Free throws in basketball game about probability Someone shoots free throws. He/She made the first one and missed the second one. From the third shot, the probability of hitting the ball equals to the free throw percentage he/she made before it. For example, if the made 87 out of 100 tries. Then the probability of making the next one is 87/100. 
What is the probability of the person making the n th? Does it matter whether he makes the n-1 th free throws?
 A: I will work out the probability of making the $(n+1)$'th throw, given that they made the $n$'th thow. Let $H_n$ be the event that they hit on the $n$'th shot, and let $K_n$ be the total number of hits after $n$ shots. 
I will first prove by induction, that after $n$ throws there is an equal probability of having any number of hits. I.e., we have $P(K_n=k) = \frac{1}{n-1}$ for all $1\le k \le n-1$.
The case $n=2$ is trivial. Assume $n\ge3$. We can get $K_{n+1}=k$ in two ways: Hitting after $k-1$ hits, or missing after $k$ hits. This means:
$$
P(K_{n+1}=k)
= \frac{1}{n-1} \cdot \frac{k-1}{n} 
+ \frac{1}{n-1} \cdot \frac{n-k}{n}
= \frac{1}{n}
$$
Note that this works even for the edge cases $k=1$ and $k=n$. Now we come back to the original problem. We want to work out:
$$
P(H_{n+1}|H_n) = \frac{P(H_{n+1} \cap H_{n})}{P(H_{n})}
$$
for $n\ge 3$. By the symmetry of the setup, we have simply $P(H_{n})=1/2$. We can calculate $P(H_{n+1} \cap H_{n})$ by splitting up in the cases for $K_{n-1}$:
$$ \begin{split}
P(H_{n+1} \cap H_{n})
&= \sum_{k=1}^{n-2} \frac{1}{n-2} \cdot \frac{k}{n-1}\cdot\frac{k+1}{n}
= \frac{1}{n(n-1)(n-2)} \left(\sum_{k=1}^{n-2} k(k+1)\right) \\
&= \frac{1}{n(n-1)(n-2)} \cdot \frac{n(n-1)(n-2)}{3} 
= \frac13
\end{split}
$$
So in the end we get
$$
P(H_{n+1}|H_n) = \frac{1/3}{1/2} 
= \frac23
$$
Note that the result is independent of $n$!
Edit: 
On antkam's suggestion, I'll prove my observation in the comments. I claim that all sequences of $n$ shots that have the same number of hits are equally likely. (This can actually be proven from the property I proved inductively above, but I'll do it the other way round). For example $P(HHMM) = P(HMMH) = P(MHMH) = \ldots$, where $H$ is a hit and $M$ is a miss. Since the first two shots are fixed, the sequences begin at the third shot. This is interesting, because while there are less ways to get a very high or low number of hits, each of those sequences are more likely because of the setup. These tendencies exactly cancel out to give the uniform distribution of $K_n$.
My precise claim is this:
$$
P(S_3S_4\cdots S_{n+1}) = \frac{k!(n-k-1)!}{n!}
$$
for any $n+1 \ge 3$, where $S_i$ is the outcome of the $i$'th shot (hit or miss), and $k$ is the total number hits. 
Let $p_n = P(S_n = H)$ and $q_n = P(S_n = M)$. Note that we can write 
$$
P(S_3S_4\cdots S_{n+1})
= \prod_{S_i=H}p_i \cdot \prod_{S_i=M}q_i
$$
We have $p_i=\frac{K_i}{i-1}$ and $q_i=\frac{i-1-K_i}{i-1}$. Since the sequence goes from the third to the $(n+1)$'th shot, we get a denominator of $n!$ when we multiply all the probabilities. 
Let's consider the numerator. If we hit on the $i$'th shot ($S_i = H$), then
$$
p_{i+1} = \frac{K_i+1}{i}, \quad q_{i+1}=\frac{i-(K_i+1)}{i}
$$
and if $S_i = M$:
$$
p_{i+1} = \frac{K_i}{i}, \quad q_{i+1}=\frac{i-K_i}{i}
$$
We see that when we hit, the numerator of $p$ goes up by 1, and when we miss, the numerator of $q$ goes up by 1. Meanwhile the other numerators are unchanged.
Note that $p_3 = q_3 = 1/2$, so they both start with a numerator of 1. Thus, multiplying all the numerators together, we get $k!(n-k-1)!$, which proves the claim. 
So how does this get us the uniform distribution of $K_n$? If $K_{n+1} = k$, then the sequence of $n-1$ shots will have $k$ hits. There are $\binom{n-1}{k}$ such sequences, all equally likely, so we rediscover the result that:
$$
P(K_{n+1}=k) = \frac{k!(n-k-1)!}{n!}\cdot \binom{n-1}{k}
= \frac{k!(n-k-1)!}{n!} \cdot \frac{(n-1)!}{k!(n-k-1)!}
= \frac{1}{n}
$$
A: Technically if he made the first one, he could not fail the next one since he did 1 out 1 and this success probability was $100\%$.
