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Say we roll two dice. The first one gives $ 1 $ and the other gives $ 2 $. What is the probability that the sum is $ 3 $, given these results?

Intuitively, I would say it’s $ \dfrac{1}{1} $. However, if we apply Bayes’ Theorem, we get $$ \frac{1}{12} \times \frac{\left( \frac{1}{36} \right)}{\left( \frac{1}{36} \right)}, $$ which gives $ \dfrac{1}{12} $. However, this doesn’t make any sense.

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Bayes' theorem states P(A|B) = P(B|A)*P(A)/P(B). Say A: sum is 3, B: first die gives 1 and second die gives 2. P(B|A)=1/2, P(A)=2/36 and P(B)=1/36. Applying Bayes' theorem gives P(A|B)=1. –  BNJMNDDNN Jul 12 '13 at 20:50
P(A|B) = P(A)P(B)/P(B) = 1/12*1/36/1/36 = 1/12. How did you get 1? –  George Jul 12 '13 at 20:55
P(B|A)P(A) isn't the same as P(A)P(B). P(B|A) is the chance that the first die gives 1 and the second gives 2, given the fact that the sum equals 3. This is a chance of 1/2. P(A) is the chance that the sum equals 3, which is 2/36. P(B) is the chance that the first die gives 1, and the second 2, this is 1/36. P(A|B) is the one you asked, which is indeed 1 as you'd expect intuitively –  BNJMNDDNN Jul 12 '13 at 20:58
So P(A)P(B) =/= P(A and B) = P(B|A)P(A)? –  George Jul 12 '13 at 21:02

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