Basketball Player

Suppose a basketball player make $N$ shots, and we know that in the first $n(\le N)$ shots he sinked $m(\le n)$ shots. If from the $(n+1)^{th}$ shot, his ratio of sink a shot is his accumulative ratio before the shot, e.g., suppose he sinked 40 shots in the first 50 shots, then his ratio of sink the next shot is 80%. What’s the probability that he will make $M$ shots finally?

Using the knowledge of permutation and combination, we can solve this problem directly. Let $E_{m,n}$ be the event that $m$ success in the first $n$ shots. It turns out to be that

Notice that

i.e., the number of sinked shots $M$ is uniformly distributed on its support given that the player only sinked 1 shot in the first 2 shots.

Let $X_k, n\le k\le N$ be the number of shots the player sinks in the first $k$ shots, then the distribution of $X_{k+1}$ conditioning on $X_{k}$ is given in Table 1. Using the similar method as we’ve done in the first 3 problems, we can easily find the first and second moment of $X_k$ which are given below:

So the variance of $X_k$ is

From the above formulas, we can know that the expectation and variance of the number of sinked shots are linear quadratic functions of $k$ respectively, and both of them increase as $k$ increases. This makes it hard for us to predict $X_k$ when $k$ is big. The 2-$\sigma$ intervals for $X_k$ is shown in the follow figure.

Since we know the distribution of the number of sinked shots given that the player sinked $m$ shots in the first $n$ shots, we can calculate the first and second moments directly, which yields the following equations: