Spin for a count, then deal that many cards

Question

You spin a wheel that stops on one of 2, 4, 6, 8 with equal chance, and call the result N. You then deal N cards from a large deck, each card equally likely to be worth any value from 1 to 13 (dealt with replacement, so each card is fresh). You add the values. What total value should you expect?

Accepted Answer

The count N is random, and each card value is random too. The tower rule keeps the two layers apart: E[	ext{total}] = E\big[\,E[	ext{total} \mid N]\,\big]. Step 1, freeze the count. If you know N = n, you deal n independent cards. Each card averages \frac{1 + 2 + \cdots + 13}{13} = \frac{91}{13} = 7, so E[	ext{total} \mid N = n] = 7\,n. Step 2, average over the count. The wheel is uniform on \{2,4,6,8\}, so E[N] = \frac{2+4+6+8}{4} = 5. Then E[	ext{total}] = E[7\,N] = 7\,E[N] = 7 	imes 5 = 35. The expected total is 35: expected number of cards (5) times average card value (7). Takeaway. …

Spin for a count, then deal that many cards

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