Five Ways to Calculate the Wait for a Successful Shot

Danielle takes independent half-court shots at the end of practice until she makes one. Each shot succeeds with probability 20%. How many attempts does she need on average? Intuitively, one make per five shots suggests an answer of five, but intuition alone is not a proof.

For a nonnegative integer-valued random variable N, the tail-sum formula says E[N] = Σᵢ₌₁∞ P(N ≥ i). Danielle needs at least i attempts exactly when she misses the first i−1 shots. Independence gives P(N ≥ i) = (4/5)^(i−1).

The expected number is therefore 1 + 4/5 + (4/5)² + ⋯. This geometric series has first term 1 and ratio 4/5, so its sum is 1/(1−4/5) = 5. The initial intuition gives the correct answer in this model.

Recursion offers another proof. On the first shot Danielle either succeeds and stops, or misses and returns to the same situation as at the start. If E denotes the expected number of attempts, conditioning on the first shot gives E = (1/5)·1 + (4/5)·(1+E).

The first case costs exactly one shot; the second costs the missed shot plus another expected E shots. Simplifying gives E = 1 + (4/5)E, hence E/5 = 1 and E = 5. The reset after a miss is what makes the recursive equation possible.

We can also start directly from the definition of expectation. The model allows Danielle to continue after any number of misses, even a million. That is not a perfect description of real fatigue, but such extreme outcomes have tiny probabilities in this model.

The probability that her first success is on attempt i is (4/5)^(i−1)(1/5): i−1 misses followed by a make. Thus E[N] = Σᵢ₌₁∞ i(1/5)(4/5)^(i−1). This is an arithmetico-geometric series because the integer coefficient grows with i.

To sum it, expand each term vertically into i identical copies. Each row is then a geometric series: the first sums to 1, the second to 4/5, the third to (4/5)², and so on. Their total is again 1/(1−4/5) = 5.

Calculus gives another route. Replace 4/5 by x while keeping the factor 1/5: the sum becomes (1/5)(1 + 2x + 3x² + ⋯). The expression in parentheses is the derivative of 1 + x + x² + ⋯ = 1/(1−x), for |x| < 1.

Differentiating gives (1/5)/(1−x)². Substituting x = 4/5 yields (1/5)/(1/5)² = 5. This evaluates the same infinite sum without rearranging it into rows.

The original intuition can also be made precise through a long-run experiment. Record many sequences of misses followed by a make, with each row representing one completed practice. The average number of attempts per practice is the total number of shots divided by the number of rows.

Each completed row contains exactly one made shot. Over a large sample, the fraction of shots made approaches 1/5, so the number of rows is approximately one-fifth of the number of shots. Their ratio approaches 5 attempts per completed practice.

For the bonus problem, Danielle shoots until she makes two consecutive shots. Two specified shots both succeed with probability 1/25, which might suggest 25 attempts, but that intuition does not account for what happens after partial progress and a miss.

Let X be the expected number of attempts from the beginning and Y the expected additional attempts after one make. On the first shot, a make costs one attempt and moves to Y; a miss costs one and returns to X. Thus X = (1/5)(1+Y) + (4/5)(1+X).

From the one-make state, the next shot succeeds with probability 1/5 and finishes the task, or misses with probability 4/5 and returns to the start. Therefore Y = (1/5)·1 + (4/5)(1+X) = 1 + (4/5)X.

Substituting this expression into the first equation gives X = 30. The intuitive answer of 25 is wrong: a miss after a first make erases the partial run, so the waiting time must account for restarting. Sawyer invites viewers to find other approaches and generalizations.