$66$ points in $100$ shots.
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 $$P(66, 100) = \frac{1}{99}$$
Among the 98 shots from 3rd to 100th we need 65 scoring shots. These can be chosen in $\binom{98}{65}$ ways. Let us compute the probability of one of such arrangements of scoring shots. The numerator contributed by the scoring shots contains all the numbers from 1 to 65 and the denominator is $99!$. Consider the numerator contributed by the non scoring shots. These are of the type $k - m_k$ where $m_k$ is the score up to the $k-1$ shots. The maximum of these numbers is 33. Note that no two of these can be equal. Thus we have 33 slots and we have 33 numbers to fill and hence all the numbers from 1 to 33 are contributed by the non scoring shots. Thus the probability of any such arrangement is the same and equals $\frac{33!65!}{99!}$. Since there are $\binom{98}{65}$ such arrangements, the required probability is $\binom{98}{65} \frac{33!65!}{99!} = \frac{1}{99}$
I think the existing answers are fine, but here's one that's shorter than the induction proof in Zubin Mukerjee's answer and maybe more elementary than the counting argument in Muralidharan's, and that (I think) gives more sense of why the thing is true.
The idea is to make a new thing that's "obviously" distributed the same way as the number of hits after $n$ shots, and that also is "obviously" equal to $1/(n-1)$.
So, suppose we have an infinite sequence of (independent) random numbers $x_1,x_2,x_3,\dots$ -- they can come from any nice smooth distribution you like, because all we'll care about is which ones are bigger than which other ones, but for concreteness let's say uniform between 0 and 1. And let's write $a_n$ for the position of $x_1$ among the first $n-1$ of these numbers when written in increasing order: so $a_n=1$ if $x_1$ is the smallest of them, and $a_n=n-1$ if $x_1$ is the largest. Equivalently: $a_n$ is the number of $\{\,x_1,\dots,x_{n-1}\,\}$ that are $\leq x_1$. Note that the probability that any two $x_j$ are equal is zero, and I'll just ignore that possibility in what follows.
First: clearly $a_n$ is equally likely to be any number from $1$ to $n-1$, because these numbers are picked independently from the same distribution: there's no reason why any of them should be more likely to be (say) 37th than any other.
Second: I claim that $a_n$ is distributed the same way as the number of hits after $n$ shots, for $n\geq2$. When $n=2$ these are both always 1. When going from $n$ to $n+1$, $a_n$ either doesn't change (if $x_n>x_1$) or increases by 1 (if $x_n<x_1$). And the probability of the latter is exactly $a_n/n$, which is the same as the probability that the number of hits increases by 1. So, by induction, the two always match.
(Why is that probability $a_n/n$? Because there are $n$ equally likely positions for $x_n$ relative to the first $n-1$ numbers, and $a_n$ of them make $x_n<x_1$ by definition of $a_n$.)
This proof is "really" the same as Zubin Mukherjee's: all the calculations are the same. (Except for the edge cases, which can be done more easily than in Zubin's answer as it currently stands: I'll leave a comment there to point out how.) Its real advantage (in my eyes) is that once you understand it, it makes it obvious that the answer is what it is.