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