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PaperclipPerfector  7mo ago (text post)   2443 thread views

Friday Fun Thread for September 6, 2024

Be advised: this thread is not for serious in-depth discussion of weighty topics (we have a link for that), this thread is not for anything Culture War related. This thread is for Fun. You got jokes? Share 'em. You got silly questions? Ask 'em.

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Before you are three closed, identical boxes. The first box contains two silver coins and nothing else. The second box contains two gold coins and nothing else. The third box contains two coins (one gold and one silver) and nothing else. You have no idea which box is which, you cannot see inside the boxes, and they are mounted to the wall so you cannot lift them to estimate their weight.

You reach your hand into one of the boxes and withdraw a gold coin. If you reach into the same box to withdraw the second coin, what is the probability that that coin is also gold?

It's 66%. You know that the box you withdrew the first coin from can't be the box with two silver coins in it, so it must be one of the other two boxes. There are exactly three coins remaining in these two boxes combined, one of which is silver and the others gold. Ergo the odds of you withdrawing a second gold coin are 2/3.

This puzzle was shared on a Facebook meme page I follow. I'm not trying to flex or anything, but I solved it instantly and the solution seems incredibly obvious to me. I was very surprised to see the comments full of people asserting that the answer is 50%. There's even a Wikipedia article about it, and it's referred to as a "paradox". One of my pet peeves is when the word "paradox" is used to refer to mathematical problems with counterintuitive solutions, or counterintuitive findings from the sciences - as opposed to contradictions in logic. Russell's paradox is a legitimate paradox: there is no good answer to the question "if a barber only shaves men who do not shave themselves, does he shave himself?" The "twin paradox" in general relativity isn't actually a paradox, but I can see how it runs counter to our human intuition of how things work in Mediocristan. But in this case, I don't think this particular puzzle even rises to the level of "mathematical problem with a counterintuitive solution": the solution seems incredibly obvious and straightforward. The word "paradox" gets used far too freely.

So what would you say is your probability of withdrawing a gold coin if everything else is the same, but the third box has one gold coin and 10 silver coins, instead of just one gold and one silver coin?

Of withdrawing another gold coin? 2/3

Why 2/3?

OP's intuition says that once you pick one gold coin, you know that you have one of two boxes, and that there are exactly 12 coins in those two boxes combined, two of which are gold, so that would put the probability of getting another gold coin at 2/12.

Wait, I am wrong.

The probability of picking the two-gold box is 1/3. The probability of picking the mixed box and finding gold on the first draw is 1/3*1/11=1/33.

11/33 vs 1/33, I am 11 times more likely to find another gold coin, not two times.

Because you know that you picked gold initially. The odds of the second coin being gold is the odds that you didn't pick 1/3 boxes with with both gold and silver coins, meaning 2/3. The only way the second coin isn't gold is that the initial choice was the box with both silver and gold coins in it, the number of silver coins in that box do not matter because of the precondition of having picked a gold coin.

Of course they matter, they increase the chance that the gold picked in round one was from the double gold box dramatically, which itself hugely increases the odds of round 2 is also gold.

They dont matter because the question is conditioned on that we already picked a box with a gold coin.

The question is what the odds are that we picked the box with both gold and silver, given that we have a box with at least a gold coin in it. There is 1/3 with gold and silver, hence the probabilty of the second coin being gold is 2/3. You could increase the amount of silver coins by infinity and it wouldn't matter. You're picking boxes, not coins.

Yes. But the fact that w already picked a box with the gold coin tells us that it was almost certainly the double gold box and therefore that the probability of the gold second coin is even higher.

No, it tells us nothing. The question is conditioned on a gold coin having been picked.

We didn't pick a box at random, the gameshow host did and revealed a gold coin.

More comments

I think this checks out but someone else will need to check for me.

It's probably good that you're not trying to flex too much about how smart you are due to finding the solution to this problem incredibly obvious, because it seems that you got the answer correct the same way that a broken clock gets the time correct twice every day. ;)

Okay, rude.

I disagree. Maybe this is the reason I "always forget" the simple route; because I'm not sure it's actually right. I did this two different ways, my renormalization route (thinking of things as a tree with info sets) and just brute reproducing the wiki entry on using Bayes to solve it.

Method 1: Renormalization

There's a 1/3 chance of picking each box, one which has a 100% chance of giving you a gold on the first draw and the other has a 1/11 chance (ignoring the option with zero chance of getting a first gold), so the chances of me being in each relevant box at the current state are 1/3 and 1/33. To renormalize, I need to multiply by the reciprocal of their sum, 1/3 + 1/33 = 12/33.

So my chance of being in the GG box is 11/12 and my chance of being in the G10S box is 1/12.

Method 2: Straight Bayes, yo

Just shutting up and calculating, reproducing the wiki article directly.

P(GG|see gold) = P(see gold|GG)*(1/3) / [P(see gold|GG)*(1/3) + 0 + P(see gold|G10S)*(1/3)]

P(GG|see gold) = (1/3) / (1/3 + 0 + 1/33)

= (1/3) / (12/33)

= 11/12

I'd say law of conditional probability is the simplest route here. P(2nd Coin Gold | 1st Coin Gold) = P(Both Gold) / P(1st Coin Gold) = 1/3 / (1/3 * 1 + 1/3 * 1/11) = 1/3 / (1/3 + 1/33) = 1/3 / (12/33) = 1/3 * 33/12 = 11/12.

This is obviously correct, I have no idea what the other people are saying.

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