Expected number of darts thrown in a game

Question

$A$ and $B$ play a game where they take turns throwing darts at a dart board. The winner is the person to hit first. $A$ hits with probability $a$ and $B$ with probability $b$.

I showed (by computing a sum) that the expected number of darts thrown in one game is $$\frac{\alpha}{a}+\frac{\beta}{b}$$ where $\alpha$ and $\beta$ are the probability $A$ and $B$ win resp.

This made me think there should be a solution using the law of total probability (since $\alpha+\beta=1$). That would suggest that the expected number of darts thrown given $A$ wins is $\frac{1}{a}$. This is nonsense, so I dismissed this as a coincidence.

However, I ran a simulation for the analogous game for three players $A$ $B$ and $C$ and found that the expected number of darts thrown in the game is

$$\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}$$

Can someone explain whats going on here? I'm expecting someone will be able to arrive at this answer (and presumably the result for $n$ players) in a much more satisfying way than direct computation.

EDIT: The result also holds for four players computationally

Answer 1

Here's an explanation for where this comes from.

Suppose that after the game ends, the losing player(s) continue to throw until they also hit the target. Now we have three random variables $X+Y=Z$ where $X$ is the number of throws in the original game, $Y$ is the number of extra throws, and $Z$ is the total number of throws.

First assume two players. Then $\mathbb E(Z)=\frac 1a+\frac 1b$, since the total number of throws is just the sum of two independent geometric variables. Also we have $\mathbb E(Y)=\frac\alpha b+\frac\beta a$, since in fact $\mathbb E(Y\mid\text{A wins})=\frac 1b$, etc. (Once A has won, the number of extra throws B needs is just a geometric random variable.) Now $Y$ and $Z$ are certainly not independent, but still linearity of expectation gives $$\mathbb E(X)=\mathbb E(Z)-\mathbb E(Y)=\frac{1-\beta}a+\frac{1-\alpha}b=\frac \alpha a+\frac \beta b.$$ Similarly for three players you have $\mathbb E(Z)=\frac 1a+\frac 1b+\frac 1c$ and $$\mathbb E(Y)=\alpha\left(\frac 1b+\frac 1c\right)+\beta\left(\frac 1a+\frac 1c\right)+\gamma\left(\frac 1a+\frac 1b\right)=\frac{1-\alpha}a+\frac{1-\beta}b+\frac{1-\gamma}c,$$ giving the same result. Doing the same for more than three is just a matter of notation.

Answer 2

Let's generalise to $n$ players with player $i$ having probability $p_i$ of hitting (your $a,b,c$) and probability $q_i$ of winning overall (your $\alpha,\beta, \gamma$).

We will also let the probability of no hits after $i$ attempts be $r_i =\prod\limits_{j=1}^i (1-p_i)$ with $r_0=1$. For $i>n$ you have $r_i=r_{i-n}r_n$.

Then $q_i = p_i r_{i-1} + q_i r_n = p_i\dfrac{r_{i-1}\,}{1-r_n}$.

The expected number of darts thrown in the game is $\mathbb E[D] = 1p_1r_0+2p_2r_1+\cdots+np_n r_{n-1}+r_n (\mathbb E[D]+n)$ $= r_0+r_1+\cdots+r_{n-1}+r_n \mathbb E[D]$ $=\dfrac{\sum\limits_{i=1}^n r_{i-1}}{1-r_n} $ $= \sum\limits_{i=1}^n \dfrac{q_i}{p_i}$, as in your hypothesis.

Answer 3

With the following approach the question is trivial. Let us consider three players with probabilities $p_1$, $p_2$, $p_3$ of hitting and $q_1$, $q_2$, $q_3$ of not hitting. And suppose they throw consecutively, for $N$ rounds, $N$ times each. Then: \begin{align*} & Np_1 &&\text{average number of times the first player will hit and win.}\\ & Nq_1p_2 &&\text{average number of times the second player will win.}\\ & Nq_1q_2p_3 &&\text{average number of times the third player will win.}\\ & N(1+q_1+q_1q_2) &&\text{average number of throws to count.}\\ & N(1-q_1q_2q_3)&&\text{average number of games played.} \end{align*}