a box has six fair dice and a single unfair die with all sixes. One die is chosen at random and rolled twice. On both rolls, six appears.
What is the probability that the selected die is the unfair one?
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Bayes's theorem says that $p(A|B) = p(B|A)p(A)p(B)^{-1}$, where $A$ is the event of picking the unfair die and $B$ is the event of rolling two 6s. Then $p(A|B) = 1 \cdot \frac{1}{7} \cdot (1 \cdot \frac{1}{7} + \frac{1}{6^2} \cdot \frac{6}{7})^{-1} = \frac{6}{7}$.
answered Feb 6, 2018 at 1:18
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$\begingroup$ how are you getting 1/7 vs 6/7 for the prior chances of picking the two dice? If you're picking dice at random, I'd tend to think that'd be 1/2 for each $\endgroup$– Glen_bCommented Feb 6, 2018 at 1:49
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1$\begingroup$ @Glen_b - the box has six fair dice and only one unfair dice, I missed that on my first time through as well. $\endgroup$– jbowmanCommented Feb 6, 2018 at 2:48