A first pick sets the ceiling for the second

Question

Pick a whole number A uniformly from \{1, 2, \dots, 8\}. Then pick B uniformly from \{1, 2, \dots, A\}. What is the expected value of B?

Accepted Answer

The second pick B depends on the first pick A, so condition on A and apply the tower rule: E[B] = E\big[\,E[B \mid A]\,\big]. Step 1, freeze the first pick. Given A, the pick B is uniform on \{1, \dots, A\}, and a uniform pick on \{1, \dots, m\} averages \frac{m+1}{2}. So E[B \mid A] = \frac{A + 1}{2}. Step 2, average over the first pick. A is uniform on \{1, \dots, 8\}, so E[A] = \frac{1 + 8}{2} = 4.5. Then E[B] = E\!\left[\frac{A + 1}{2}\right] = \frac{E[A] + 1}{2} = \frac{4.5 + 1}{2} = 2.75. So E[B] = \frac{11}{4} = 2.75. Takeaway. For a two-stage pick where the first sets the range of the…

A first pick sets the ceiling for the second

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