Free throws until the first miss

Question

A free-throw shooter misses each shot independently with probability 0.25. Let N be the number of shots up to and including the first miss. What are the mean and variance of N?

Accepted Answer

The trick is to label the event you are waiting for as the "success." Here the success is a miss, with probability p = 0.25 per shot. Then N (shots until the first miss) is geometric with mean 1/p and variance (1-p)/p^2: \mathbb{E}[N] = \frac{1}{0.25} = 4, \qquad \operatorname{Var}(N) = \frac{0.75}{0.25^2} = \frac{0.75}{0.0625} = 12. Standard deviation \sqrt{12} \approx 3.46. On average the shooter makes 3 in a row and misses on the 4th attempt. Takeaway. Relabel the awaited event as the success (p = 0.25): the wait has mean 1/p = 4 and variance (1-p)/p^2 = 12.

Free throws until the first miss

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