Basketball Expectation Value question

Question

Context

Here is the problem I was working on: A player is shooting free throws. They make the first, miss the second, and from then onwards, the probability that they make a shot is the fraction of total shots taken so far that they have made. What is the expected score after 100 shots?

Attempt

Let $E_{i}$ be the expected number of free throws made after $i$ attempts.

$E_{i} = p_{i}(1 + E_{i-1}) + (1-p_{i})(E_{i-1}) = p_{i} + E_{i-1}$

Now, out of nowhere, I had the brilliant idea to set $p_{i}=\frac{E_{i-1}}{i-1} \implies E_{i} = \frac{E_{i-1}}{i-1}+E_{i-1}=\frac{i}{i-1}E_{i-1}$

And, as it turns out, this is correct. I then went on to simply say $E_{100} = \frac{100}{99}E_{99}=\frac{100}{99}\frac{99}{98}...\frac{4}{3}E_{3}$.

The question gives us enough information to determine that $E_3 = 1.5$, and so $E_{100}=50$.

My Question

Why can we say that $p_i=\frac{E_{i-1}}{i-1}$? Isn't this the average probability? Since it is the average probability, why can we work with it? I somehow had the intuition that we can, but I can't justify it.

Answer 1

The problem is a slight, uh, uncredited modification of problem B1 from the 2002 Putnam Competition, and solving that problem gives us enough to go on to solve this problem along the way.

We can prove that after taking $n$ shots, each score between $1$ and $n-1$ has a probability of $\frac1{n-1}$, by induction on $n$. The expected value of this distribution is $\frac12 n$, and so the expected value for $n=100$ in particular is $50$. (A possible shortcut for the expected value is to observe a symmetry between shots made and shots missed, so the probability question asked on the Putnam is more interesting.)

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To justify the calculation suggested in the question, it helps to distinguish between the random variables and their expected values. Let $\mathbf X_i$ be the actual score after $i$ shots, so that $E_i = \mathbb E[\mathbf X_i]$. Then the identity between $E_{i-1}$ and $E_i$ can be stated more precisely: conditioned on $\mathbf X_{i-1}$, the value of $\mathbf X_i$ is equal to $\mathbf X_{i-1} + 1$ with probability $\frac{\mathbf X_{i-1}}{i-1}$ and equal to $\mathbf X_{i-1}$ otherwise, so we have $$ \mathbb E[\mathbf X_i \mid \mathbf X_{i-1}] = \mathbf X_{i-1} + \frac{\mathbf X_{i-1}}{i-1} = \frac{i}{i-1} \mathbf X_{i-1}. $$ In other words: no matter what the score is after $i-1$ shots, once it is known, then the expected score after $i$ shots is $\frac{i}{i-1}$ times that score.

By the law of total expectation, $\mathbb E[\mathbf X_i] = \mathbb E[\mathbb E[\mathbf X_i \mid \mathbf X_{i-1}]]$, and thererefore $$ E_i = \mathbb E[\mathbf X_i] = \mathbb E[\mathbb E[\mathbf X_i \mid \mathbf X_{i-1}]] = \frac{i}{i-1} \mathbb E[\mathbf X_{i-1}] = \frac{i}{i-1} E_{i-1}. $$ In other words, the law of total expectation is what justifies using the expected value calculation using the expected value of the probability in place of the probability.

Now that we have shown this, it is okay to do the calculation $E_{100} = \frac{100}{99} \cdot \frac{99}{98} \cdots \frac 43 E_3$, or even to keep going and say $E_{100} = \frac{100}{99} \cdot \frac{99}{98} \cdots \frac 32 E_2 = \frac{100}{2} E_2$. (This saves us a minor calculation at the end, since it is certain that $E_2 = 1$.)

Answer 2

Why can we say that $p_i={E_{i−1}\over i−1}$ ? Isn't this the average probability? Since it is the average probability, why can we work with it? I somehow had the intuition that we can, but I can't justify it.

Indeed, $p_i$ is the expected probability of success for the $i$-th shot.

Since this is a Bernoulli variable, it is the expected success rate for that shot. So immediately we have that: the expected count for successes over $i$ shots equals the sum of the expected success for that shot and the expected count for successes over the prior $i-1$ shots.

$$\mathsf E_i = p_i+\mathsf E_{i-1}$$

Since the probability of success for a shot is the ratio of prior successes to prior shots, therefore the expected probability of success for that shot is the ration of the expected count of prior successes to the count of prior shots:

$$p_i = {\mathsf E_{i-1}\over i-1}$$

So, since the first two shots are a success and a failure, we have, for $i\geq 3$:$$\begin{align}\mathsf E_i &={i\over i-1}\mathsf E_{i-1}\\ &= \prod_{k=3}^i{k\over k-1}\\&={i\over 2}\end{align}$$