ASF Spoilers Thread

bellenuit · 31 May 2015

Cynic, correct on all which is self evident.

They are referred to as self descriptive numbers (I didn't want to give them that name in the problem as it would be too easy to search the answers on Google).

More on them here.

http://en.wikipedia.org/wiki/Self-descriptive_number

cynic · 31 May 2015

bellenuit said:
Cynic, correct on all which is self evident.

They are referred to as self descriptive numbers (I didn't want to give them that name in the problem as it would be too easy to search the answers on Google).

More on them here.

http://en.wikipedia.org/wiki/Self-descriptive_number

Thanks for that bellenuit.

Perhaps it's just that I'm particularly fond of numbers, but I found that problem to be by far the easiest you've presented thus far.

galumay · 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?

My initial logic is that its the same as if you had never added a marble, ie you put in a red marble and took out a red marble, therefore the remaining marble could either be red or blue - 50% chance.

skc · 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?

Nice question. The answer is 1/3. As there are only 2 marbles, it's easy to work out by just drawing a "possibility tree".

Gets a bit more complicated if there were a few more in the bag.

cynic · 12 June 2015

skc said:
Nice question. The answer is 1/3. As there are only 2 marbles, it's easy to work out by just drawing a "possibility tree".

Gets a bit more complicated if there were a few more in the bag.

Funnily enough, I'm arriving at 25% and I thought I was applying the same methodology.

Edit: I think I've spotted my mistake and now agree with your answer.

skc · 12 June 2015

cynic said:
Funnily enough, I'm arriving at 25% and I thought I was applying the same methodology.

You need to remove the Blue/Blue scenario... because it is not a valid one as it was given that we removed a red ball.

So there are only 3 possible "branches", of which only one ends up having a red in the bag.

galumay · 12 June 2015

skc said:
Nice question. The answer is 1/3. As there are only 2 marbles, it's easy to work out by just drawing a "possibility tree".

Gets a bit more complicated if there were a few more in the bag.

Once again it seems the answer is totally counterintuitive! I have no doubt you are right - but for the life of me cant see how!

As I said, it seems to me its as if nothing has changed, there was a 50% chance of the ball being red or blue, you put one in and take out one the same colour, the remaining ball still has a 50% chance of being either colour.

I think this is the same as the similar problem that I couldnt get my head around - the one that many mathematicians and scholars agreed with my incorrect answer!

cynic · 12 June 2015

galumay said:
...
As I said, it seems to me its as if nothing has changed, there was a 50% chance of the ball being red or blue, you put one in and take out one the same colour, the remaining ball still has a 50% chance of being either colour.
...

Your thinking would only be correct if you could be certain that you took out the same red that you put in.

cynic · 12 June 2015

skc said:
You need to remove the Blue/Blue scenario... because it is not a valid one as it was given that we removed a red ball.
...

Yes! That was where I slipped up!

skc · 12 June 2015

galumay said:
Once again it seems the answer is totally counterintuitive! I have no doubt you are right - but for the life of me cant see how!

As I said, it seems to me its as if nothing has changed, there was a 50% chance of the ball being red or blue, you put one in and take out one the same colour, the remaining ball still has a 50% chance of being either colour.

I think this is the same as the similar problem that I couldnt get my head around - the one that many mathematicians and scholars agreed with my incorrect answer!

Hang on a second... I am wrong! I forgot the bold part.

P.S. My daughter is sick so I only got 4 hour or so sleep last night... good enough excuse?

Then it should be 50/50?!

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?

cynic · 12 June 2015

skc said:
Hang on a second... I am wrong! I forgot the bold part.

P.S. My daughter is sick so I only got 4 hour or so sleep last night... good enough excuse?

Then it should be 50/50?!

Now I definitely do not agree with you, unless by "Red" as opposed to "red" it means the same marble was removed.

bellenuit · 12 June 2015

Actually the answer is 2/3.

Let's name the initial marbles R and B and the added marble Ra

There is a 50% chance at the beginning that the bag contains R and a 50% chance that it contains B.

After Ra is added, there is now a 50% chance that the bag contains R + Ra and a 50% chance that it contains B + Ra.

A marble is drawn from the bag (at this stage assume any colour). So there are now 4 possibilities for what's inside the bag and outside the bag.

A. If the bag contained R + Ra (overall a 50% chance), then you could have:
A1. R inside and Ra outside, or
A2. Ra inside and R outside.

There is a 50% chance for either of these outcomes (if A is true) and since there is just a 50% chance for A being true, then there is a 25% chance for A1 and a 25% chance for A2 overall.

B. If the bag contained B + Ra (overall a 50% chance), then you could have:
B1. B inside and Ra outside or
B2. Ra inside and B outside.

As before there is a 50% chance for either of these outcomes (if B is true) and since there is just a 50% chance for B being true, then there is a 25% chance for B1 and a 25% chance for B2 overall.

So there are in total four possible outcomes, each with an equal 25% probability (giving us 100%).

But since we are only considering the situation where a Red is drawn first, then we are left with just 3 possible outcomes, each with equal probability: A1, A2 and B1. Two of these have a Red left in the bag (A1 and A2) and the other a Blue left in the bag (B1), so the probability of a Red marble being left in the bag is 2/3 or 66%.

cynic · 12 June 2015

bellenuit said:
Actually the answer is 2/3.

This actually matches the intended outcome of SKC's original calculation (and my revised one). It's been a pretty wild week.

bellenuit · 12 June 2015

cynic said:
This actually matches the intended outcome of SKC's original calculation (and my revised one). It's been a pretty wild week.

I thought your answers might have been just an inadvertent mix up of the colours, but as there was no detailed workout given, I couldn't tell.

That puzzle is attributable to Lewis Carroll.

keithj · 13 June 2015

bellenuit said:
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.

View attachment 63013

The crucial hint given implies the radius of the cylinder & the radius of the sphere are not required, so ANY parameters should give the required answer.

So by making the radius of the drilled hole infinitely small, the height of the cylinder becomes equal to the diameter of the sphere. And therefore the volume of the full sphere less the volume of an infinitely small cylinder is the answer.... (4/3 x pi x 1 x 1 x 1) - (2 x pi x 0 x 0)

However, upon checking with other examples it becomes clear that it's a bit more complex. And google quickly led to the napkin ring problem.

bellenuit · 13 June 2015

keithj said:
The crucial hint given implies the radius of the cylinder & the radius of the sphere are not required, so ANY parameters should give the required answer.

So by making the radius of the drilled hole infinitely small, the height of the cylinder becomes equal to the diameter of the sphere. And therefore the volume of the full sphere less the volume of an infinitely small cylinder is the answer.... (4/3 x pi x 1 x 1 x 1) - (2 x pi x 0 x 0)

Correct, keithj. The remaining volume is purely dependent on the height of the cylinder, not on the radius of the sphere. As the radius of the sphere increases, the radius of the cylinder must also increase so that the height remains 2 cms. Therefore to solve the problem, you take, as you have done, the extreme case where the cylinder shrinks to zero radius, in which case the volume of the remainder is the same as the volume of the sphere itself. This sphere has a radius then of 1cm.

So the answer is: 4π/3 cubic cms (as r**3 = 1**3 = 1)

Thanks for the reference to the Wikipedia Napkin Ring problem as I would have been in strife if anyone asked for proof that the remaining volume is constant (for any given height of the cylinder) irrespective of the sphere radius.

This is the link for those who wish to pursue it and you will find a good graphic there too.

https://en.wikipedia.org/wiki/Napkin_ring_problem

keithj said:
However, upon checking with other examples it becomes clear that it's a bit more complex.

I am confused by what you mean here. Are you saying that you have found examples where this isn't the case?

cynic · 14 June 2015

I arrived at a different answer.

Basically I calculated the result as being equal to the volume of the sphere less the volume of the cylinder and also less a volume attributable to the two removed caps at the poles.

If my trigonometry and condensation skills have served me correctly, the answer is something akin to:

pi X ((4 X R X R X R X ((1/3)-((1/90) X inverse cosine (1/R)))- (4 X ((R X R) -1)))

Edit: I'm still pondering bellenuit's explanation and am as yet undecided either way.

cynic · 14 June 2015

I calculated the radius of the cylinder (using pythagorus theorem) as being equal to sqrt[(R X R) - 1].

The increase in expansion of the cylinder radius (essential to maintenance of cylinder height in the event of expansion of the sphere radius) would certainly increase the volumes of the removed caps and cylinder.

I am still undecided as to whether or not these changes exactly compensate.

cynic · 14 June 2015

Okay.

I've looked at the Wikipedia article on the Napkin ring problem and the explanation in the proof about the volume being directly related to cylinder height regardless of sphere radius do indeed make sense.

So I'm okay with it now, so please feel free to disregard my rather convoluted answer.

keithj · 14 June 2015

bellenuit said:
I am confused by what you mean here. Are you saying that you have found examples where this isn't the case?

No. My initial assumption was that the formula was simply 4/3 x pi x (h/2)^3. i.e the same as a sphere for your example. Trying an example with a non-zero hole radius soon killed that theory and hence my comment that it was a bit more complex.

The wiki page gives the correct formula of pi x h^3 / 6.

I too was heading in the same direction as cynic (Vol of sphere - vol of cylinder - 2 x vol of end cap) when the wiki page pointed me towards napkins.

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