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A casino game has two dice, each with faces numbered $1$ to $6$. One of them is fair but the other is biased such that a $6$ is twice as likely to appear on top as any one of the other faces. One of the dice is rolled $10$ times, and a $6$ appears on top exactly four times. What is the probability that the die was the biased one?

I get the probability of the biased die as $\frac 2 7$ but from there I am not sure what to do.

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Let $A$ be the event: "is rolled $10$ times, and a $6$ appears on top four times".

Then $$ P(A|biased) = \binom {10}4 (2/7)^4 (5/7)^6\\ P(A|unbiased) = \binom {10}4 (1/6)^4 (5/6)^6\\ $$

Then write $$P(biased|A) = \frac{P(A, biased)}{P(A)} = \frac{P(A|biased) P(biased)}{P(A|biased) P(biased) + P(A|unbiased) P(unbiased)} $$

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  • $\begingroup$ Can you please explain why this works and your thinking/plan? $\endgroup$ – louie mcconnell Oct 26 '14 at 1:20
  • $\begingroup$ this is the bayes property. $\endgroup$ – mookid Oct 26 '14 at 8:47
  • $\begingroup$ do you have a link? $\endgroup$ – louie mcconnell Oct 26 '14 at 21:04
  • $\begingroup$ of course: en.wikipedia.org/wiki/Bayes%27_theorem $\endgroup$ – mookid Oct 26 '14 at 21:06
  • $\begingroup$ I got u +1, but i'm not the OP. $\endgroup$ – louie mcconnell Oct 26 '14 at 21:23

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