Two dice set the count, then draw that many tickets

Question

You roll two fair six-sided dice and let $N$ be their sum. You then draw $N$ tickets, each ticket showing a value equally likely to be $1, 2, 3, 4, 5,$ or $6$ (drawn independently). You add the ticket values.

What total should you expect?

Accepted Answer

The count N comes from two dice, and each ticket is random. Split the layers with the tower rule: E[	ext{total}] = E\big[\,E[	ext{total} \mid N]\,\big]. Step 1, freeze the count. If N = n, you draw n independent tickets, each averaging \frac{1+2+3+4+5+6}{6} = 3.5. So E[	ext{total} \mid N = n] = 3.5\,n. Step 2, average over the count. The sum of two fair dice averages E[N] = 3.5 + 3.5 = 7. Therefore E[	ext{total}] = E[3.5\,N] = 3.5\,E[N] = 3.5 	imes 7 = 24.5. The expected total is 24.5, that is \frac{49}{2}: expected count (7) times average ticket (3.5). Notice you never needed the full d…

Two dice set the count, then draw that many tickets

Your answer

Discussion

Learn the concepts

Related questions