<s>[INST] $66$ 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?
 [/INST] Here is a slightly shorter version of Zubin Mukerjee's solution, with a (I believe) simpler argument for the edge cases $P(1,N)$ and $P(N-1,N)$. The arguments are effectively the same up to that point.
We denote the probability that  we have a score of $x$ after $N$ shots as $P(x,N)$.
The problem states that
  $$ P(1,2)=1,$$
hence
  $$ P(0,2) = P(2,2) = 0.$$
We also have that the probability that we score on the $N$-th shot with a previous score $x$ is $\dfrac{x}{N-1}$.
We can write $P(x,N)$ as the sum of the chance of scoring on the $N$-th shot with a previous score of $x-1$, and the chance of not scoring the $N$-th shot with a previous score of $x$, or
  $$
    P(x,N) = P(x-1,N-1) \frac{x-1}{N-1}
             + P(x,N-1)\left(1 - \frac{x}{N-1}\right).
  $$

We conjecture that, for $N>1$,
  $$ P(x,N) =
  \begin{cases}
    \frac1{N-1} \quad & 1\leq x \leq N-1,\\
    \,\,\, 0 & \text{otherwise}.
  \end{cases}
  $$
Base Case
Above, we see that $P(1,2) = 1$ and $P(0,2) = P(2,2) = 0$, which satisfies the conjecture.
Induction
Assuming the conjecture, we have 
  $$ P(x,N-1) =
  \begin{cases}
    \frac1{N-2} \quad & 1\leq x \leq N-2,\\
    \,\,\, 0 & \text{otherwise}.
  \end{cases}
  $$
For $2\leq x \leq N-2$,
  $$
    \begin{aligned}
    P(x,N) &= P(x-1,N-1) \frac{x-1}{N-1}
             + P(x,N-1)\frac{N-x-1}{N-1}\\
           &= \frac1{N-2}\cdot\frac{x-1}{N-1}
             + \frac1{N-2}\cdot\frac{N-x-1}{N-1} \\
           &= \frac1{N-1}.
    \end{aligned}
  $$
Edge cases (alternative argument) - for $x=1$,
  $$
    \begin{aligned}
    P(1,N) &= P(0,N-1) \frac{1-1}{N-1}
             + P(1,N-1)\frac{N-1-1}{N-1}\\
           &= 0 + \frac1{N-2}\cdot\frac{N-2}{N-1}\\
           &= \frac1{N-1}.
    \end{aligned}
  $$
For $x=N-1$,
  $$
    \begin{aligned}
    P(N-1,N) &= P(N-2,N-1) \frac{N-2}{N-1}
             + P(N-1,N-1)\frac{N-(N-1)-1}{N-1}\\
           &= \frac1{N-2}\cdot \frac{N-2}{N-1} + 0\\
           &= \frac1{N-1}.
    \end{aligned}
  $$
Finally, it should be clear that $P(0,N) = P(N,N) = 0$ (since for $N=2$ we have already scored one and missed one).
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