Solution
A Repeated Face
Show the problem again
A fair six-sided die is rolled repeatedly until some face appears for the second time (any face that has already appeared). What is the expected number of rolls, including the final one? Give a reduced fraction (a decimal rounded to two or three places is also accepted).
Worked solution
The answer is 1223/324, about 3.775. Use E[N] = Σ_{k≥0} P(N > k). The event N > k means the first k rolls are all different, which has probability 6·5·...·(6 − k + 1)/6^k. So E[N] = 1 + 1 + 5/6 + 20/36 + 120/216 + 360/1296 + 720/7776, and the terms stop at k = 6 because seven rolls must repeat a face. Over the common denominator 7776 the sum is 29352/7776 = 1223/324. This is the birthday problem with a six-day year: the first repeat comes, on average, before the fourth roll is finished, well short of the seven rolls the pigeonhole principle allows.
Source: Birthday-problem variant in wide circulation; no traceable original source. Statement written for AxiomIQ.