# Expected number of targets hit by 10 shooters, 2 bullets each, 20 targets

Consider a competition with 10 clay shooters. Each shooter has equal ability and they all use identical shotguns. Each shotgun is loaded with two bullets, therefore each shooter can shoot twice. One throws 20 clay targets simultaneously, each shooter picks a target at random and shoots at it once, all the shooters shooting at the same time. Then each shooter picks one of the remaining targets at random, again independently of the other shooters, and shoots at it. This ends the shooting.

What is the expected number of targets to get hit?

I don't know how to answer question like this because the question doesn't give what is the probability of a shooter can hit the target.

Let's say the first shooter hit a target, then the second shooter hit $\frac{19}{20}$ of a target because I think he has a $\frac{1}{20}$ chance of shooting at the same target. The third hunter, well, if there's one target is hit so far, that's a $\frac{1}{20}$ chance, and then the third shooter hit $\frac{19}{20}$ of a target. If there are two targets get hit, then the third shooter hit $\frac{18}{20}$ of a target, but this looks pretty messy and I'm not even sure this is the correct approach. Is there any way to solve it?

• I think you are expected to assume perfect accuracy.The thing that makes this problem less routine is the fact that presumably each shooter uses her two shells on different targets. Aug 16, 2014 at 12:08
• @AndréNicolas My guess is the expected number of targets = the expected number of 1st shoot + the expected number of 2nd shoot since the event is independent. How if we find the expected number of 1st shoot first and use the same approach to find the expected number of 2nd shoot? Is this possible? Aug 16, 2014 at 12:14
• @AndréNicolas Do you think the below answer is correct? Anyway, I'm waiting for your answer to make me more sure :) Aug 16, 2014 at 12:29
• It seems that I misinterpreted the question. I thought of the two shots as fired simultaneously. Aug 16, 2014 at 17:17
• @AndréNicolas Sorry for my bad English. Did has edited the question for me. So, do you have a clue (an answer maybe) to solve this question? :) Aug 16, 2014 at 17:38

Assume that there are $t$ targets and $s$ shooters. The first target is aimed at by the first shooter for her first shot with probability $1/t$ and hit with probability $p$ hence the first target is not hit by the first shot of the first shooter with probability $$q(t)=1-p/t.$$ By independence, the first target is not hit by the first shot of any shooter with probability $q(t)^s$. Thus, the number $T$ of targets not hit by any first shot is such that $$E(T)=t\cdot q(t)^s.$$ Assume now that $t\gt s$, hence $T\geqslant1$ almost surely. Applying once more the reasoning above, one sees that the mean number of targets not hit after two rounds of shots is $r=E(T\cdot q(T)^s)$, that is, $$r=E\left(T\cdot\left(1-\frac{p}T\right)^s\right).$$ Unfortunately, $r$ seems to depend on the full distribution of $T$, not only on its mean $E(T)$, and the hypothesis that $t=2s$ does not seem to help.

Exact computations when $t=4$ and $s=2$ yield $$r=4-4p+\tfrac12p^2+\tfrac16p^3+\tfrac1{24}p^4,$$ a formula which does not point to the possibility of huge simplification when $t=2s$ for some general $s$.

• Let's analyse the first shoot only. In my opinion, using your notation, clearly this is binomial process and the probability that no one will shoot at that particular target should be $(1-1/t)^s$, therefore the expected number of target with no one shooting at them is $t(1-1/t)^s$. How come you get $p/t$? Aug 17, 2014 at 4:20
• @Venus Sorry but if you are not reading what I write, communication (and progress) will be difficult... First, we KNOW that the number of targets hit at the first shot is NOT binomial, a reason is explained in a comment to the other answer. Second we KNOW that the expected number of targets not hit at the first shot is $tq(t)^s$, this is what my answer explains. I nowhere said that the expected number of targets not hit was $p/t$ (and I nowhere studied the mean number of targets not aimed at, only the mean number of targets not hit).
– Did
Aug 17, 2014 at 8:40
• The reason why I think this is binomial is from the point of view of a particular target the probability that someone will shoot at the target is $1/t$. We have $s$ independent trials, so this is binomial probability. Besides, you wrote in your answer $tq^s$, it is clearly the expected value of binomial probability. I only disagree about the $q(t)$ because I think it should be $(1-1/t)^s$. CMIIW Aug 17, 2014 at 11:03
• No, there are $s$ independent trials on the same target hence "the point of view of a particular target" is to be hit or not hit, which is Bernoulli, not binomial. What I wrote in my answer is $tq(t)^s$ (not $tq(t)^2$...), which happens to be the expectation of an unrelated binomial distribution, yes, as well as the expectation of an unrelated Poisson distribution as well as the expectation of a host of other unrelated distributions. That E(T)=E(Binomial) does not imply that T is binomial.
– Did
Aug 17, 2014 at 11:09
• Isn't the expected number shooting at any target $p/t$ and the probability that no one will shoot at that particular target $(1-1/t)^s$? Aug 17, 2014 at 11:26

This problem is much more nuanced than I'd originally anticipated. The problem is that neither shot can be viewed as a set of independent trials, and thus not as a binomial process.

Letting $p\left(T_{i} = m\right)$ be the probability that the $i$th target is selected by $m$ shooters, and $p\left( H|m\right)$ be the probability that a target is hit if selected by $m$ shooters, then we can construct the tree $$\sum\limits_{{m_1}} {p\left( {{T_1} = {m_1}} \right)\left[ {\left. {\begin{smallmatrix} {p\left( {H|{m_1}} \right)}&{{n_1} = 1} \\ {1 - p\left( {H|{m_1}} \right)}&{{n_1} = 0} \end{smallmatrix}} \right|\,\,\,\sum\limits_{{m_2}} {p\left( {{T_2} = {m_2}|T_1 = m_1} \right)} \left[ {\left. {\begin{smallmatrix} {p\left( {H|{m_2}} \right)}&{{n_2} = 1} \\ {1 - p\left( {H|{m_2}} \right)}&{{n_2} = 0} \end{smallmatrix}} \right|\,\, \ldots } \right]} \right]}$$ by tracing back through the tree over all sums, which yields the probability of each tuple $\left( {{n_1},{n_2},{n_3}, \ldots ,{n_{20}}} \right)$ with $n_i = 0,1$ representing the number of hits on target $i$.

The problem is that $n_1,n_2,\ldots$ aren't IID and hence their sum isn't binomially distributed. I don't know what the conditional distribution looks like.

Let $p$ be the probability of shot success.

The expected value as a function of $p$ has the following form for some low numbers of shooters, with ${\rm number\ of\ targets = 2 \cdot number\ of\ shooters}$: $$\begin{array}{*{20}{l}} {}&{\rm after\ first\ shot}&{\rm after\ second\ shot} \\ {{n_{{\text{tgt}}}} = 2,\,\,{n_{{\text{s}}}} = 1}&p&{2p} \\ {{n_{{\text{tgt}}}} = 4,\,\,{n_{{\text{s}}}} = 2}&{2p - \tfrac{1}{4}{p^2}}&{4p - \tfrac{1}{2}{p^2} - \tfrac{1}{{2 \cdot 3}}{p^3} - \tfrac{1}{{{2^3} \cdot 3}}{p^4}} \\ {{n_{{\text{tgt}}}} = 6,\,\,{n_{{\text{s}}}} = 3}&{3p - \tfrac{1}{2}{p^2} + \tfrac{1}{{{2^2} \cdot {3^2}}}{p^3}}&\begin{gathered} 6p - {p^2} - \tfrac{{11}}{{{3^2} \cdot 5}}{p^3} - \tfrac{{23}}{{{2^3} \cdot 3 \cdot {5^2}}}{p^4} + \\ \,\,\,\,\,\,\,\,\,\tfrac{{71}}{{{2^5} \cdot {3^2} \cdot {5^2}}}{p^5} + \tfrac{{17 \cdot 37}}{{{2^6} \cdot {3^4} \cdot {5^2}}}{p^6} \\ \end{gathered} \\ {{n_{{\text{tgt}}}} = 8,\,\,{n_{{\text{s}}}} = 4}&{4p - \tfrac{3}{4}{p^2} + \tfrac{1}{{{2^4}}}{p^3} - \tfrac{1}{{{2^9}}}{p^4}}&\begin{gathered} 8p - \tfrac{3}{2}{p^2} - \tfrac{{17}}{{{2^3} \cdot 7}}{p^3} - \tfrac{{433}}{{{2^8} \cdot {7^2}}}{p^4} + \tfrac{{43 \cdot 307}}{{{2^7} \cdot 3 \cdot 5 \cdot {7^3}}}{p^5} + \\ \,\,\,\,\,\,\,\,\,\tfrac{{66533}}{{{2^7} \cdot {3^2} \cdot {5^2} \cdot {7^3}}}{p^6} + \tfrac{{13 \cdot 61 \cdot 113}}{{{2^{10}} \cdot {3^2} \cdot {5^3} \cdot {7^3}}}{p^7} - \tfrac{{11 \cdot 109 \cdot 5059}}{{{2^{14}} \cdot {3^3} \cdot {5^3} \cdot {7^3}}}{p^8} \\ \end{gathered} \end{array}$$ $$\begin{array}{*{20}{c}} {{n_{{\text{tgt}}}} = 10,\,\,{n_{{\text{s}}}} = 5}&{5p - {p^2} + \tfrac{1}{{2 \cdot 5}}{p^3} - \tfrac{1}{{{2^3} \cdot {5^2}}}{p^4} + \tfrac{1}{{{2^4} \cdot {5^4}}}{p^5}} & {\hspace{140pt}} \end{array}$$

Hopefully somebody else will take a crack at it.

• I do agree with your approach but why didn't you separate the event (1st & 2nd shoot)? In my opinion, the 1st shoot we have n=10. Aug 16, 2014 at 12:27
• I used $n = 20$ because I assumed all 20 targets were shot at "at the same time". I've amended the solution. Note that the binomial process for $n_1$ in the first-shot-then-conditional-second-shot case would be $p\left( n_1\right) = B\left( n_1, \mu = \frac{p}{10}, n = 10\right)$.
– COTO
Aug 16, 2014 at 12:31
• Your edited version seems correct. +1. Could you make a complete answer? At least, you give me number so then I can match it using your approach. Aug 16, 2014 at 12:32
• "The first target has a 2/20 probability of being selected by the first shooter in either shot" Why? By the same reasoning, if there were only 2 targets, the first target would be selected by the first shooter in either shot with full probability?
– Did
Aug 16, 2014 at 13:35
• @Did Read on; I handle the conditional case later (below the line). In the original case, since the language was used "first and second shot are both at the same time", I assumed the selection of the two was independent. Think of it as each shooter only being able to take one shot, but with a two-barreled gun that can be simultaneously aimed at two targets.)
– COTO
Aug 16, 2014 at 13:39