Dgdrxrt

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?

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 $P(K_n=k) = frac1n-1$ for all $1le k le n-1$.

The case $n=2$ is trivial. Assume $nge3$. We can get $K_n+1=k$ in two ways: Hitting after $k-1$ hits, or missing after $k$ hits. This means:
$$
P(K_n+1=k)
= frac1n-1 cdot frack-1n
+ frac1n-1 cdot fracn-kn
= frac1n
$$

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:
$$
P(H_n+1|H_n) = fracP(H_n+1 cap H_n)P(H_n)
$$

for $nge 3$. 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 $K_n-1$:
$$ beginsplit
P(H_n+1 cap H_n)
&= sum_k=1^n-2 frac1n-2 cdot frackn-1cdotfrack+1n
= frac1n(n-1)(n-2) left(sum_k=1^n-2 k(k+1)right) \
&= frac1n(n-1)(n-2) cdot fracn(n-1)(n-2)3
= frac13
endsplit
$$

So in the end we get
$$
P(H_n+1|H_n) = frac1/31/2
= frac23
$$

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:
$$
P(S_3S_4cdots S_n+1) = frack!(n-k-1)!n!
$$
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
$$
P(S_3S_4cdots S_n+1)
= prod_S_i=Hp_i cdot prod_S_i=Mq_i
$$
We have $p_i=fracK_ii-1$ and $q_i=fraci-1-K_ii-1$. 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
$$
p_i+1 = fracK_i+1i, quad q_i+1=fraci-(K_i+1)i
$$

and if $S_i = M$:
$$
p_i+1 = fracK_ii, quad q_i+1=fraci-K_ii
$$

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 $binomn-1k$ such sequences, all equally likely, so we rediscover the result that:
$$
P(K_n+1=k) = frack!(n-k-1)!n!cdot binomn-1k
= frack!(n-k-1)!n! cdot frac(n-1)!k!(n-k-1)!
= frac1n
$$

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 $100%$.