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PaperclipPerfector  11mo ago (text post)   2594 thread views

Friday Fun Thread for September 6, 2024

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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.

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

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

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

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