Adaptive scoring

A basketball player takes 100 free throws. She scores on her first throw and misses her second. From throw 3 onward, her probability of scoring equals the fraction of throws she's made so far.

For example, after 40 throws with 23 made, the probability of scoring on throw 41 is $23/40$ . Step through the first 20 throws to see how the probability adapts — notice how a hot streak raises your next-throw probability, creating a self-reinforcing loop:

Score Distribution

Throw by Throw

Throw 1: score ✓, Throw 2: miss ✗. From throw 3, P(score) = made/thrown.

What do you think?

After 100 throws, what is P(exactly 50 baskets)?

Run the simulation and check the distribution. It looks flat — every score from 1 to 99 is equally likely.

Basketball Free Throws

Starting small

Don't let the number 100 intimidate. Start with the smallest case and look for a pattern.

After $n = 3$ throws (first throw: score, second throw: miss):

The third throw has probability $1/2$ of scoring (1 make in 2 throws).

Outcome of throw 3	Baskets after 3	P
Score	2	1/2
Miss	1	1/2

So $P_{3,1} = 1/2$ and $P_{3,2} = 1/2$ . Both equally likely! And $1/2 = 1/(n-1)$ with $n = 3$ .

After $n = 4$ throws:

Computing P(4, k) for each k

P_{4,1} = P(\text{miss} \mid (3,1)) \times P_{3,1} = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}

To have 1 basket after 4 throws: we needed 1 basket after 3 (probability 1/2), then missed throw 4 (probability 1 - 1/3 = 2/3).

Step 1 of 4

After 4 throws: $P_{4,1} = P_{4,2} = P_{4,3} = 1/3$ . The pattern holds!

Uniform Distribution Pattern

n = 4 throws

33.3%

k=1

33.3%

k=2

33.3%

k=3

After 4 throws: P(4, k) = 1/3 for all k = 1, 2, ..., 3

Each possible basket count is equally likely!

Slide the slider to see: for every $n$ , the distribution $P_{n,k} = 1/(n-1)$ is perfectly flat. This is not coincidence — it's provable by induction.

The induction proof

Uniform Score Distribution

After $n$ throws (with the first a score and second a miss, and adaptive probability from throw 3 onward), the number of baskets $k$ satisfies $P_{n,k} = \frac{1}{n-1}$ for each $k = 1, 2, \ldots, n-1$ .

Base case: $n = 3$ gives $P_{3,1} = P_{3,2} = 1/2 = 1/(3-1)$ . ✓

Inductive step: Assume $P_{n,k} = 1/(n-1)$ for all $k$ . We need to show $P_{n+1,k} = 1/n$ .

Induction Step

P_{n+1,k} = P(\text{miss} \mid (n,k)) \cdot P_{n,k} + P(\text{score} \mid (n,k-1)) \cdot P_{n,k-1}

Law of total probability: either we had k baskets after n throws and missed, or we had k-1 baskets and scored.

Step 1 of 5

The cancellation makes it work. The "miss" term contributes $1 - k/n$ and the "score" term contributes $(k-1)/n$ . Together they sum to $(n-1)/n$ , which is independent of $k$ . This uniformity is maintained at every step.

So $P_{n+1,k} = 1/n$ for all valid $k$ , completing the induction. For $n = 100$ :

$P_{100,50} = \frac{1}{99}$

What do you think?

Does this mean each score from 0 to 100 is equally likely?

Why is this surprising?

The adaptive probability creates a self-reinforcing process. When the player is doing well (high scoring rate), she's more likely to keep scoring. When she's doing poorly, she's more likely to keep missing. This is a Polya urn process — it's like adding a red ball to an urn every time you draw red, and a blue ball every time you draw blue. Watch multiple games unfold simultaneously:

Many Games at Once

Each path shows baskets scored over 50 throws. Watch how they spread out.

Intuitively, you'd expect this reinforcement to create extreme outcomes — lots of games with very high or very low scores. And indeed, each individual score is just 1/99 ≈ 1%. But the remarkable fact is that no score is more likely than any other. The reinforcement doesn't create peaks; it creates perfect uniformity.

Pólya Urn Model

Start with $r$ red and $b$ blue balls. Draw a ball, note its color, return it along with one new ball of the same color. This is equivalent to the basketball scoring process (with $r = 1$ "score" and $b = 1$ "miss" initially). The limiting fraction of red balls is uniformly distributed on $[0,1]$ .

Competition appearances

This problem appears in quant interviews (Two Sigma, DE Shaw), in Putnam competition problems about sequences with adaptive probabilities, in Pólya urn theory as a foundational example of stochastic processes, and in machine learning where the Chinese Restaurant Process and Dirichlet-Multinomial models are generalizations.

Don't be intimidated by large numbers. Start with the smallest case, compute, look for a pattern, and prove it by induction. The number 100 was a red herring.

After 10 throws (score first, miss second), P(exactly 5 baskets)? (decimal like 0.11)

1/4

The takeaway

Two techniques drive the solution: start small and find patterns before tackling the full problem, then use induction to confirm the pattern. The uniformity comes from a perfect cancellation in the law of total probability. In competitions, when the problem gives you a large number (100, 1000, etc.), it's almost always a hint to solve the small case first.