# Two rifleman shooting at a target where the game ends when two hits are observed.

The problem is as following:

Person A and B are shooting at a target. Independently of who is shooting, the probability that the shot results in a hit is $p$, and each shot is independent from each other. They will shoot one by one, in the order $A,B,A,B,\dots$ until two hits are observed. Find the probability that it's the same person who got those hits.

My attempts:

One thing that I've noted is that the problem emphasizes independence, which is good if we would need to simplify things. My idea is like this:

$$P(A{\text{ hits twice| Two hits}}) + P(B{\text{ hits twice| Two hits}})$$

I've been trying several times, by arriving at the conclusion that it's $p^2(1-p)(2-p)$ by reasoning that the above leads to "A hit, B miss, A hit" or "A miss, B hit, A miss, B hit". My only concern is that the game could continue several rounds, i.e. "A miss, B miss, A miss, B miss, " so it feels like we should use the binomial/Poisson distribution somehow (or geometric, or an infinite sum, etc).

The answer should be as following:

$$\frac{{1 - p}}{{2 - p}}$$

Does anyone know how this problem can be tackled?

• There is no distinction between persons $A$ and $B$. One of the two will eventually hit first. Then, what is the probability that the next successful hit is an even number of hits after that? – Michael Apr 4 '16 at 20:05

The geometric solution to complement Henning's:

Following the first hit, add up the probabilities of an odd number of misses, and a hit:

$$P = \sum_{j=0}^{\infty} (1-p)^{2j+1}p = \frac{p(1-p)}{1-(1-p)^2} = \frac{1-p}{2-p}.$$

The number of initial misses is immaterial to the solution, but we do need to get one hit before we can get the second hit. The problem is a bit ill-defined without achieving that first hit, so it's expedient (and reasonable) to assume that one of the players got it.

But note that the probability of $n$ misses followed by a hit is

$$P'(n) = (1-p)^{n}p$$

and the sum over all $n$ is

$$P' = \sum_{j=0}^{\infty} p(1-p)^j = \frac{p}{1-(1-p)} = 1.$$

So if we hand-wave over the notion of "an infinite number of misses, followed by a hit" then we see that the result is exactly the same if we account for the initial misses. Each of the terms in $P$ (the top equation of my answer) is multiplied by the infinite number of terms shown in $P'$, but the terms of $P'$ add up to $1$.

• This makes perfect sense. I'm just a little bit confused how the sum takes care of the "first hit". To me, it sums the odd number of misses and one hit only. Have I missed something? – Artem Apr 4 '16 at 20:55
• You need to get the first hit before you get the second hit. How long it takes to get to that first hit is immaterial to the solution, because, once achieved, we only care that the same player make the second hit, and we can treat them the same because they're equally skilled. – John Apr 4 '16 at 21:08
• Edited my answer with some more clarification (or maybe I just made things more confusing!) haha – John Apr 4 '16 at 21:26

Without loss of generality suppose that $A$ scores the first hit. What you're interested in is then then probability that $A$ also scores the next hit after that.

And then we can simplify the question to, "in a game where the first to score any hit wins, what is the chance $q$ of the second player winning?"

We can try to compute this by summing a series (which turns out to be geometric), but it is quicker to observe that if the first player misses then the situation is symmetric, just with the player swapped, and the probability of the now-first player to win must be $1-q$. So $q$ has to satisfy $$q = p\cdot 0 + (1-p)\cdot (1-q)$$ which is easily solved.

Yet a different way of phrasing it is: Right after one hit, let $q$ denote the probability we are after, that the same person scores the second hit. Two cases will succeed:

• A miss followed by a hit (immediate succes)
• Two misses and we are back to the same situation with probability $q$ again

Thus $$q = (1-p)p+(1-p)^2q$$ which can then be solved for $q$.

For B to score both hits and win:

1. B has to score the first hit
2. Given that B scored the first hit, B has to score the next hit

Let the probability in #1 be $p_B$

For #2, it's the same as starting fresh and waiting for #1 to happen (A will shoot first, followed by B)

$P(\text{B scores the next hit | B scored the first hit}) = p_B$

$P(\text{B scores first two hits}) = p_B^2$

Now let's look at A winning with first two hits:

1. A scores the first hit.
2. Given that A scored the first hit, A scores the second hit.

Let #1 be $p_A$ . #2 is like B shooting first and A scoring the first hit (which is symmetric of #2 in B's case), so the probability is $p_B$

Probability A scores both hits and wins = $p_Ap_B$

P(The winner scores both hits) = P(A scores both hits) + P(B scores both hits)

$= p_Ap_B + p_B^2 = p_B (p_B + p_A)$

But $p_A + p_B = 1$ so the probability is simply $p_B$

So now the problem comes down to just solving for the probability of B making the first hit. So B can score the first hit on the 2nd, 4th, 6th... any even numbered shot

$p_B = (1-p)p + (1-p)^3p + (1-p)^5p + ...$

$p_B = (1-p)p * ( 1 + (1-p)^2 + (1-p)^4 + ..)$

$p_B = (1-p)*p*\frac{1}{1 - (1-p)^2}$

Which comes out to $\frac{1-p}{2-p}$