two stacks containing the same number of cards.

eg worst possible scenario is 1st, 2nd, 3rd & 4th contains 1,2,3 and 4 upward facing cards, which means the 5th & 6th contain none.

Does this solution satisfy the the terms of the problem statement ?


Or the trivial solution is to deal 52 stacks each containing 1 card.....   ???


----------



## bellenuit (19 May 2015)

cynic said:


> Are both towns at the same level?




Not necessarily.



> Do both stacks have to add up to the entire deck?




Yes


----------



## bellenuit (19 May 2015)

keithj said:


> Deal 6 equal stacks each containing 7 cards.
> 
> There is guaranteed to be at least two stacks containing the same number of cards.
> 
> ...




Its meant to be just two stacks in total. Though I must concede that you have had him deal two stacks each with the same number of upward facing cards (0). Technically that is a solution, but I think pushing the boundaries of what the wording said:

_The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards._

A couple of marks for lateral thinking though.


----------



## keithj (19 May 2015)

bellenuit said:


> Its meant to be just two stacks in total. Though I must concede that you have had him deal two stacks each with the same number of upward facing cards (0). Technically that is a solution, but I think pushing the boundaries of what the wording said:
> 
> _The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks *have exactly the same number* of upward facing cards._
> 
> A couple of marks for lateral thinking though.



hmmm... thought so .

I think the bolded bit above is the key.  You'd expect exactly 2 decks with an equal number of upturned cards to be exactly 5 in each, however....








... by dealing 26 cards into one stack & 26 into the other stack, but flipping each 2nd card, you'd end up with around half the face up cards now facing down, and around half the face down ones now facing up...

... and closer to an equal number of face up cards in each stack.


getting closer ?


----------



## bellenuit (19 May 2015)

keithj said:


> getting closer ?




Depends where you are going.

Just to clarify for all doing this. He deals out all the 52 cards into two separate heaps so that each have the same number of upward facing cards. There need not be the same number of cards in each heap.


----------



## skc (19 May 2015)

bellenuit said:


> The question. One Monday morning the train makes its outbound journey and the following day it returns. Without the driver intentionally trying to do so, will the train (using its exact middle point as reference) ever be at the same point on the tracks at the same clock time on that Tuesday's return journey as it was on Monday's outward journey?[/B]




Yes... Visualise that you have 2 trains both departing at the same time, one from either town, running towards each other. They will definitely run into each other at one moment in time through the journey.

This would have been harder to work out but the fuse question before was too similar.


----------



## bellenuit (19 May 2015)

skc said:


> Yes... Visualise that you have 2 trains both departing at the same time, one from either town, running towards each other. They will definitely run into each other at one moment in time through the journey.




Correct



> This would have been harder to work out but the fuse question before was too similar.




It was the fuse problem that made me remember this problem.


----------



## galumay (19 May 2015)

I am thinking to turn the deck upside down so there are 42 facing up and only 10 face down, but it doesnt seem to help!

EDIT - Ok, got it. I wont post the solution yet. Will be interested to see if anyone works it out without cheating. (like me!)


----------



## cynic (20 May 2015)

One solution:

If the blind man were to tear each card into two pieces and pile them separately he would have two piles with the same number of card pieces facing up as facing down.

Another solution:

Deal out ten cards and then turn the ten card pile upside down.

Edit: Quick explanation:- The number of upside down cards in the remainder of the deck will equal ten less the upside cards dealt into the pile of ten. Hence the right way up cards in the ten pile equals the number of upside down cards in the remaining deck. Hence the need to turn the ten pile upside down.


----------



## bellenuit (20 May 2015)

cynic said:


> One solution:
> 
> Deal out ten cards and then turn the ten card pile upside down.
> 
> Edit: Quick explanation:- The number of upside down cards in the remainder of the deck will equal ten less the upside cards dealt into the pile of ten. Hence the right way up cards in the ten pile equals the number of upside down cards in the remaining deck. Hence the need to turn the ten pile upside down.




Correct. That's it.


----------



## skc (20 May 2015)

cynic said:


> Deal out ten cards and then turn the ten card pile upside down.




Delightfully simple. 

I can't believe I didn't work it work. 

BTW, Bellenuit... if you have more of these can you please post them earlier in the day (say around lunch). It's been affecting my sleep lately


----------



## barney (20 May 2015)

This is an oldie but still a goody if you don't know it ....... if you have a brain like mine it won't be immediately obvious 

I know these are simple, but ........ Welcome to my world


*Three friends check into a hotel.  They are staying in the same room.

They pay $30 to the manager and go to their room.

The manager suddenly remembers that the room rate is $25 and gives $5 to the bellboy to return to the people.

On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person.

Now each person paid $10 and got back $1.

So they paid $9 each, totalling $27. The bellboy has $2, totalling $29.

Where is the missing $1?*


----------



## cynic (20 May 2015)

skc said:


> Delightfully simple.
> 
> I can't believe I didn't work it work.
> ...




That's exactly how I felt about that fuse problem!

Based upon the current trend, I suspect that bellenuit is going to have us experiencing more of the same throughout the coming weeks.


----------



## skyQuake (20 May 2015)

barney said:


> This is an oldie but still a goody if you don't know it ....... if you have a brain like mine it won't be immediately obvious
> 
> I know these are simple, but ........ Welcome to my world
> 
> ...




Manager has 25, bellboy has 2, = $27
Each person has $-9, = $-27

Here's a few from a Children book that were surprisingly tricky!


----------



## bellenuit (20 May 2015)

skyQuake said:


> Manager has 25, bellboy has 2, = $27
> Each person has $-9, = $-27
> 
> Here's a few from a Children book that were surprisingly tricky!
> View attachment 62644




I think for the first one, the water will be heavier by the equivalent weight of water that his finger displaces.

First take a detached object that is floating. When something floats in water that is equal or less dense than water, it displaces the equivalent amount of water to its own weight and the total weight (if it were a container like the picture) would be the original weight plus the weight of the floating object (or plus the weight of the displaced water, which is the same). This is Archimedes Principle. The upward push from the water is equivalent to the downward push of the object. If the object is denser than water, the amount of water it displaces if fully submerged, weighs less than the object itself, so the upward pressure is less than the downward pressure hence it sinks.

In this case the finger doesn't float but is being held up by the hand. But the amount of pull pressure the hand needs to hold the finger at that point is reduced by the push up from the water and that push up is equivalent to the amount of water the finger displaces. If the finger were detached and assuming it were denser than water it would sink as in the example above. The pull exerted by the hand to stop it sinking is equivalent to the difference between the weight of the submerged finger and the weight of the displaced water.

It is very hard to judge the "pull" needed in practice as most of the effort you would feel is in holding up your complete arm and hand and the finger is a negligible portion of that. But if instead of your finger, imaging most of your body submerged in water and you are holding on to a bar of some sort. It should be obvious that it takes less effort to stay holding the bar when your are nearly fully submerged in water than if you were doing it completely out of water. That's because of the assistance the water is giving you keeping you almost afloat with you only having to contribute a small bit of extra effort to so that the displaced water's upward pressure equals you body's downward pressure.


----------



## trainspotter (20 May 2015)

First one is a NO - it is called displacement. As long as his finger does not touch the bottom or sides then it will NOT be heavier.

Second - Because it was typed

Third - The elephant. Same colour dress as the bag and we all KNOW elephants LOVE to colour match their accesories


----------



## bellenuit (20 May 2015)

trainspotter said:


> First one is a NO - it is called displacement. As long as his finger does not touch the bottom or sides then it will NOT be heavier.




If the finger was held up not by the hand but was severed and held up by a very slender thread that someone was holding in their hand so that the finger doesn't sink. It still doesn't touch the sides or bottom, but would clearly add to the weight of the glass.

The picture shows the glass not being full, so I take it we can assume that the displaced water doesn't overflow.


----------



## cynic (20 May 2015)

Answer to question 3: The butler did it!

Answer to question 2: The butler did it!!

Answer to question 1: Ask the bleeding butler, she must know because she's always doing it!!!


----------



## skyQuake (20 May 2015)

1. First one is indeed Yes. It does weight more - the finger has taken the place of the water and also "fills in" for the weight of the water. But since the water is still there, the glass weights more.
TL;DR - The total volume of finger+water in the glass has increased, thus weight increases.

I'll give an extra clue with 2&3 - pay close attention to the picture. This one ain't abstract


----------



## skc (20 May 2015)

skyQuake said:


> 1. First one is indeed Yes. It does weight more - the finger has taken the place of the water and also "fills in" for the weight of the water. But since the water is still there, the glass weights more.
> TL;DR - The total volume of finger+water in the glass has increased, thus weight increases.
> 
> I'll give an extra clue with 2&3 - pay close attention to the picture. This one ain't abstract




2. It's a 4 ring binder but Bear's paper only has 3 holes.

3. Cat with the missing high heel.


----------



## cynic (20 May 2015)

skc said:


> 2. It's a 4 ring binder but Bear's paper only has 3 holes.




I knew it!! The butler has a three ring binder!!!


----------



## cynic (20 May 2015)

skyQuake said:


> 1. First one is indeed Yes. It does weight more - the finger has taken the place of the water and also "fills in" for the weight of the water. But since the water is still there, the glass weights more.
> TL;DR - The total volume of finger+water in the glass has increased, thus weight increases.
> 
> I'll give an extra clue with 2&3 - pay close attention to the picture. This one ain't abstract




The elephant is the butler!!!


----------



## Ves (20 May 2015)

Love this sort of thing.  Haven't posted any responses,  because usually when I've had the chance to check someone else has long figured it out.  

Just a thought.... not sure if it's a problem for others,  but it's hard to open this thread and not be exposed to spoilers  (ie.   someone else's thoughts or even the answer).

Would it be possible to have two separate threads - one exclusively for the logic puzzles / riddles  and another for discussions of the processes and eventual answers? Shouldn't be too hard to cross-reference between the two with the right headings or post references  (Tech/A used to do it in his charting threads).   I hope that something like this would keep it in the spirit of community collaboration,  because it's great to see,  especially in the often maligned general chat section of ASF.


----------



## skyQuake (20 May 2015)

skc said:


> 2. It's a 4 ring binder but Bear's paper only has 3 holes.
> 
> 3. Cat with the missing high heel.




Correct!

Well done ASF, you have collectively passed the recommend requirement of 9 and over

Lets see how you fare with the next one


----------



## cynic (20 May 2015)

Ves said:


> Love this sort of thing.  Haven't posted any responses,  because usually when I've had the chance to check someone else has long figured it out.
> 
> Just a thought.... not sure if it's a problem for others,  but it's hard to open this thread and not be exposed to spoilers  (ie.   someone else's thoughts or even the answer).
> 
> Would it be possible to have two separate threads - one exclusively for the logic puzzles / riddles  and another for discussions of the processes and eventual answers? Shouldn't be too hard to cross-reference between the two with the right headings or post references  (Tech/A used to do it in his charting threads).   I hope that something like this would keep it in the spirit of community collaboration,  because it's great to see,  especially in the often maligned general chat section of ASF.




Good idea!


----------



## cynic (20 May 2015)

I think I know the answer to #5, but I'll wait for creation of that other thread.

Edit: and #6 also.

2nd edit: and #5 as well.


----------



## skyQuake (20 May 2015)

Link to the spoilers thread
https://www.aussiestockforums.com/forums/showthread.php?t=29872&


----------



## bellenuit (20 May 2015)

*You have a box containing a number of red and blue balls. You also have an unlimited supply of red and blue balls outside the box at your disposal.

You take two balls from the box at random. If they are the same colour you put back in a red ball. If they are different colours you put back in a blue ball. Thus after each go, you reduce the number of balls in the box by 1 and will eventually have just 1 ball in the box which ends the game.

Assuming you are told the number of blue and the number of red balls in the box at the outset, can you determine the colour of the last remaining ball. You can assume there are at least two balls in the box to begin with.*


----------



## bellenuit (20 May 2015)

A famous classic puzzle.

*You are a criminal found guilty of a serious crime and have been sentenced to death. The judge, being a puzzle freak, decides to give you a chance of avoiding death.

He shows you three closed boxes on a table. You are told two are empty, but the third contains a royal pardon. You are to point to one of the boxes and if it is the one with the pardon inside you will be freed, otherwise you will be put to death.

You point to one of the boxes. The judge, knowing which box contains the pardon, opens the lid of one of the remaining boxes, one which he knows for sure doesn't have the pardon and shows you that it is empty.

He then tells you that he will allow you to change your mind if you want. You can either stick with the box you originally pointed to or you can pick the other unopened one instead. Assuming you do not want to die, does changing you choice increase your chance of freedom or make no difference. Explain why.*


----------



## skyQuake (20 May 2015)




----------



## bellenuit (21 May 2015)

Just to keep it moving along.

*There are 4 closed boxes each containing a ball of a different colour (one ball per box). There is a competition to see who can guess the correct coloured ball in each box. 123 people participate.

When the boxes are opened, it turns out that:
43 people have guessed none of the box contents correctly,
39 have guessed 1 box correctly, 
31 have guessed 2 boxes correctly.

How many have guessed 3 boxes and how many have guessed all 4 boxes correctly?

(Note: it should be obvious from the figures, but guessing 2 correctly means exactly 2. Ditto for 3 and 4. So no arguments such as if you guessed 2 correctly you also must have guessed 1 correctly. They are not cumulative.*


----------



## pixel (21 May 2015)

bellenuit said:


> Just to keep it moving along.
> 
> *There are 4 closed boxes each containing a ball of a different colour (one ball per box). There is a competition to see who can guess the correct coloured ball in each box. 123 people participate.
> 
> ...




43+39+31=113
Leaves 10 to get them all right.


----------



## galumay (21 May 2015)

pixel said:


> 43+39+31=113
> Leaves 10 to get them all right.




Pixel, the spoiler thread is the place to post your answers, it was set up so people dont see the answers when they open this thread - before they have had a chance to solve them.

(makes me think about a potentially more elegant solution, some forums have a 'spoiler text' option that means you can only read that segment of the post when you hover the mouse over it - that would save having a separate thread but also hide the answers people posted.)


----------



## bellenuit (21 May 2015)

A hard one to put your brain in top gear. It does require some math though, but not much more than knowing the circumference of a circle is 2πR (π being the mathematical constant pi, in case it doesn't display correctly on your device).

*You, a Greek mathematician, are swimming in a perfectly round lake when you see a savage cyclops eying you from the shore. He has manoeuvred to the closest point on the shore to where you currently are and there is nothing he would like more than to eat you. You know the  cyclops can't swim, but he can run 4 times faster than you can swim. However, you can run much faster than the cyclops, so you know if you can get to the shore before he gets to you, you can outrun him and escape.

You are a good swimmer and can maintain your maximum speed indefinitely even if it involves sharp changes in direction (e.g. no loss of speed if you switch direction). The same applies to the cyclops. He can run 4 times faster than you can swim and switch direction without loss of pace.

Is it possible for you to get to the shore and escape assuming he always tries to get to the point on the shore where you are trying to swim to? If so, explain how.*


----------



## bellenuit (22 May 2015)

I'm updating the last paragraph of this question as it was phrased badly and may lead people astray.

*You, a Greek mathematician, are swimming in a perfectly round lake when you see a savage cyclops eying you from the shore. He has manoeuvred to the closest point on the shore to where you currently are and there is nothing he would like more than to eat you. You know the cyclops can't swim, but he can run 4 times faster than you can swim. However, you can run much faster than the cyclops, so you know if you can get to the shore before he gets to you, you can outrun him and escape.

You are a good swimmer and can maintain your maximum speed indefinitely even if it involves sharp changes in direction (e.g. no loss of speed if you switch direction). The same applies to the cyclops. He can run 4 times faster than you can swim and switch direction without loss of pace.

Is it possible for you to get to the shore and escape assuming he is very smart and always tries to get to that point on the shore that is closest to where you currently are? If so, explain how.*


----------



## skyQuake (22 May 2015)

Last Slylock I got.

If you enjoyed, I highly recommend buying the books. For kids/partner/yourself(?)


----------



## bellenuit (25 May 2015)

Denise's Birthday Puzzle (Cheryl Part 2)

*Albert, Bernard and Cheryl became friends with Denise, and they wanted to know when her birthday is. Denise gave them a list of 20 possible dates.

17 Feb 2001, 16 Mar 2002, 13 Jan 2003, 19 Jan 2004

13 Mar 2001, 15 Apr 2002, 16 Feb 2003, 18 Feb 2004

13 Apr 2001, 14 May 2002, 14 Mar 2003, 19 May 2004

15 May 2001, 12 Jun 2002, 11 Apr 2003, 14 Jul 2004

17 Jun 2001, 16 Aug 2002, 16 Jul 2003, 18 Aug 2004

Denise then told Albert, Bernard and Cheryl separately the month, the day and the year of her birthday respectively.

The following conversation ensues:

Albert: I don’t know when Denise’s birthday is, but I know that Bernard does not know.
Bernard: I still don’t know when Denise’s birthday is, but I know that Cheryl still does not know.
Cheryl: I still don’t know when Denise’s birthday is, but I know that Albert still does not know.
Albert: Now I know when Denise’s birthday is.
Bernard: Now I know too.
Cheryl: Me too.

So, when is Denise’s birthday?

To clarify: neither Albert, Bernard or Cheryl know anything else at the start apart from the fact that Albert has been told the month, Bernard the day (meaning the number of the day), and Cheryl the year.*


----------



## bellenuit (27 May 2015)

*An Saudi sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. After hearing his advice, they jump on the camels and race to the city as fast as they can. So what advice did the wise man offer to them?*


----------



## bellenuit (27 May 2015)

*You have 12 billiard balls that look identical in every respect, except one is slightly different in weight (could be heavier or lighter than the other 11). You have a simple balance (one that stays level if both trays contain exactly the same weight, but will tip one way or the other if they are different).

You have to determine by no more than 3 weighings, which ball is different and whether it is heavier or lighter than the others. How can you do it?*


----------



## bellenuit (31 May 2015)

Not as easy as it looks.......

*Interesting Numbers

Give me a 7 digit number with the following properties:

The 1st (leftmost) digit gives the number of digit 0s in the number.
The 2nd digit gives the number of digit 1s in the number.
The 3rd digit gives the number of digit 2s in the number.
The 4th digit gives the number of digit 3s in the number
etc. until finally ...
The 7th digit gives the number of digit 6s in the number.

For example, 1210 is a 4 digit number with those same properties*

If you are adventurous, you can also gives answers for 8, 9 and 10 digit numbers.

It should be obvious that there isn't an answer for every possible number of digits. For example, there isn't a 1 digit number that could have those properties.

And for those of you who have nothing better to do, give a 16 digit number (in hexadecimal) that has those properties. Hexadecimal, for those not familiar with it, is to the base 16 rather than base 10 of our decimal system. It uses A, B, C, D, E and F to represent 10 to 15 in the number scale.


----------



## bellenuit (11 June 2015)

*A bag contains one marble and there is a 50% chance that it is either Red or Blue.

A Red marble is added to the bag, the bag is shaken and a marble is removed. This removed marble is Red.

What is the probability that the remaining marble is Red?*


----------



## luutzu (12 June 2015)

bellenuit said:


> *A bag contains one marble and there is a 50% chance that it is either Red or Blue.
> 
> A Red marble is added to the bag, the bag is shaken and a marble is removed. This removed marble is Red.
> 
> What is the probability that the remaining marble is Red?*




50?


----------



## galumay (12 June 2015)

for answers please use the spoilers thread

https://www.aussiestockforums.com/forums/showthread.php?t=29872&


----------



## trainspotter (12 June 2015)

bellenuit said:


> *A bag contains one marble and there is a 50% chance that it is either Red or Blue.
> 
> A Red marble is added to the bag, the bag is shaken and a marble is removed. This removed marble is Red.
> 
> What is the probability that the remaining marble is Red?*




Does not compute ?


----------



## bellenuit (12 June 2015)

bellenuit said:


> *A bag contains one marble and there is a 50% chance that it is either Red or Blue.
> 
> A Red marble is added to the bag, the bag is shaken and a marble is removed. This removed marble is Red.
> 
> What is the probability that the remaining marble is Red?*




To clarify, in case there is confusion. The initial marble has a 50% chance of being Red and a 50% chance of being Blue. This is in case someone is thinking that the chance of it being red or blue is 50% (25% each) and the chance of it being another colour is 50%.


----------



## bellenuit (13 June 2015)

*You have a perfect solid sphere upon which you designate two opposite points as poles (like the earth's North and South poles). You drill a perfect cylindrical hole from one pole to the other (e.g. the hole goes right through the centre). The length of the hole is 2 cm. This is the length of the edge of the hollow cylinder created by the hole, not the diameter of the sphere (or if you were to put the sphere in a vice with what were the pole ends touching the arms of the vice, this would be the separation of those arms). See image below. 

What is the remaining solid volume of the sphere?

(Crucial) hint: You have been given enough information to solve this problem.

For those whose schooldays are a distant memory, the volume of a sphere is (4πr**3)/3 where π is Pi and r**3 means the radius r cubed.*


----------



## bellenuit (19 June 2015)

An easy one for those still up this late......

*You have 4 pieces of paper. On one side of each piece of paper is a number, on the other a letter. The pieces of paper are laid out in front of you and the sides facing up shows these 4 letters/numbers: E K 4 9

What is the minimum number of these and which ones must you turn to prove the following rule is true: "if a piece of paper has a vowel on one side, the other side must have an even number".

Which must you turn and why?*

Answers in the spoilers thread please.


----------



## bellenuit (19 June 2015)

A harder one........

*You have a 99 X 99 squared draught board and an unlimited supply of draughts. The colour of the squares or of the draughts are not important for this game. Initially the board is clear of draughts. You play the following game against a single opponent. Each player must in turn place a draught on the board, but it can only be in an empty square that doesn't have a draught in any of the 8 squares surrounding it (e.g. doesn't have a draught in any of the immediately adjacent squares either horizontal, vertical or diagonal to the square you want to place your draught in). The first player that is unable to place a draught loses.

You get to go first. What game strategy should you use to ensure you always win.*

Answers in the spoilers thread please.


----------



## trainspotter (19 June 2015)

TAKE a look at the two wheels in the image above. Which one is moving faster?


----------



## bellenuit (23 June 2015)

Another in the same vein as the Blue Eyed Slaves.

*Prison Numbers

Two of the kingdom's finest logicians are caught embezzling the kingdom's coffers and are sentenced to life imprisonment in separate solitary confinement. The king, a puzzle addict, decided to give them a chance of freedom, and posed the following challenge to them.

Each was separately asked to think of an integer greater than or equal to zero and tell it to the jailer who reported both numbers to the king. The king then had a messenger convey this message to both prisoners:

"Each day at 5pm, the jailer will come to your separate cells and give you the opportunity to tell him what number the other prisoner thought of and if you are correct, you will be freed. You can only arrive at the answer through logical deduction and guessing or cheating in any way is out of the question (there are horrible consequences for doing so). If you have logically deduced the number the other prisoner thought of you must tell the jailer, otherwise you just say you do not know. In the latter case the jailer will tell you, after he has heard the other prisoner's response, whether the other prisoner is to be freed or not that day."

With both thinking this is an impossible challenge, the messenger then added: "To assist you, the king has said that the sum of both numbers is either 10 or 13."

On the 5th day, both prisoners are freed. What were the two numbers that they thought of? How did you arrive at this conclusion?*

Answers in Spoilers thread please.


----------



## bellenuit (26 June 2015)

A hardish one for the weekend.

*Box of Marbles of Four Colours* 

*A father comes home and gives his maths prodigy daughter Mary a box of marbles as a present. "This box contains marbles of 4 different colours; red, green, blue and yellow. There are more red than green, more green than blue and more blue than yellow" said the father. "How many of each colour are there?" asks Mary. "You tell me" says the father and adds "I can tell you that in total there are fewer than 18 marbles and that the product of the four numbers is the same as our house number" Mary thinks for a while and then says "I need to know a bit more. Is there more than one yellow marble?" As soon as the father replies, Mary says "I've worked it out" and correctly states the number of each colour.

Knowing her own house number and whether or not there was more than one yellow marble, Mary found the problem trivial. 

Based on the above information only, you can now work out the number of marbles of each colour. How many of each are there and give your bullet proof reasoning for you to arrive at those numbers?*

Answers in the Spoilers thread please.


----------



## SirRumpole (18 July 2015)

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.



May 15, May 16, May 19



June 17, June 18



July 14, July 16



August 14, August 15, August 17



Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.



Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.



Albert: Then I also know when Cheryl’s birthday is.



So when is Cheryl’s birthday?


----------



## bellenuit (19 July 2015)

SirRumpole said:


> Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is.




SirR, this has been posed already. See #5 on this thread.


----------



## SirRumpole (19 July 2015)

bellenuit said:


> SirR, this has been posed already. See #5 on this thread.




My apologies


----------



## cynic (19 July 2015)

SirRumpole said:


> My apologies




I got really excited when I logged in to and saw an update to this thread thinking "Oh goody! Another delightful puzzle!!".

As soon as I realised it was a repeat, my excitement was quickly replaced by disappointment. 

However, I am sincerely appreciative that yourself and others have willing put in the time and effort to offer ASF  some terrific puzzles.


----------



## SirRumpole (19 July 2015)

cynic said:


> I got really excited when I logged in to and saw an update to this thread thinking "Oh goody! Another delightful puzzle!!".
> 
> As soon as I realised it was a repeat, my excitement was quickly replaced by disappointment.
> 
> However, I am sincerely appreciative that yourself and others have willing put in the time and effort to offer ASF  some terrific puzzles.




You're welcome Cynic,

Maybe the last word on the ABC puzzle


----------



## bellenuit (21 July 2015)

This (old) one appeared on a newspaper blog today, so some of you may have seen it. For those who haven't, here goes:

*Increasing the Ratio of Women in the Population.

A king (dirty old leech) decided that there should be proportionally more women in his kingdom than the current 50%. He knows that the probability of a girl at birth is exactly 50% and that males and females have on average exactly the same lifespan (in his kingdom anyway), so he assumes this strategy will work.

He orders that couples should strive to have as many children as possible, but only under the following conditions:

1. If they have a female baby, they are to have no more children.
2. If they have a male baby, they are to continue to have children until condition 1 is satisfied.

Question 1: Will that strategy increase, reduce or make no difference to the proportion of women in the population.

Question 2: With only recourse to edicts of the above kind (when to reproduce and not reproduce etc.) and without recourse to unsavoury methods such as infanticide, forced emigration or things similar, what, if any, would be the optimum policy to adopt to achieve an increased proportion of women in the population. (I know this part is going to cause a lot of controversy).*

Answers in the spoilers thread please and the reasoning for your conclusions are required.


----------



## bellenuit (28 July 2015)

An easy one that can be solved with no math, just some logic.

*A plane flies in a straight line from Perth to Sydney, then back in a straight line from Sydney to Perth. It travels with a constant engine speed (constant air speed) and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from Perth to Sydney?*

Answers in Spoiler thread please.


----------



## bellenuit (29 July 2015)

This one is not so easy......

*Four people with different physical ailments are at the edge of a wide raging river that they must cross to escape from a bushfire quickly heading towards them. The river is spanned by a rickety old rope supported timber batten bridge that can only hold two people at a time.

Because of their physical ailments, the four cannot cross the bridge at the same rate. A can do it in 1 minute, B in 2, C in 5 and D in 10 minutes. 

It is nighttime and they have just one dim flashlight between them. Without the flashlight, trying to cross the dilapidated bridge would mean certain death. Since two crossing must share the flashlight, they can only cross at the pace the slowest of the two is capable of.

What is the quickest time all four can get safely across the river and how?*

Answers in Spoilers thread please.


----------



## bellenuit (29 July 2015)

*You have the four 3-link chains shown on the left. Your task is to join them together to form the 12-link chain on the right. Each link has a single cut through it that can be prised open to join (or unjoin) two links and then is prised closed.

What is the minimum number of links that must be prised open to allow the 12-link circle to be formed. Explain method used.*

Answers in Spoilers thread please


----------



## bellenuit (3 August 2015)

Five similar puzzles that I came across today in increasing complexity. You are to find the missing value in each of the following images (answers in Spoilers Thread).......

Note: For all puzzles, all answers are whole numbers (no fractions). The instructions are a bit ambiguous, but I also understand that to solve the puzzles you do not need to deal with fractions.

*Number 1*


----------



## bellenuit (3 August 2015)

*Number 2*


----------



## bellenuit (3 August 2015)

*Number 3*

Note: I can confirm that having now solved Number 3, that 1, 2 and 3 can be solved without resorting to using fractions in interim results. I assume the same applies to 4 and 5.


----------



## bellenuit (3 August 2015)

*Number 4*


----------



## bellenuit (3 August 2015)

*Number 5*


----------



## bellenuit (11 August 2015)

A fairly easy one.




*You have a board marked into a grid of 20 X 20 squares. You also have rectangular dominoes that are two grid squares in size, like the one shown above. Clearly you could cover the total board surface with 200 of these dominoes if they are laid flat, each covering two squares ((20 * 20) /2).

However, two diagonally opposite corner squares are cut from the board as shown above. Is it possible to place 199 dominoes flat down on the board so that they completely cover the remaining surface of the board? 

If yes, prove it. If no, prove it.*

Answers in Spoilers thread.


----------



## bellenuit (31 August 2015)

Another easy chessboard problem......

*You have an infinitely large chessboard (one that has an infinite number of rows and columns) where each square has sides of 2cm. You throw a perfectly round 1cm diameter coin onto the chessboard. Assuming it always lands flat, what is the probability that it will land straddling two or more squares.*

Answers in Spoilers


----------



## bellenuit (13 September 2015)

A few easy ones.

No. 1

*Anne and Jessica were working on the computer along with their friends Sandy and Nicole.  Suddenly, I heard a crash and then lots of shouts.  I rushed in to find out what was going on, finding the computer monitor on the ground, surrounded with broken glass!   

Jessica saying, "It wasn't me!"
Sandy saying, "It was Nicole!"
Anne yelled, "No, it was Sandy!"
With a straight face Nicole said, "Sandy's a liar." 

Only one of them was telling the truth, so who knocked over the monitor?*

No. 2

*How can you cut a traditional circular cake into 8 equal size pieces, with only 3 cuts?*


----------



## cynic (13 September 2015)

bellenuit said:


> A few easy ones.
> 
> No. 1
> 
> ...




It occurs to me that No. 1. can serve as the basis for a further puzzle, if, instead of having only one girl telling the truth, posing the question of whodunnit when only one of the girls is lying.


----------



## bellenuit (28 September 2015)

This one requires a fair bit of thinking....

*Draw a side view of this wooden object

This is the top and front view of a real three dimensional wooden object. Real implies it has thickness and is not, for instance, two flat plates of zero thickness at right angles. Any hidden sides are shown with dotted lines (and there are none shown on these two views implying all sides are visible). The inner square represents a square hole running through the object (it is not just a square drawn on the object). Its lines are solid because the edges of the square are visible and not hidden.*





*You are to draw a view of the object looking at it from the left hand side (with any hidden sides shown with dotted lines). 

To help clarify, the following answer would be false.*




*Apart from the fact it is not a proper left side view, it is false because if it were that shape, the original two drawings would be represented as follows to show the hidden edges and they are obviously not.*





Answers in Spoilers


----------



## noco (29 September 2015)

Not sure if it is the right thread but it makes for interesting reading.


*1. The Post Office
Get ready to imagine a world without the post office. They are so deeply in 
financial trouble that there is probably no way to sustain it long term. Email, 
Fed Ex, and UPS have just about wiped out the minimum revenue needed to keep the 
post office alive. Most of your mail every day is junk mail and bills.

2. The Cheque
Britain is already laying the groundwork to do away with cheque by 2018.  It costs 
the financial system billions of dollars a year to process cheques.  Plastic 
cards and online transactions will lead to the eventual demise of the 
cheque.  This plays right into the death of the post office.  If you 
never paid your bills by mail and never received them by mail, the post office 
would absolutely go out of business.

3. The Newspaper
The younger generation simply doesn't read the newspaper.  They certainly don't 
subscribe to a daily delivered print edition.  That may go the way of the 
milkman and the laundry man.  As for reading the paper online, get ready to 
pay for it.  The rise in mobile Internet devices and e-readers has caused 
all the newspaper and magazine publishers to form an alliance.  They have 
met with Apple, Amazon, and the major cell phone companies to develop a model 
for paid subscription services.

4. The Book
You say you will never give up the physical book that you hold in your hand and turn 
the literal pages  I said the same thing about downloading music from 
iTunes.  I wanted my hard copy CD.  But I quickly changed my mind when 
I discovered that I could get albums for half the price without ever leaving 
home to get the latest music.  The same thing will happen with books.  
You can browse a bookstore online and even read a preview chapter before you 
buy.  And the price is less than half that of a real book.  And think 
of the convenience!  Once you start flicking your fingers on the screen 
instead of the book, you find that you are lost in the story, can't wait to see 
what happens next, and you forget that you're holding a gadget instead of a 
book.

5. The Land Line Telephone
Unless you have a large family and make a lot of local calls, you don't need it 
anymore.  Most people keep it simply because they've always had it.  
But you are paying double charges for that extra service.  All the cell 
phone companies will let you call customers using the same cell provider for no 
charge against your minutes.

6. Music
This is one of the saddest parts of the change story.  The music industry is 
dying a slow death.  Not just because of illegal downloading.  It's 
the lack of innovative new music being given a chance to get to the people who 
would like to hear it.  Greed and corruption is the problem.  The 
record labels and the radio conglomerates are simply self-destructing.  
Over 40% of the music purchased today is "catalogue items," meaning traditional 
music that the public is familiar with.  Older established artists.  
This is also true on the live concert circuit.  To explore this fascinating 
and disturbing topic further, check out the book, "Appetite for 
Self-Destruction" by Steve Knopper, and the video documentary, "Before the Music Dies."

7. Television Revenues
To the networks are down dramatically.  Not just because of the economy.  
People are watching TV and movies streamed from their computers.  And 
they're playing games and doing lots of other things that take up the time that 
used to be spent watching TV.  Prime time shows have degenerated down to 
lower than the lowest common denominator.  Cable rates are skyrocketing and 
commercials run about every 4 minutes and 30 seconds.  I say good riddance 
to most of it.  It's time for the cable companies to be put out of our 
misery.  Let the people choose what they want to watch online and through Netflix.

8. The "Things" That You Own
Many of the very possessions that we used to own are still in our lives, but we may 
not actually own them in the future.  They may simply reside in "the 
cloud."  Today your computer has a hard drive and you store your pictures, 
music, movies, and documents.  Your software is on a CD or DVD, and you can 
always re-install it if need be.  But all of that is changing.  Apple, 
Microsoft, and Google are all finishing up their latest "cloud services."  
That means that when you turn on a computer, the Internet will be built into the 
operating system.  So, Windows, Google, and the Mac OS will be tied 
straight into the Internet.  If you click an icon, it will open something 
in the Internet cloud.  If you save something, it will be saved to the 
cloud.  And you may pay a monthly subscription fee to the cloud 
provider.  In this virtual world, you can access your music or your books, 
or your whatever from any laptop or handheld device.  That's the good 
news.  But, will you actually own any of this "stuff" or will it all be 
able to disappear at any moment in a big "Poof?"  Will most of the things 
in our lives be disposable and whimsical?  It makes you want to run to the 
closet and pull out that photo album, grab a book from the shelf, or open up a 
CD case and pull out the insert.

9. Joined Handwriting (Cursive Writing)
Already gone in some schools who no longer teach "joined handwriting" because nearly 
everything is done now on computers or keyboards of some type (pun not intended)

10. Privacy
If there ever was a concept that we can look back on nostalgically, it would be 
privacy.  That's gone.  It's been gone for a long time anyway..  
There are cameras on the street, in most of the buildings, and even built into 
your computer and cell phone.  But you can be sure that 24/7, "They" know 
who you are and where you are, right down to the GPS coordinates, and the Google 
Street View.  If you buy something, your habit is put into a zillion 
profiles, and your ads will change to reflect those habits..  "They" will 
try to get you to buy something else.  Again and again and again.

All we will have left that which can't be changed.......are our "Memories".

Logic is dead.
Excellence is punished.
Mediocrity is rewarded.
and 
Dependency is to be revered.

When people rob banks they go to prison.
When they rob the taxpayer they get re-elected






*


----------



## bellenuit (5 January 2016)

One to get you going for 2016.

*The countdown to the new year:
10 9 8 7 6 5 4 3 2 1 = 2016

Your task is to fill in the gaps between those 10 sequential numbers on the left of the = sign so that the equation = 2016. You can use +, -, * and /. You can also use parenthesis to ensure there is no ambiguity in the order the calculations are performed, although if omitted the standard order of calculation; *, / before +, -; will be assumed.

The countdown order must be maintained and you cannot concatenate adjacent numbers (for example, you cannot treat the three numbers 7 6 5 as the single number 765) 

So, for example, a solution could look like this....

10 * (9+8) -7............. =2016

The answer isn't unique as there are several ways to make the equation work.
*

Answers in spoilers thread please.


----------



## bellenuit (13 January 2016)

Apparently >80% of people get this wrong.

*Jack is looking at Anna and Anna is looking at George. Jack is married and George is not.

Question:

On the information given above, is a married person looking at an unmarried person?

A. Yes
B. No
C. Cannot be determined*

Answers in Spoilers please.


----------



## bellenuit (25 January 2016)

A very hard one to keep you busy over the Australia Day period. It requires poetry and math.

*Can you express this equation in words in the format of a Limerick?*




Answers (and there can be more than one obviously) in Spoilers please.


----------



## SirRumpole (31 January 2016)

I'm sure the people here will solve this in an instant !

'No winning answers yet' to British spy agency's Christmas puzzle challenge

http://www.abc.net.au/news/2016-01-...o-british-spy-agency-christmas-puzzle/7127758


----------



## bellenuit (3 February 2016)

A shadowy puzzle

*You have a light source that throws light out in all directions. You are to construct a closed room around that light source using straight perpendicular walls which meet each other at corners. The shape you end up with must be such that every wall is either partly or fully shaded from the light source (e.g.  every wall of the room is either partly or fully shaded from the light source by other walls in the room).

There is no limit to the number of walls you require to do this, but those that can do it with the least number of walls are the best solutions.

There is no trick to the question. We are just looking for a two dimensional shape (as if looking down from above). You assume the position of the light source is fixed and design your shape from that, so you are NOT required to produce a shape that subsequently must have all walls partly/fully shaded if the light source is move to another position. Also, it is a continuous shape, where the walls join each other at their ends only with the first wall meeting up with the last to form a complete closed shape with the light source within it.*


----------



## bellenuit (4 February 2016)

bellenuit said:


> A shadowy puzzle
> 
> *You have a light source that throws light out in all directions. You are to construct a closed room around that light source using straight perpendicular walls which meet each other at corners. The shape you end up with must be such that every wall is either partly or fully shaded from the light source (e.g.  every wall of the room is either partly or fully shaded from the light source by other walls in the room).
> 
> ...




Some additional information....

I hope I haven't confused people by describing the walls as perpendicular. I meant perpendicular to the ground, not perpendicular to each other. They do not have to meet at 90 degree angles. The reason I specified perpendicular (to the ground) was that some people who were attempting to solve the problem when it was posed in the paper I read had asked if the walls could be at other than 90 degrees to the ground and such solutions were excluded by the problem poser.


----------



## bellenuit (14 March 2016)

*Sudoku with a Twist*

*An interesting one for Sudoku fans. It is a standard Sudoku puzzle, except it has no pre-given numbers. Instead, additional information is provided by a sub-puzzle within the larger puzzle. Familiarity with Kakuro certainly helps.*




*It is a standard Sudoku puzzle in that each 9-cell row (9 in total), each 9-cell column (9 in total) and each 9-cell 3X3 box with a bold outline (9 in total) must contain the digits 1 through 9 without repetition within the row, column or 3X3 box. The numbers within a row, column or 3X3 box can be in any order.

Additionally, cells are grouped together and delineated by dotted lines. Each group has a number at the top leftmost corner, the total for that group. All the cells within a group must add up to the groups total and no number can be repeated within a group.

Like all Sudoku puzzles, there is only one valid solution.*

Answers in Spoilers thread please.


----------



## bellenuit (14 March 2016)

Pi Day

I haven't tried this myself yet, but in honour of today being Pi day, I thought I would get it posted before end of day. Every year has Pi Day on 3.14 (March 14th) as it represents Pi to 2 decimal places in US date format. This year is also once in a century unique because when the year is added in we get Pi to 4 decimal places, 3.1416. Anyway, the problem:




Answers in Spoilers thread please.


----------



## bellenuit (6 June 2016)

An easy one......




*You have a 4 square X 4 square board with 4 pieces: 1, 2, 3 and T. The grey areas represent unoccupied space.

You can move a piece (only one at a time allowed) by sliding (not removing it) on the board to any other unoccupied space, that need not be adjacent, so long as it is only slid through unoccupied space (e.g. you cannot push another piece out of the way).

Your task is to move the big piece T to the bottom right hand corner in just 5 moves (without changing its orientation). 

For easy nomenclature, imagine the board as a 4 X 4 grid with rows numbered A, B, C and D and columns numbered 1, 2, 3, 4. For the big piece T, represent where it is moved to by the position of its bottom right hand square, which currently sits in B3.

So to move piece 3 to the bottom left hand square write: 3 to D1.*

Answers in Spoilers thread please.


----------



## bellenuit (20 June 2016)

*Anyone for Tennis?

Q1. The men’s singles event at Wimbledon is a knockout tournament with 128 players. How many matches are there in total (extra points for those that can show how they arrived at the answer with the least amount of effort).

Q2. What is the minimum number of times you need to hit the ball with your racket in order to win a set in tennis, assuming that the set goes its full course (e.g. your opponent doesn't forfeit the set for any reason or the set isn't abandoned for any reason like weather etc)?*


----------



## bellenuit (1 August 2016)

An easy one.

*Flipping For Your Life*

*You and a friend have been captured on a desert island ruled by a crazed mathematical despot.

You will be locked in separate cells in the island’s prison, and then set the following task:

Every minute for an hour you will each flip a coin. The flips are simultaneous, and after each flip you will make a prediction as to whether the other person’s flip was heads or tails. You cannot hear the other person's prediction nor use any means to learn of the other's prediction.

So, you both make 60 flips and 60 predictions.

The despot rules that he will kill the two of you if on any one of the 60 predictions you are both correct. (In other words, you both flip, both predict the result of the other person’s flip, and are both right). To escape with your lives at least one of you must predict wrongly each time.

You are given ten minutes to think up a survival strategy before being taken to the cells. Once you are in the cells you cannot communicate with each other, although you are obviously able to see the results of your own flips.

Can you guarantee your survival, and if so, how?*

Answers in Spoilers thread please.


----------



## bellenuit (13 September 2016)

Two puzzles today.

*1) Volumes I, II and III of a dictionary are stacked vertically side by side on a shelf (no gaps between them) in left to right order and with their spines visible in the normal way. The thickness of all the pages together in each volume is 6cm, and the thickness of each of the back and front covers of each volume is 0.5cm.

What is the horizontal distance from the first page of Volume I to the final page of Volume III?

2) The following statement is correct

< F is the first and the seventh letter of this sentence.>

Using the sentence above as a model, fill in the gap in the following sentence to make it correct:

<C is the first and the [...] letter of this sentence.>*

Answers in Spoilers please...


----------



## bellenuit (26 September 2016)

I haven't tried these myself yet, so I don't know how difficult they are.

*Next in Series*

*What is the next number in each of the following 3 series?

1) 23, 9, 20, 14, 14, 9, 20, 6, ...

2) 2.1, 3.5, 3.3, 2.3, 1.3, 2.4, 2.5, 2.6, 1.8 ...

3) 1, 2, 9, 12, 70, 89, 97, 102 ...



4) Why might Henry I be an appropriate way to end the following series?

Mary I; George III, Henry III, James II, George IV, Charles I, ...*

In relation to question 4, I don't know whether one needs a knowledge of British royalty to solve, but I would assume no.

Answers in Spoiler thread please.


----------



## bellenuit (10 October 2016)

This looks like a good one.

*3 friends (A, B and C) are playing table tennis. They play the usual way: two play at a time, the winner stays on, and the loser waits his/her turn again. At the end of the day, they summarise the number of games that each of them played:

A played 10

B played 15

C played 17.

Who lost the second game and why do you conclude that?*

Answers in Spoiler thread please.


----------



## bellenuit (13 November 2016)

*How many Red Balls (only), including the one already there, are needed to make it balance? Worked out logic to be included, not just guesses.
*

Answers in Spoilers please.


----------



## bellenuit (30 July 2017)

Cat Puzzle

*A straight corridor has 7 closed doors along one side. Behind one of the doors sits a cat. Your mission is to find the cat by opening the correct door. Each day you can open only one door. If the cat is there, you win. You are officially smarter than a cat. If the cat is not there, the door closes, and you must wait until the next day before you can open a door again.

If the cat was always to sit behind the same door, you would be able to find it in at most seven days, by opening each door in turn. But this mischievous moggy is restless. Every night it moves one door either to the left or to the right. 

How many days do you now need to make sure you can catch the cat?

(First some clarifications. The 7 doors are in a line, so if the cat is behind the first or the last door, it has only one option for where it can move during the night. Otherwise, each night it decides randomly whether to move to the left or to the right.)
*
Answers in Spoilers thread please.


----------



## peter2 (4 September 2019)

peter2 said:


> *Puzzle*: For those technically inclined and like a challenge.
> Please read and enjoy the latest (Half yearly report 29/8/19) Directors Review of Operations describing the progress or lack of it in the development of their MEMS wafer. It is truly a literary masterpiece of creativity.
> https://www.asx.com.au/asxpdf/20190829/pdf/4480g6dpsm87yb.pdf




Posted this in the AKP thread but thought a few of the smarter technophiles might like to read it also. Please post your brief plain English translations in the AKP thread.


----------



## noirua (3 May 2020)

Educational Videos: Lots of puzzles, lots and lots and very many lots: https://www.pedagonet.com/Maths/tricks.htm


----------



## basilio (25 May 2020)

Watch how an ex investment bank analyst ( who really hated his lucrative but miserable job) responds to a sudoku puzzle with just 2 given numbers in the 81 square block.

All in real time and a mind blower .

https://www.theguardian.com/lifeand...become-internet-sensation-with-sudoku-channel
https://slate.com/human-interest/2020/05/the-miracle-sudoku-is-an-absolutely-riveting-video.html


----------

