Assume that there are two dies, one with five faces and one with six.
If one were to choose a die at random and then roll a 3, what is the likelihood that one rolled the five-faced die rather than the six-faced die?
At first glance, intuition suggests that the number of faces must affect the likelihood. For example, if one die had five faces, and the other had one million faces, rolling a 3 would make it seem unlikely that one had rolled the million-faced die and landed on one of only 5 faces shared with the other die.
However, it also seems that the problem is independent of the number of faces on either die and is wholly contingent on the action of choosing one of two die at random.
Is the likelihood 50%, or does the number of faces factor into the likelihood?
This is a question for conditional probability. Let $A$ be the event that the five sided die is rolled and let $B$ be the event that $3$ is rolled.Then by Baye's Theorem $$ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^c)P(A^c)}=\frac{(1/5)(1/2)}{(1/5)(1/2)+(1/6)(1/2)}=\frac{6}{11} $$ while $$ P(A^c\mid B)=\frac{5}{11} $$ so it is more likely for the dice to have been five sided given the info that the roll came up $3$.