A die picks the bias, then eight flips

Question

A spinner picks one of five bias settings, $0.1, 0.3, 0.5, 0.7, 0.9$ , each equally likely, and that becomes the coin's probability of heads $p$ . You then flip the coin $8$ times.

How many heads should you expect?

Accepted Answer

The bias is chosen first, then eight flips run at it. Condition on the chosen setting and apply the tower rule: E[	ext{heads}] = E\big[\,E[	ext{heads} \mid p]\,\big]. Step 1, freeze the setting. Given bias p, eight flips average 8\,p heads. Step 2, average over the setting. The five settings are equally likely, so their mean is E[p] = \frac{0.1 + 0.3 + 0.5 + 0.7 + 0.9}{5} = \frac{2.5}{5} = 0.5. Therefore E[	ext{heads}] = E[8\,p] = 8\,E[p] = 8 	imes 0.5 = 4. You expect 4 heads. The settings are symmetric around 0.5, so their average is 0.5 and the flips behave like fair-coin flips on avera…

A die picks the bias, then eight flips

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