If a pitcher has a $\frac{1}{2}$ probability of throwing a clean inning, what is their expected WHIP? 
If Jacob has a $\frac{1}{2}$ probability of throwing a clean inning in any given inning, what is his expected WHIP?

Some definitions and simplifying assumptions:

*

*OBP, or on-base percentage, is a hitter’s chance of reaching base rather than making an out in any given plate appearance.

*WHIP, or walks+hits per inning pitched, is (for the purposes of this problem) the number of hitters who reach base against a pitcher per inning.

*An inning ends upon the achieving of 3 outs.

*A clean inning is when three outs are achieved without any batters
reaching base.

*If Jacob starts an inning, he completes that inning.

*Every batter has the same OBP against Jacob.

*Every batter either reaches base or produces one out.

*The outcome of each plate appearance is independent from the outcome of any other plate appearance.

As to my work, I know that the expected OBP against Jacob can be calculated as follows: $(1-x)^3=\frac{1}{2}$, and then $x=\frac{2-\sqrt[3]{4}}{2}\approx 0.206$. But I don’t then know how best to meld that into the “innings” framework to compute the expected WHIP. However, I do also feel like I might be missing something obvious, so if I am, pointing that out to me would also be helpful.
 A: Let $E_i$ denote the expected number of additional hitters who will reach base given that there have been $i$ outs thus far. The desired quantity is $E_0$ since we want to know how many hitters will reach base over one inning (which starts with $0$ outs).
If $x$ is the (common) $\mathsf{OBP}$ of each batter, then the following equations hold:
$$\begin{align}
E_0 &= x (1 + E_0) + (1 - x) E_1 \\
E_1 &= x (1 + E_1) + (1 - x) E_2 \\
E_2 &= x (1 + E_2) + (1 - x) \cdot 0.
\end{align}
$$
The reason for the $x(1 + E_i)$ term is that with probability $x$ a hitter will reach base, and that adds $1$ to the count and puts the pitcher in the same situation of only having $i$ outs that they were in before that batter hit (hence $E_i$).
The reason for the $(1 - x)E_{i+1}$ term is that with probability $1 - x$ the count remains the same, and the pitcher is now in the situation that they have one more out to their advantage. Of course, the last equation has a $0$ there because if we already have two outs, and the batter is put out, then the pitcher’s inning is over, and there is nothing to add.
Working backwards, we find that $E_2 = x/(1 - x)$, $E_1 = 2 x/(1 - x)$, and $E_0 = 3x/(1 - x)$. Since you have already done the work to find that $x = 1 - 2^{-1/3}$, we have
$$E_0 \approx 0.779763.$$
A: The probability of a clean inning is $p^3$ where $p$ is the probability of getting an individual batter out. It's given that $p^3=\frac12$, so $p=\sqrt[3]{\frac12}$. The expected number of batters the pitcher has to face in order to get one out is $\frac1p$. The expected number of batters the pitcher has to face in an inning ($3$ outs) is $\frac3p$, so the expected number of batters to get on base in an inning is $\frac3p-3=3(\sqrt[3]2-1)$, which agrees with shoteyes' answer.
