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Say I have 8 fair dice and one trick die that always rolls "1."

I pick one die out of the nine and roll it three times, getting three 1's in a row.

What is the probability the dice I picked was fair?

My thought process:

I want to use conditional probability in the following way:

$P(F | 3x1)$ = $P(F \cap3x1)/P(3x1) $.

I know that if I roll a fair die 3 times, there are $6^3$ possibilities, but if I roll the trick die three times, there is only one possibility. How do I put these ideas together?

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    $\begingroup$ Is this true? For example, (fair, 1, 1, 2) lives in F, but does not live in $F \cap 3x1$ $\endgroup$ – user581882 Sep 23 '19 at 21:52
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The probability you pick a fair die is $7/8$, and the probability (given that) you roll three $1$s is $\left( \frac{1}{6}\right)^3$.

The probability you pick the trick die is $1/8$ and the probability (given that) you roll three $1$s is 1.0.

Multiply out and take the ratio:

$$\frac{\frac{1}{8}}{\frac{7}{8} \left( \frac{1}{6} \right)^3 + \frac{1}{8}} = \frac{216}{223} = 0.96861.$$

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