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