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My question is a follow up to the famous problem 'Basketball Scores' from the green book. The question can be found here:

A person shoots basketball 100 times. First time he scores a point and second time he misses it. For the following shots, the probability of him scoring a point is equal to number of points scored before this shot divided by number of shots taken before this shot, for ex: if he is into his 21st shot and he has scored 13 points in the first 20 shots then the probability of him scoring in 21st shot is 13/20. What is the probability of scoring 66 points in the 100 shots (including the first two)

The general solution is for any $1 \le x \le n-1$:

$$P(x, n) = \frac{1}{n - 1}$$

where for the specific problem ($n=100$ and $x=66$) is $\frac{1}{99}$.

I solved the problem using the factorial approach. I also understand how the following recurrence is formed and the answer is verified using induction in this answer, in which we need to know the answer.

My question is, how do I solve for $P(x, n)$ using the following 2-variable recurrence relation without knowing the answer?

$$P(x,n) = P(x-1,n-1)\left(\frac{x-1}{n-1}\right) + P(x,n-1)\left(\frac{n-1 - x}{n-1}\right)$$

I know that we can use generating functions to solve for single variable recurrences. How exactly do I extend that here and arrive at the result that $P(x, n) = \frac{1}{n - 1}?$ Any other method to solve this recurrence or any intuitive reason explaining why the answer is independent of k (without using factorials) would also be appreciated.

Amir
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Devansh Agarwal
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  • Do you want to solve the recurrence relation without induction? – Amir Feb 26 '24 at 14:30
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    Yeah. Basically, in order to "solve" it using induction, I would require to know the answer, right? That's how induction works. My question is how do I get to the answer in the first place? How do I actually SOLVE this recurrence? – Devansh Agarwal Feb 26 '24 at 19:41
  • So, it is a nice question, though it may be difficult to answer. – Amir Feb 26 '24 at 19:44
  • What would the initial condition for this recurrence be? Based on the final answer it'd seem to be $P(x,2)=1$. – Semiclassical Feb 26 '24 at 21:30
  • @Semiclassical You are correct. You can go through the link attached to read the original problem. The base case is $P(x, 2) = 1$. – Devansh Agarwal Feb 27 '24 at 06:16

1 Answers1

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Here I explained how the two-dimensional recurrence can be solved directly without knowing the answer.

Let us first fix $x=1$. Then, we have the following recurrence:

$$P(1,n) = P(1,n-1)\left(\frac{n-2}{n-1}\right).$$

The solution of this is given by

$$P(1,n) = \frac{c}{n-1},$$

where from $P(1,2)=1$, we have $c=1$.

Then, for $x=2$, we have the following recurrence:

$$P(2,n) = \frac{1}{(n-2)(n-1)} + P(2,n-1)\left(\frac{n-3}{n-1}\right)$$

The solution of this is given by

$$P(2,n) = \frac{2c+n}{(n-1)(n-2)},$$

where from $P(2,3)=P(1,2) \times \frac{1}{2} +0=\frac{1}{2}$, we have $c=-1$.

Next, for $x=3$, we have the following recurrence:

$$P(3,n) = \frac{2}{(n-2)(n-1)} + P(3,n-1)\left(\frac{n-4}{n-1}\right)$$

The solution of this is given by

$$P(3,n) = \frac{6c+n^2-5n+6}{(n-1)(n-2)(n-3)},$$

where from $P(3,4)=P(2,3) \times \frac{2}{3} +0=\frac{1}{3}$, we have $c=0$.

This can be continued in the same way.

As shown above, our problem reduces to solve the following one-dimensional recurrence:

$$f(n) = \frac{k-1}{(n-2)(n-1)} + f(n-1)\left(\frac{n-1-k}{n-1}\right)$$

for fixed $k$ and $n \ge k+1$ with $f(k+1)=\frac{1}{k}$ (set $n=k+1$ in the above).

Defining

$$g(n)=f(n)(n-1),$$

we obtain

$$\color{blue}{g(n)(n-2) = (k-1) + g(n-1)(n-1-k)}$$

with $\color{blue}{g(k+1)=1}$ for $\color{blue}{n \ge k+1}$.

From here, the general solution of the above is given by

$$g(n) = \frac{(c-1)\Gamma (n-k)}{\Gamma (2-k)\Gamma (n-1)}+1.$$

Now from $g(k+1)=1$, we obtain $c=1$, which gives $g(n)=1$, and finally we obtain $$f(n)=\frac{1}{n-1}.$$

Thus, for any $k$ and $n\ge k+1$, we have

$$P(k,n) =f(n)=\frac{1}{n-1}.$$

Amir
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