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.