# Probability of picking a fair dice given certain rolls

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?

• Is this true? For example, (fair, 1, 1, 2) lives in F, but does not live in $F \cap 3x1$ – theta Sep 23 at 21:52

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