Problem 69. Aziz and Basketball
Aziz is practicing basketball free throws. His first shot goes in, and his second shot misses. From then on, the probability of his \((n+1)\)’th shot going in is proportional to his performance in the previous \(n\) shots (i.e., if 2 out of the first 5 shots go in, the 6th shot has a 2/5 chance of going in).
What is the probability that Aziz will make exactly 25 successful shots out of 50 shots?
The answer is 1/49.
We start by noting that after \(n \geq 2\) shots, the probability of exactly 0 or \(n\) successful shots is 0. We prove below that for any \(1 \leq k < n\) the odds that \(k\) out of \(n\) shots are successful is \(1/(n-1)\). We prove this by induction:
Base case: For \(n=2\), it is trivial to see that the conditions hold.
Induction step: Assume that the condition holds for \(n \geq 2\) shots and any \(1 \leq k < n\). That is, if we denote the probability of exactly \(k\) goals in \(n\) shots by \(p(k, n)\), the induction hypothesis states that \(p(0, n) = p(n, n) = 0\) and for \(0 < k < n\) we have \(p(k, n) = 1/(n-1)\). Based on this, we prove that \(p(k,n+1)\) = \(1/n\) for any \(1 \leq k \leq n\). We consider the following three cases:
\(k=1\): Either we have zero goals in the first \(n\) shots (probability \(p(0,n) = 0\)) and the \((n+1)\)’th shot is successful (probability 0) or we have one goal in the first \(n\) shots (probability \(p(1,n) = 1/(n-1)\)) and the \((n+1)\)’th shot is unsuccessful (probability \((n-1)/n\)). Therefore we have: \[p(1, n+1) = p(0, n) \cdot 0 + p(1, n) \cdot (n-1)/n = 0 + 1/(n-1) \cdot (n-1)/n = 1/n.\]
\(k=n\): Either we have \(n-1\) goals in the first \(n\) shots (probability \(p(n-1,n) = 1/(n-1)\)) and the \((n+1)\)’th shot is successful (probability \((n-1)/n\)) or we have \(n\) goals in the first \(n\) shots (probability 0) and the \((n+1)\)’th shot is unsuccessful (probability 0). Therefore we have:
\[p(n, n+1) = p(n-1, n) \cdot (n-1)/n + p(n, n) \cdot n-n/n = 1/(n-1) \cdot (n-1)/n + 0 = 1/n.\]
\(1 < k < n\): Either we have \(k-1\) goals in the first \(n\) shots (probability \(p(k,n)) = 1/(n-1)\)) and the \((n+1)\)’th shot is successful (probability \((k-1)/n\)) or we have \(k\) goals in the first \(n\) shots (probability \(p(k,n)=1/(n-1)\)) and the \((n+1)\)’th shot is unsuccessful (probability \((n-k)/n\)). Therefore we have:
\[p(k, n+1) = p(k-1, n) \cdot (k-1)/n + p(k, n) \cdot (n-k)/n = 1/(n-1) \cdot (k-1)/n + 1/(n-1) \cdot (n-k)/n = 1/n.\]
Therefore, the proof also holds for \(n+1\) shots and any \(1 \leq k < n\) and thus holds for all natural numbers greater than or equal to 2.