1
$\begingroup$

a box has six fair dice and a single unfair die with all sixes. One die is chosen at random and rolled twice. On both rolls, six appears.

What is the probability that the selected die is the unfair one?

$\endgroup$
3
$\begingroup$

Bayes's theorem says that $p(A|B) = p(B|A)p(A)p(B)^{-1}$, where $A$ is the event of picking the unfair die and $B$ is the event of rolling two 6s. Then $p(A|B) = 1 \cdot \frac{1}{7} \cdot (1 \cdot \frac{1}{7} + \frac{1}{6^2} \cdot \frac{6}{7})^{-1} = \frac{6}{7}$.

$\endgroup$
  • $\begingroup$ how are you getting 1/7 vs 6/7 for the prior chances of picking the two dice? If you're picking dice at random, I'd tend to think that'd be 1/2 for each $\endgroup$ – Glen_b Feb 6 '18 at 1:49
  • 1
    $\begingroup$ @Glen_b - the box has six fair dice and only one unfair dice, I missed that on my first time through as well. $\endgroup$ – jbowman Feb 6 '18 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.