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