A fair die is rolled repeatedly until the first 6 appears. You are then told that every roll in the sequence came up even. What is the expected number of rolls?

A fair die is rolled repeatedly until the first 6 appears. You are then told that every roll in the sequence came up even. What is the expected number of rolls?

Approach: Compute the joint probability that the run has length n and every roll is even, then divide the two sums rather than reasoning about a die restricted to even faces.

3/2. Let N be the number of rolls. P(N = n, all even) = (2/6)^{n-1} * (1/6), since the first n-1 rolls must be 2 or 4 and the last must be the 6. Summing over n gives P(all even) = (1/6)/(1 - 1/3) = 1/4, and E[N * 1{all even}] = (1/6) * sum_{n>=1} n (1/3)^{n-1} = (1/6) * (1 - 1/3)^{-2} = 3/8. So the conditional expected value is (3/8)/(1/4) = 3/2. The common wrong answer is 3, from treating the rolls as uniform on {2, 4, 6}, which is a different experiment. Conditioning the original process on the whole path being even favours short paths, which is why 3/2 sits below the unconditional geometric mean of 6.

Follow-up: Given that every roll came up even, what is the probability that the run lasted exactly two rolls?

Key concepts: conditional expectation, geometric distribution, expected value.