probability - Shooting Free Throws

Shaq is shooting free throws. He makes his first shot, then misses his second.

Confidence is a huge factor in how well Shaq plays. This means that for each subsequent shot, the probability he makes it is equal to the fraction of shots he has made so far.

For example, there is a $\frac12$ chance he makes his third shot. If he makes that, there is a $\frac23$ chance he will make his fourth.

After 101 shots (including the first two), what is the probability that Shaq sank less than 21 baskets?

^{Side note: Symmetry implies that Shaq will make half of his free throws on average, which matches his career average of 52.7% pretty closely.}

Answer

After $n$ shots, it is equally likely that Shaq has made any number of shots from $1$ to $n-1$.

Proof by induction:

If $n=2$, then Shaq has made $1$ shot, which is the entire range from $1$ to $2-1$.

Otherwise:

• For Shaq to make $1$ shot out of $n$, he must make $1$ shot out of $n-1$ and then miss the next shot. This has a probability of $\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.


• For Shaq to make $n-1$ shots out of $n$, he must make $n-2$ shots out of $n-1$ and then make the next shot. This has a probability of $\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.



• For Shaq to make $k$ shots out of $n$, with $1
  

So the probability that Shaq makes $m$ shots out of $n$, for $1\le m\le n-1$, is always $\frac{1}{n-1}$. The probability that Shaq makes between $1$ and $20$ shots inclusive out of $101$ is $\frac{20}{100}=\boxed{\frac{1}{5}}$.