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PaperclipPerfector  1yr ago (text post)   2762 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.

Two intuitive ways to formulate this come to mind.

First, law of conditional probability. P(A | B) * P(B) = P(A and B), or alternatively, P(A | B) = P(A and B) / P(B). P(2nd Coin Gold | 1st Coin Gold) = P(Both Coins Gold) / P(1st Coin Gold). P(1st Coin Gold) is 50% or 1/2, the probability of selecting the box with both gold coins (1/3) plus the probability of selecting the mixed coin box times one half (1/3 * 1/2 = 1/6). Probability both coins are gold is 1/3, as it can only occur if you pick the one box of three with two gold coins. 1/3 / (1/2) = 2/3.

Second, just thinking of all the possible outcomes before any coins are picked. Let's arbitrarily designate one coin in each box X, and the other Y. So boxes one to three contain coins Gold X and Gold Y, Gold X and Silver Y, and Silver X and Silver Y, respectively. There are only six possible outcomes for first and second pick, comma separated:

  1. Gold X, Gold Y
  2. Gold X, Silver Y
  3. Silver X, Silver Y
  4. Gold Y, Gold X
  5. Silver Y, Gold X
  6. Silver Y, Silver X

If you picked gold first, you know you can't be in outcomes 3, 5, or 6. Out of outcomes 1, 2, and 4, two of those have gold as the second pick. Thus, 2/3 once again.

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