# Expectation/Statistics/Binomial Distribution Question

So I'm studying for an exam and stumbled upon this problem:

Assume that the number of games, $X$ that a baseball relief pitcher will save in $n$ games follows a binomial distribution, i.e. $X \sim \text{Binomial}(n,p)$. In a five-game play-off series between equally matched teams, the number of games played is a r.v., $N$, with pmf:

\begin{array}{ccl} p_N(n) = \binom{n-1}{2} ( \frac{1}{2} ) ^{n-1} &,& n=3,4,5 \\ 0 &,& \text{otherwise} \end{array}

Find the expected number of saves this relief pitcher will make in this series.

I am not so sure if I did it correctly. So what I did was create a distribution table: \begin{array}{c|c|c|c} n & 3 & 4 & 5 \\ \hline p_N(n) & 1/4 & 3/8 & 3/8 \\ \end{array} And then calculate the expectation from there: \begin{array}{c} E(N) = (3)(1/4) + (4)(3/8) + (5)(3/8) = 4.125 \end{array} So I am just not sure if I did the question correctly. But that kinda seems weird to me seeing that it was stated in the given that it follows a binomial distribution and yet have not been used.

I'm still struggling with this question. Any help would be appreciated. Thanks so much!

• You need to find $p_X(s)$ which is the pmf of $X$. Then you would need to find $p_N(p_X(s))$ which would be the pmf of all the games played. After that you can find $E$ as simply the first derivative of the found pmf at $1$ Jun 19, 2018 at 15:49
• Your solution is correct. At the moment I don´t see a way to write the pmf as a binomial distribution. Jun 19, 2018 at 16:24

What you calculated, $\operatorname{E}[N]$, is the expected number of games played in a 5-game playoff series. But what you want to know is the expected number of saved games by the relief pitcher. Conditioned on the number of games played, the number of games saved follows a binomial distribution. What this means is that, for example, if only $N = 3$ games are played, then the number of games saved by the pitcher follows a binomial distribution with parameters $N = 3$ and $p$, where $p$ is the chance of the pitcher saving any single game in the series. Since your question does not appear to furnish this probability, the desired expected value will be a function of $p$.
More formally, $X \mid N \sim \operatorname{Binomial}(N,p)$, where $$\Pr[N = n] = \begin{cases}\binom{n-1}{2} 2^{-(n-1)}, & n \in \{3,4,5\} \\ 0, & \text{otherwise.} \end{cases}$$ Then, by the law of total expectation (also called the law of iterated expectation), $$\operatorname{E}[X] = \operatorname{E}[\operatorname{E}[X \mid N]] = \operatorname{E}[Np] = p \operatorname{E}[N] = \frac{33}{8}p.$$
Now, let's process this solution a bit. If the relief pitcher saves every game he plays, i.e. $p = 1$, then the expected number of games saved equals the expected number of games played, since then "games played" = "games saved." And this is reflected in the above calculation when we set $p = 1$. At the other extreme, when the pitcher never succeeds, $p = 0$, then the expected number of games saved is $0$ no matter how many games are played, and this again is reflected in the above formula.
Another aspect to this question is the unconditional (or marginal) distribution of $X$, that is to say, what is the probability mass function of $X$ itself, when we do not condition on $N$? This is a separate computation, and as we have seen, it is not required to answer the original question. We would compute $$\Pr[X = x] = \sum_{n=3}^5 \Pr[X = x \mid N = n]\Pr[N = n].$$ This is simply the law of total probability. The expression $\Pr[X = x \mid N = n]$ corresponds to the binomial PMF, since it is conditional on $N$. So we could write the above as $$\Pr[X = x] = \sum_{n=3}^5 \binom{n}{x} p^x (1-p)^{n-x} \binom{n-1}{2} \frac{1}{2^{n-1}},$$ and since the sum has only three terms, we could just write it out explicitly, but you will see that this does not simplify into any particularly simple or familiar distribution.