ASF Spoilers Thread

[galumay]

9.) the official answer is very poor! "Honey doesnt come from flowers". I prefer the fact that there is no waste stream, once the pollen has been converted to honey the rest of the flower should come out as waste. Not one of his best!

13.) rubber shrinks when you heat it.

14.) it happened "here" but no glass on road?

15.) brown car with concrete splashed on it, blue paint from car & letter box on it and panel damage!

 

[bellenuit]

Your whole explanation makes no sense to me! Not saying its wrong, just that I still dont get it!!

If i have picked 1 box out of 100 i have a 1% chance of being correct, if the judge opens 98 of the 99 left and proves they all have nothing in them, we now have 2 boxes, 1 with the pardon, 1 without. Its patently obvious that which ever box you choose has a 50% chance of containing the pardon, so switching cant change your odds.

Your reasoning makes not the slightest sense to my dumb head!!

OK, read the wiki and all i can say is i am happy to be wrong along with so many geniuses and mathematicians! Still dont believe it though.

Just because there are two choices doesn't mean each choice is of equal probability.

Think of it this way. If you believe that staying with your first choice gives you a 50/50 chance of picking the right box, then if you were to play that game 1000 times, you are saying that you would expect to pick the correct box out of the 100 five hundred times on average. You know each time you play that at least 98 of the other 99 boxes are empty. Whether the judge opens 98 of those boxes which HE knows before hand are empty and shows them to you or he just leaves them all closed, why would your chances of winning be different?

I think you are confusing this problem with randomly opening the other boxes and each time one is found without a pardon then the probability of all the rest (including yours) having the pardon increases. By the time you have opened all but one of the remaining boxes and still haven't found one with a pardon, then obviously yours and that box is down to a 50/50 chance. Every time a box was opened at random, it had a 1/100 chance of having the pardon, and if it had, then that was it. What was left would then all have had zero chance. If it didn't have the pardon, all the remaining boxes had an increased chance.

But this problem is different. The judge knows which boxes do not have a pardon and opens only those. Unlike the other scenario there was never a possibility that opening one of those boxes could have revealed the pardon, so there was never a possibility of the remaining boxes suddenly having a zero chance.




Each of those boxes have a 99 out of 100 chance of

 

[keithj]

Here's a simple explanation:
Imagine now instead that instead of revealing a box,
The judge lets you
a) keep your initial choice
b) open BOTH remaining boxes.

b) is now a no brainer (switching) 66% vs 33%

b) is also the same as giving you both boxes (with one already open)

Here's 3 other ways to think about it....

1) You could try the experiment 100 times & see how many times you're successful by switching.

2) What if you used a pack of cards - you have a 1/52 chance of picking Ace of Spades (& a 51/52 chance of it being elsewhere) . Get your butler to sneak a look at your card - if it's the Ace then he should remove 50 of the remaining cards, otherwise just leave the Ace there by itself for you to optionally choose.

It should be clear that switching will have a 51/52 chance of being correct.

Try that 100 times too!

3) Think of it as 2 separate events. The 1st event has a 1/52 chance of success. The 2nd event (ie the optional switch) has a 100% chance of success (but only if the 1st event proved unsuccessful). So the odds of success by switching are 51/52 x 100%. While the odds of success by NOT switching will remain at 1/52.

 

[galumay]

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

If they guessed 3 correctly, then they know the fourth as well, so 10 picked all 4 boxes. (and all 3 boxes by definition.)

 

[bellenuit]

If they guessed 3 correctly, then they know the fourth as well, so 10 picked all 4 boxes. (and all 3 boxes by definition.)

Yes, or to put it another way, it is impossible to just guess exactly 3 correctly, so 0 guessed and the remaining 10 guessed all 4. Too easy.

 

[pixel]

Yes, or to put it another way, it is impossible to just guess exactly 3 correctly, so 0 guessed and the remaining 10 guessed all 4. Too easy.

sorry Belle,
I hadn't seen this as the solution thread. Shall put my guesses here in future.

 

[pixel]

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.

Answer: I don't think it's possible.
Start in the middle of the lake and swim away from where the Cyclops stands. That will give you the shortest distance to shore, which is the farthest away from Cyclops. He will have to run a half circle, the length of which is 3.1415... times the radius that you need to swim. But he runs 4-times as fast as you swim.

 

[skc]

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.[/B]

So say I am at near 12 o'clock position of the round lake, but quite close to the shore... if I turn around and head towards the 6 o'clock shore, does the cyclop run towards that point?

If so, can I not simply swim slower? Say I swim at a speed such that I am only 1/4 of the way across the lake when the cyclop has already made it to the opposite shore. At which time I simply swim at maximum speed back to the 12 o'clock edge. I only need to swim a distance of 0.5R when the cyclop needs to cover 3.14159R, or >6x as far.

 

[pixel]

So say I am at near 12 o'clock position of the round lake, but quite close to the shore... if I turn around and head towards the 6 o'clock shore, does the cyclop run towards that point?

If so, can I not simply swim slower? Say I swim at a speed such that I am only 1/4 of the way across the lake when the cyclop has already made it to the opposite shore. At which time I simply swim at maximum speed back to the 12 o'clock edge. I only need to swim a distance of 0.5R when the cyclop needs to cover 3.14159R, or >6x as far.

yes, that should work. Provided the cyclops is stupid enough to run at maximum speed to that far 6 o'clock point before I even made it half-way to the middle of the lake.

A smart cyclops would only need to stay on the side of the lake and move to the point that you're closest to, never overtake you.

 

[galumay]

Answer: I don't think it's possible.

Agreed, as long as he is a smart cyclops!

 

[bellenuit]

So say I am at near 12 o'clock position of the round lake, but quite close to the shore... if I turn around and head towards the 6 o'clock shore, does the cyclop run towards that point?

If so, can I not simply swim slower? Say I swim at a speed such that I am only 1/4 of the way across the lake when the cyclop has already made it to the opposite shore. At which time I simply swim at maximum speed back to the 12 o'clock edge. I only need to swim a distance of 0.5R when the cyclop needs to cover 3.14159R, or >6x as far.

Sorry, that was bad phrasing on my behalf. He will only try to get to the point you are heading to if you are swimming away from the centre. It's hard to think of all contingencies. Just assume he is smart and will always stay as close or try to get as close on the shoreline as he possibly can to where you currently are. So if you are heading towards the centre, he will stay or head to the point if not already there that is closest to where you currently are in case you backtrack. If heading away from the centre he will try and get to the point that you seem to be heading to. I'll add something to the original to make this clearer.

 

[bellenuit]

I'll post the answer about midday EST Friday.

 

[galumay]

mmm...i have been doing some fiddling with my basic maths and a ruler, compass and pencil!

Seems to me there is a circle that I can swim in where I will be able to move faster than the cyclops relative to the circle he has to run around. That means i can get opposite him, while closer to the bank than the center.

The bit of maths i cant quite get my head around is whether that circle will mean i am close enough to the bank to make it before the cyclops - i suspect it is, but i cant be certain.

I need to make a sprint to the edge of less than 3.89m to beat him to the bank if he is directly opposite. (based on a lake of 10m diameter).

 

[cynic]

Presuming that I start in the centre of the lake. I'd first swim PI/8 lake radius in the opposite direction to the Cyclops. In that time the Cyclops will have covered 1/4 lake circumference in his efforts to draw closer. At that point I'd change direction and swim PI/8 radius towards the shore opposite Cyclops new position. Again he'd cover 1/4 circumference in that time. I'd continue this exercise until such time as the distance between myself and the shore opposite the Cyclops position was slightly less than PI/8 lake radius, at which point I'd safely swim to shore knowing that the Cyclops won't be able to reach me in time to catch me.

 

[Tisme]

Problem #168/#169

Cyclops has the specific advantage of 4/ ¶. So the swimmer has to swim >4/ ¶ away from the Cyclops. I'm guessing that because the swimmer doesn't have his trig tables nor a handy calculus ready reckoner he's going to go three dimensional and duckdive into the water and swim far enough (away from Cyclops) concealed to reduce the distance to the shoreline to < ¶r/4....?

 

[keithj]

1. Swim to centre of lake
2. swim in a straight line directly away from cyclops
3. when cyclops moves towards closest point on shore swim slightly more than 180 degrees away from him. This change in direction should be sufficient to force cyclops to change direction.
4. Repeat step 3 until you have swum 1/4 distance to shore. This zigzagging forces cyclops to oscillate about the point furthest away from you. At the 1/4 radius distance the angular speed you can swim is the same as the speed cyclops can run, so any more zigzagging will prove pointless
5. Swim directly to shore safe in the knowledge that you can swim 3/4 of the radius faster than he can run 1/2 the circumference.

As an alternative to steps 3 & 4. Simply swim out to a distance slightly less than 1/4 of the radius and then swim in a circle until you are directly opposite cyclops, and then do step 5.

These methods work because your angular speed is greater than cyclops angular speed if you remain inside the 1/4 radius and consequently you can get opposite him.

 

[bellenuit]

Both Cynic and Keith may be right, but I would like to see the math fully worked out so that we can be 100% certain. I think in both cases they will have swum into the area between the two inner circles in the answer below, which then allows them to get out safely.

The answer is he can get out safely. This is one proof, but as I said, zig zag methods may also prove successful.



Looking at the three circles in the diagram, the outer circle represents the lake. It has radius R metres. Lets assume you can swim at speed V metres per second.

We know that if you were at the centre of the lake and swam directly to the side, the cyclops could always get there before you, because it would take you R/V seconds to get to the shore and the maximum time he needs to run around half the circumference of the lake is πR/4V seconds. Since π is 3.14 approx, then π/4 is less than 1, hence the cyclops will always get there first.

In the time the cyclops takes to run a half circumference, πR/4V seconds, you can swim πR/4 metres. This is 0.79R and is represented by the distance D in the diagram (not to scale). So it is clear if you can manage to to be anywhere less than D metres from the shore when the cyclops is at the exact opposite side of the lake (180 degrees removed), you can get to the shore before he can get to your point of exit. The innermost circle of radius A represents all points where you would NOT make it to the shore before the cyclops if he were to start 180 degrees removed from you. A is obviously R-D, or 0.21R. So if you are anywhere outside that innermost circle and the cyclops is 180 degrees removed from you, you can get to the shore before he gets to you. So how do you manage to do this?

The time taken for the cyclops to fully complete a circle is 2πR/4V or πR/2V seconds. The circle with radius B represent the circle that you can fully swim around in the same time it takes the cyclops to run around the lake. The time you take to swim around that circle is 2πB/V seconds and if this is the same time as it takes the cyclops to run around the lake, we get:

2πB/V = πR/2V or B = R/4 or 0.25R

So if you are on that circle with radius B, you can swim around it in the same time it takes the cyclops to run around the lake. Obviously if you are on a smaller circle (one with radius less than B) you can swim around that smaller circle faster than the cyclops can run around the lake.

Although I have drawn it that way, it wasn’t known until we worked the above out that the circle with radius A (where A = 0.21R) is inside the circle with radius B (where B is 0.25R). This is fortunate as it means that if we swim to anywhere in the area between the two inner circles, because we are within the circle with radius B we can swim in a circular fashion around the lake faster than the cyclops can run around it, so we will be able to get to a point where he is 180 degrees removed from where we are, but because we are outside the circle with radius A, we can then get to the shore before he can get around to the point we exit.

So to escape, you just need swim to just outside the A circle and then swim on an imaginary circle around the centre until we are 180 degrees removed from where the cyclops is. At that point we then swim directly to the shore and will be able to get out before he makes back around.

 

[skyQuake]

Just swim away from him in the opposite direction he is facing.

Being a cyclops he lacks depth perception and thus you'll get to at least 0.5 radius before he notices. By then its too late

 

[skc]

Last Slylock I got.

a). The kids knew the cookies were Oatmeal?

b). Tilt the drum and pour it until the fluid surface makes a perfect diagonal across the height of the drum

c). The message would take 5 million years to reach the aliens. Although, it's questionable what speed "intergalactic radio message" travels at.

d). A 3 prong plug on the computer vs 2-hole power point from the wall.

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

I have to say that the questions are not logically watertight enough for an adult... but I'd love it as a kid. I have a friend's 8yr old boy birthday coming up...

 

[keithj]

This is one proof, but as I said, zig zag methods may also prove successful.

I think the proof is simply that the angular velocity of the swimmer is always greater than the cyclops provided he stays within 1/4 of the radius, and he can afford to swim slightly further out with each zig. The swimmer would have to consult trig tables to calculate exactly what angle is optimum at each zig & zag, but in practice the swimmer would just swim more parallel with the shore until the cyclops reversed.

The zigzags would get shallower as the distance from the centre increased. On the whole I think the zigzag method would be more efficient than the swimming in a circle method (and it would have the benefit of p***ing the cyclops off ).

And many thanks to you & others for providing these puzzles .