# A basic intuition on a probability problem

Two players take turns shooting at a target, with each shot by player $i$ hitting the target with probability $p_i$, $i=1,2$. Shooting ends when two consecutive shots hit the target. Let $\mu_i$ denote the mean number of shots taken when player $i$ shoots first, $i=1,2$.

Now, I have calculated that $$\mu_1 - \mu_2 = \frac{p_2 - p_1}{p_1+p_2-p_1p_2-2}=\frac{q_2-q_1}{1+q_1q_2}$$

where $q_i=1-p_i, i=1,2$

i.e. if the player with higher winning probability starts the game then the mean number of shots taken is higher than the case when the the player with lower winning probability starts the game. I don't understand the intuition behind this.

• I also believe OP's formula is off, but I find something opposite in direction from Christian, still in OP's direction. – alex.jordan May 30 '13 at 23:05
• Actually I think my own answer and possibly others are overcomplicating this. If the better player starts, then the better player's first shot does not have opportunity to end the game. So the better player serves to end the game sooner if they come second. – alex.jordan May 30 '13 at 23:27
• @Christian Blatter is your formula correct. If $p_1=1$ then from your formula $\mu_1-\mu_2=p_2-1$ whereas from the discussion of alex.jordan it is coming as $1-p_2$ – RIchard Williams May 31 '13 at 1:28
• @prasenjit: I had missed the phrase "two consecutive shots". – Christian Blatter May 31 '13 at 8:21

## 4 Answers

Since the correctness of the OP's formula for $\mu_1-\mu_2$ has been questioned here is a roundup of the argument that leads to it:

Consider $\mu_i$ as the expected number of remaining shots when it is $P_i$'s turn and there is no immediately foregoing hit, and similarly let $\lambda_i$ be the expected number of remaining shots when it is $P_i$'s turn and there is an immediately foregoing hit. Then we have the following equations: \eqalign{\mu_1&=1+p_1\lambda_2+(1-p_1) \mu_2 \cr \mu_2&=1+p_2\lambda_1+(1-p_2) \mu_1 \cr \lambda_1&=1+p_1\cdot 0+(1-p_1)\mu_2\cr \lambda_2&=1+p_2\cdot 0+(1-p_2)\mu_1\ .\cr} Solving this system for the $\lambda_i$, $\>\mu_i$ leads to $$\mu_1-\mu_2={p_1-p_2\over 1+q_1q_2}\ ,$$ where $q_i=1-p_i$.

The fact that $p_1>p_2$ implies $\mu_1>\mu_2$ can intuitively be explained as follows: Since a hit by the weaker player is necessary to end the game, letting him begin gives him an extra chance to shoot and will, on the average, shorten the game.

For the intuition, take an extreme case, where the better player is almost sure to hit, and the other player is quite a bit less accurate. If the bad player goes first, there are two chances for an end by the time $3$ shots have been taken. If the good player goes first, an end by the time $3$ shots have been taken is less likely.

• +1 We want to give the weaker shooter more opportunities. – leonbloy May 30 '13 at 17:30

The argument below should also give you the intuition as to why if $p_1 > p_2$, we have $\mu_1 > \mu_2$. What the argument says is that the onus is on the shooter who shoots given that the other shooter has hit the target. We hence want this person to have a higher probability of actually hitting the target.

If player $1$ starts first, then the expected number of shots fired is given by \begin{align} \mu_1 & = \underbrace{2p_1p_2}_{2 \text{ shots if both shoot on their first attempt}} \\ & + \underbrace{p_1(1-p_2)(2+\mu_1)}_{\text{If $1$ hits and $2$ fails, need to restart with $1$ and $2$ shots have gone}}\\ & + \underbrace{(1-p_1)(1+\mu_2)}_{\text{If $1$ doesn't hit, then restart with $2$ and $1$ shot is already gone}} \end{align} Similarly, if player $2$ starts first, we then have $$\mu_2 = 2p_2p_1 + p_2(1-p_1)(2+\mu_2) + (1-p_2)(1+\mu_1)$$ Subtract the two equations, do some nice algebra and you will be able to isolate the $\mu_1 - \mu_2$ term.

Another way would be to solve for $\mu_1$ and $\mu_2$ explicitly and compute $\mu_1 - \mu_2$.

If the better player starts, then the better player's first shot does not have opportunity to end the game. So the better player serves to end the game sooner if they come second. That is, if the better player plays first, the game is expected to last longer.

I'm not sure that your formula is right though. Consider the special case where $p_1=1$, so that it all rests on player 2, with probability of success $p=p_2$ and failure $q=1-p$. We compute \begin{align}\mu_1&=2p+4qp+6q^2p+\cdots\\&=2p\sum_{n=1}^\infty nq^{n-1}\\&=\frac{2p}{(1-q)^2}=\frac{2}{p_2}\end{align} And \begin{align}\mu_2&=2p+3qp+5q^2p+\cdots\\&=p+p\sum_{n=1}^\infty (2n-1)q^{n-1}\\&=p+p\sum_{n=1}^\infty n\sqrt{q}^{n-1}-p(2\sqrt{q})\sum_{n=1}^\infty nq^{n-1}\\&=p+p\frac{1}{(1-\sqrt{q})^2}-p(2\sqrt{q})\frac{1}{(1-q)^2}\\ &=p+p\frac{(1+\sqrt{q})^2}{(1-q)^2}-\frac{2\sqrt{q}}{p}\\ &=p+\frac{(1+\sqrt{q})^2}{p}-\frac{2\sqrt{q}}{p}\\ &=p+\frac{1+q}{p}=p_2+\frac{2}{p_2}-1 \end{align}

So $\mu_1-\mu_2=1-p_2$. This is not consistent with your formula when $p_1=1$.