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

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