Frank is shooting free throws. He makes his first free throw and misses his second free throw. For n≥3, the probability of making the nth free throw is equal to the proportion of free throws he made during his first n−1 attempts. How many free throws can Frank expect to make in 100 attempts?
We know, \(P(3) = 1/2\)
Let \(X = 0\) if miss and \(X = 1\) if throw.
\[ \begin{aligned}
P(n) &= \frac{S}{n-1} \\ S &= (n-1)P(n)
\end{aligned}$$
$$P(n) = S + X_{n-1}\]
where \(S = \sum X \text{ if } X=1\)
$$ \begin{aligned} \mathrm{E}(P_{n+1}) = \mathrm{E}(S/n) \
\end{aligned} $$