Self-Reinforcing Free Throws

She makes shot 1, misses shot 2, and from then on her probability of making the next shot equals her current fraction of makes.

1. Initial state

After two shots:

Makes

Misses

Shot 3:

P(make) = 1 / 2

If shot 3 is made:

2 makes, 1 miss → P(next make) = 2 / 3

If shot 3 is missed:

1 make, 2 misses → P(next make) = 1 / 3

This is a classic Pólya urn process: success reinforces future success, and failure reinforces future failure.

Among shots 3 through 100, there are 98 additional shots.

probability = 1/99

Surprisingly, the final total number of makes is uniformly distributed over 1, 2, ..., 99.

Probability of exactly 50 makes

1 / 99

≈ 1.0101%

Let X be the number of makes among shots 3–100. Then X ∈ {0,1,...,98}.

A Pólya urn starting with one “make” and one “miss” has a Beta-binomial distribution:

P(X = k) = C(98,k) · B(k+1, 99-k) / B(1,1)

Using factorials:

C(98,k) · k! · (98-k)! / 99! = 98! / 99! = 1 / 99

So every value k = 0,...,98 is equally likely. Since the first shot was already a make,

total makes = 1 + X

Therefore the total is uniform over 1,...,99. To finish with exactly 50 makes, we need X = 49.

P(total makes = 50) = 1 / 99 ≈ 1.01%