Two Distinct Faces

Question

On average, how many rolls of a fair die does it take until you have seen two distinct faces?

Accepted Answer

The first roll is free: whatever it shows becomes your first distinct face. From then on, each roll differs from it with probability frac56, so the additional wait is geometric with mean frac65: mathbbE = 1 + frac65 = frac115 = 2.2. These are exactly the first two stages of the coupon-collector decomposition (frac66 + frac65); the full six-face version continues +frac64+frac63+frac62+frac61 to reach 14.7. The trap is answering 2, implicitly assuming the second roll can't repeat the first. It can, with probability frac16, and the geometric wait prices in arbitrarily long repeat streaks: there is still a frac136 chance you're unfinished after three rolls. Why interviewers ask it: it's the smallest instance of "split a waiting time into stages with different rates, then add expectations." If you articulate "stage k succeeds at rate p_k, expectations add by linearity," you've demonstrated the tool that prices the coupon collector, the birthday-repeat problem, and most waiting-time games. It also makes a clean quick market: the distribution is 1 + mathrmGeom(5/6) with mode 2 and a thin right tail, so you'd quote tight around the low end, fair value 2.2.

Two Distinct Faces

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