$66$ points in $100$ shots. I just received a probability problem from a friend via a text and by the time I took to read it, I was sent a solution as well - which is confusing.. The question goes like this.. 

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 confusing solution:

$\dfrac1{(n-1)} \ge \dfrac1{99}$

Can someone please explain the logic behind this?

Let $P(x, n)$ be the probability that you have $x$ points after $n$ throws. We are given that $$P(1,2) = 1$$
You are also given information about the probability of scoring at any step, if you know the number of points previously scored. In particular:
$$P(\text{scoring from } x-1, n-1 ) = \frac{x-1}{n-1}$$
$$P(\text{scoring from } x, n-1 ) = \frac{x}{n-1}$$

Now, note that in order to have $x$ points after $n$ throws, we can either have $x-1$ points after $n-1$ throws, then score a point, or have $x$ points after $n-1$ throws, then not score a point. 
We can write this as follows:
$$P(x,n) = P(x-1,n-1)P(\text{scoring from } x-1, n-1 ) + P(x,n-1)(1 -
P(\text{scoring from } x, n-1 ))$$
This is the same as
$$P(x,n) = P(x-1,n-1)\left(\frac{x-1}{n-1}\right) + P(x,n-1)\left(1 -\left(\frac{x}{n-1}\right)\right)$$
$$P(x,n) = P(x-1,n-1)\left(\frac{x-1}{n-1}\right) + P(x,n-1)\left(\frac{n-1 - x}{n-1}\right)$$
Call this last result "potato", because we will refer to it later as such.

Statement:
For all $n > 1$ and $1 \leq x \leq n-1$, 
$$P(x,n) = \frac{1}{n-1}$$
Proof:
Our base case is $n=2$, for which we are given that $$P(1,2)=1$$
For the inductive step, we assume that for some $k \geq 2$, the following is true for all $x$ between $1$ and $k-1$, inclusive:  $$P(x,k) = \frac{1}{k-1}$$
Now suppose $y$ satisfies $2 \leq y \leq k-1$, and we wish to find $P(y, k+1)$. By the previous result "potato", we can write
$$P(y,k+1) =  P(y-1,k)\left(\frac{y-1}{k}\right) + P(y,k)\left(\frac{k - y}{k}\right)$$
But we know from our inductive assumption that $P(y,k) = P(y-1,k) = 1/(k-1)$. Therefore, 
$$P(y,k+1) = \left(\frac{1}{k-1}\right)\left(\frac{y-1}{k}\right) + \left(\frac{1}{k-1}\right)\left(\frac{k - y}{k}\right)$$
$$P(y,k+1) =\frac{y-1}{k(k-1)} + \frac{k-y}{k(k-1)}$$
$$P(y,k+1) = \frac{k-1}{k(k-1)}$$
$$P(y,k+1) = \frac{1}{k}$$
We are finished for the cases when $2 \leq y \leq k-1$, but we still have to consider the edge cases $y=1$ and $y=k$.
After $k+1$ throws, in order to have only $1$ point, we must miss every throw from the third to the $(k+1)^\text{th}$. The probability of missing the third throw is $1/2$. Given this, the probability of missing the fourth throw is $2/3$. Given this, the probability of missing the fifth throw is $3/4$, and so on. We get a telescoping product:
$$P(1, k+1) = \prod_{j=1}^{k-1} \frac{j}{j+1} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{k-2}{k-1} \cdot \frac{k-1}{k} = \frac{1}{k}$$
Similarly, after $k+1$ throws, in order to have exactly $k$ points, we must get a point on every throw from the third to the $(k+1)^\text{th}$. The probability of getting a point on the third throw is $1/2$. Given this, the probability of getting a point on the fourth throw is $2/3$. Given this, the probability of getting a point on the fifth throw is $3/4$, and so on. We once again have a telescoping product:
$$P(k, k+1) = \prod_{j=1}^{k-1} \frac{j}{j+1} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{k-2}{k-1} \cdot \frac{k-1}{k} = \frac{1}{k}$$
This completes the inductive step, so we've finally shown that for all $n >1$ and $1 \leq x \leq n-1$, $$\boxed{P(x,n) = \frac{1}{n-1}\,}$$

In particular, your question asked for $$P(66, 100) = \frac{1}{99}$$
