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PaperclipPerfector  12d ago (text post)   2137 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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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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