Streaky Free Throws: Success Probability Equals Running Average
A basketball player is shooting free throws. She makes her first shot and misses her second. From the third shot onward, the probability that she makes a shot equals the fraction of her previous shots she has made so far. (So her third shot succeeds with probability $1/2$; if it succeeds, her fourth shot succeeds with probability $2/3$, otherwise with probability $1/3$; and so on.)
What is the probability that she makes exactly 50 of her first 100 shots?
Hints
- Try small cases. After 3 shots, she has made 1 or 2 shots, each with probability $1/2$. After 4 shots, compute the distribution of the number made.
- Conjecture: after $n$ shots, the number made is uniform on $\{1, 2, \ldots, n-1\}$, each with probability $1/(n-1)$. Prove it by induction.
- Inductive step: $P(k \text{ of } n+1) = \frac{1}{n-1}\left[\frac{k-1}{n} + \frac{n-k}{n}\right] = \frac{1}{n}$. Then plug in $n = 100$.
Worked Solution
How to Think About It: The rule "probability of success equals the current success fraction" is exactly the Polya urn: start with one make-ball and one miss-ball, draw one at random, and put it back with another ball of the same type. The distribution of the number of makes has a famous, very clean form, and the way to discover it is to compute a couple of small cases and then prove the pattern by induction.
Quick Estimate: After three shots she has 1 or 2 makes with probability $1/2$ each. After four shots: 1, 2 or 3 makes, and a quick check gives $1/3$ each. The pattern "uniform on $1, \ldots, n-1$" suggests $1/99$ at $n = 100$.
Formal Solution:
*Step 1 -- Set up.* Let $M_n$ be the number of makes after $n$ shots, $n \ge 2$. We have $M_2 = 1$. Given $M_n = k$, shot $n+1$ succeeds with probability $k/n$, so $M_{n+1} = k+1$ with probability $k/n$ and $M_{n+1} = k$ with probability $(n-k)/n$.
*Step 2 -- Claim.* For every $n \ge 2$ and every $k \in \{1, \ldots, n-1\}$,
$$P(M_n = k) = \frac{1}{n-1}.$$
*Step 3 -- Induction.* The base case $n = 2$ is $P(M_2 = 1) = 1$. Assume the claim for $n$. For $k \in \{1, \ldots, n\}$,
$$P(M_{n+1} = k) = P(M_n = k-1)\,\frac{k-1}{n} + P(M_n = k)\,\frac{n-k}{n}.$$
For $2 \le k \le n-1$ both terms use the inductive hypothesis: $\frac{1}{n-1}\cdot\frac{(k-1) + (n-k)}{n} = \frac{1}{n-1}\cdot\frac{n-1}{n} = \frac{1}{n}$. For $k = 1$: only the second term, $\frac{1}{n-1}\cdot\frac{n-1}{n} = \frac{1}{n}$. For $k = n$: only the first term, $\frac{1}{n-1}\cdot\frac{n-1}{n} = \frac{1}{n}$. So $M_{n+1}$ is uniform on $\{1, \ldots, n\}$, completing the induction.
*Step 4 -- Evaluate.* With $n = 100$, $P(M_{100} = 50) = \dfrac{1}{99}$.
*Remark.* Any specific count from 1 to 99 has the same probability $1/99$; the answer does not depend on the number 50 at all.
Answer: $P(\text{exactly 50 of 100}) = \dfrac{1}{99} \approx 0.0101$.
Intuition
This is a Polya urn in disguise: each make adds a "make" ball and each miss adds a "miss" ball, and the next shot draws a ball at random. Polya urns have the striking property that the number of successes is uniformly distributed, so after 100 shots every count from 1 to 99 is equally likely and the answer is $1/99$ regardless of which count you ask about. Self-reinforcing processes like this are the canonical model of path dependence and "rich get richer" dynamics, and the uniform law is why early luck in such processes never washes out.