Free throws until the shooter finally misses

Question

A basketball player makes each free throw independently with probability $0.8$ , so a miss has probability $0.2$ . She shoots until she misses one.

On average, how many shots does she take to reach the first miss (counting the miss)?

Accepted Answer

The trick is to name the right "success." Here the event we are waiting for is a miss, which happens with probability p = 0.2 per shot. Making a basket is the "failure" that keeps the streak going. Each shot costs 1. With probability p = 0.2 it is the miss we are waiting for; otherwise we reset: E = 1 + (1-p)\,E \quad\Longrightarrow\quad E = \frac{1}{p} = \frac{1}{0.2} = 5 	ext{ shots.} Takeaway. Waiting-time problems are about whichever outcome you are counting toward; here the miss has probability 0.2, so the expected wait is 1/0.2 = 5.

Free throws until the shooter finally misses

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