Shooting competition. What is the best option?

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Hi everybody.
I am fighting with the following problem:

In a shooting competition, a team is composed by 2 shooters, who must shoot alternately.
They have probabilities p1 and p2 to hit the target, being p1>p2.
In order to be classified, a team must shoot 3 times and hit the target at least 2 consecutive times.

a) Which one of the two shooters must shoot two times?
b) Generalize the question in the case that they must get at least
two consecutive hits with a total of 2k+1 shoots.

These are my answers:

a)
To me the answer is clear.
The probability for a team of being classified is greater if the worst shooter (that who has probability p2 of hitting the target) shoots in the first and the last turns.

b)
Intuitively, I think the answer is the same. It is better that the worst shooter shoot in the first turn.
But how can I prove my guess.
I´d be grateful for any hint.

Thanks.
 

Answers and Replies

  • #2
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My approach for the case a) is like this:

If we consider a success (S) to hit the target and a failure (F) not to hit it, then our sample space is composed for all possible series of 3 letters with possible values S and F:

[tex]
\Omega=\left\{SSS,SSF,SFS,FSS,SFF,FSF,FFS,FFF\right\}
[/tex]

Let [tex]A_{i}[/tex] be the event of having a S both at positions i and i+1. Then the event of a team being classified is [tex]C=\cup^{3}_{i=1} A_{i}[/tex].

If p is the probability of success at odd positions and q the probability of success at even positions, then

[tex]p(A_{i})=pq[/tex]
[tex]p(A_{i}\cap A_{j})=p(SSS)=p^{2}q[/tex], i[tex]\neq[/tex]j

and therefore

(1)
[tex]pC=p(\cup^{3}_{i=1} A_{i})=2pq-p^{2}q[/tex]

If the first shooter is the one with probability [tex]p_{1}[/tex] , this implies [tex]p= p_{1}[/tex] and [tex]q= p_{2}[/tex]. Let's call this option A.
The other option (the worst shooter is the first) is [tex]p= p_{2}[/tex] and [tex]q= p_{1}[/tex]. Let's call this option B.

In (1), the first term is symmetric in p and q and has the same value for both options.
The second term is not and thus

[tex]pC_{A}=2p_{1}p_{2}-p^{2}_{1}p_{2}[/tex]

[tex]pC_{B}=2p_{1}p_{2}-p^{2}_{2}p_{1}[/tex]

so

[tex]pC_{A}<pC_{B}[/tex]

that is, B is the best option to be classified.

I’ve tried to generalize this for the case of 2k+1 shootings, but it becomes a messy.

Can anyone help me ?
 
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  • #3
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One way is using matrix powers and some ideas from Markov theory. At any point in time it only matters whether the team has already been classified or else the success/failure of their previous shot, so denote the states {C,S,F} and say they begin in state F. The state transition matrix can be written as

[tex]P_k=\begin{bmatrix} 1 & p_k & 0 \\ 0 & 0 & p_k \\ 0 & 1-p_k & 1-p_k \end{bmatrix}[/tex]

for each shooter, where the (ij)-th entry is the probability of moving from state j to state i. If shooter 1 goes first then the probability of being classified after (2k+1) rounds is

[tex]q_1 = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} (P_1P_2)^k P_1 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}[/tex]

and similarly for [itex]q_2[/itex]. The matrix powers [itex](P_1P_2)^k[/itex] and [itex](P_2P_1)^k[/itex] can be evaluated explicitly by eigenvalue methods such as diagonalization or the Perron formula.

Not sure if there's a simpler / more elegant way to do it!
 
  • #4
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Thanks a lot, bpet.
This is a very interesting way to solve the problem.
Markov theory gives a completely new point of view to the classical problems of probability.

But without using Markov ideas, could anyone think of a simpler way to solve it?
 
  • #5
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To solve it via an explicit expression for [itex]q_1(k)[/itex] would likely still involve solving 3 simultaneous recurrences, which is essentially equivalent.

With the aid of a symbolic calculator we get

[tex]q_1-q_2 = 2^{-k}p_1p_2(p_1-p_2)(b^k-c^k)/a[/tex]

where [itex]a=\sqrt{(1+p_1p_2(-6+4p_1+4p_2-3p_1p_2))}[/itex], [itex]b=1-p_1p_2-a[/itex], [itex]c=1-p_1p_2+a[/itex]. Since a, b, c are all non-negative and symmetric in [itex]p_1,p_2[/itex] with b<c then it follows that [itex]q_1\le q_2[/itex].

It would be good if there's a simpler way. Perhaps by considering the recurrence relations its possible by induction?
 
  • #6
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Thanks again bpet.
I'll work with your two answers.
 
  • #7
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Hi everybody.
This post is only to tell you that I have found an altenative way to solve this problem.
Using only a classical probabilistic reasoning it is possible to obtain a recurrent expression for the probability of a team of being classified in n rounds.
From it, it is then possible to prove that for odd values of n, it is better that the worst player start the game. (And also, that for even values, it doesn't matter).
That is what I call the beauty of maths !!!
Thanks for your help.
 

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