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A and B are involved in a duel. The rules of the duel are that they are allowed to pick up their guns simultaneously. If one or both are hit, the duel ends. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with prob $p_A$ and each shot B will hit A with prob $p_B$.

Compute: a) The probability that A is not hit. This is just $1-p_B$ for each trial. So if we assume there are $n$ duels then B has to be hit. Summing over the $n$ duels gives the probability as $$ \sum_{i=1}^{n} (1-p_B)^{n-i} p_A$$

b) The probability that both duelists are hit. I said this would be $p_A p_B$ for any one duel

c) the prob that the duel ends after the nth round of shots. It can end in 3 ways: A hits B, B hits A or they both hit each other. So I believe we want $(1-p_A)^{n-1}p_A + (1-p_B)^{n-1}p_B + (1-p_Ap_B)^{n-1}p_Ap_B$

d) The cond. prob that the duel ends after the nth round of shots given that A is not hit. So using the def of cond. prob I get $$(1-p_B)^{n-1}p_A/\sum_{i=1}^{n} (1-p_B)^{n-i} p_A$$, where P(duel ends after nth round and A not hit) is numerator and P(A not hit) as in a)

e) the cond, prob that the duel ends after the nth round given that both duelists are hit. I used Bayes to write this as P(A and B both hit| duel ends on nth round)P(duel ends on nth round)/P(A and B both hit) I said the first term in the numerator was 1/3. If the duel ends on the nth trial, then either A hit B, B hit A or they both hit each other => 1/3. Then I just substituted what we had already in the previous parts.

I don't have the answers to these questions. Can someone make any comments about them? Many thanks

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2 Answers 2

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(a) Let $S$ be the event "$A$ is not hit" (ever), and let $x$ be the probability of $S$. The event $S$ can happen in two disjoint ways: (i) The game ends in one round, and $A$ is not hit or (ii) the game lasts more than $1$ round, and $A$ is not hit.

The probability of (i) is $p_A(1-p_B)$.

The game lasts more than $1$ round, if $A$ and $B$ both miss on their first shots. This has probability $(1-p_A)(1-p_B)$. Given that the game lasts more than one round, the probability $A$ is not hit is $x$, since essentially the game starts again. So the probability of (ii) is $(1-p_A)(1-p_B)x$. It follows that $$x=p_A(1-p_B)+(1-p_A)(1-p_B)(x).$$ Solve this linear equation for $x$.

(b) Let $H$ be the event both duellists are hit. Again, as in the solution of (a), "both hit" could happen (i) in the first round or (ii) later. The probability of (i) is $p_Ap_B$. By reasoning similar to the reasoning of (a), the probability of (ii) is $(1-p_A)(1-p_B)(y)$. Thus to find $y$ we solve the linear equation $$y=p_Ap_B+(1-p_A)(1-p_B)y.$$

(c) I interpret the question as meaning that the game ends exactly at the $n$-th round. For this to happen, there must be $n-1$ misses by both, and then a hit.

The probability of a hit on any round is $1$ minus the probability of a double miss. So our required probability is $$(1-p_A)^{n-1}(1-p_B)^{n-1}\left[1-(1-p_A)(1-p_B)\right.$$

(d) This has the standard conditional probability structure. Let $S$ be the event $A$ is not hit, and let $E_n$ be the event the game ends in the $n$-th round. We want $\Pr(E_n|S)$. We can compute this mechanically, using $$\Pr(E_n|S)=\frac{\Pr(E_n\cap S)}{\Pr(S)}.$$ We computed the denominator earlier. The numerator is easy, $A$ must miss $n-1$ times and then hit, while $B$ must miss $n$ times. After computing mechanically, you might want to search for a more conceptual argument.

(e) This can be analyzed in a way very close to the analysis in (d).

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I had an exam question in college very similar to this. It was: Two players, $A$ and $B$ flip individual coins with probabilities $p_A$ and $p_B$ of getting heads. The players flip one at a time and a winner is determined as the first player to flip heads on their respective coin. Given that player $A$ goes first, what is the probability that he/she wins the game. I solved it using an infinite sum but didn't come up with the shortcut you provided above and only found out about it after talking to a classmate. Hopefully I don't forget this technique again as it is very useful. –  Patrick Dec 2 '12 at 20:47
For c), I also have $$P = (1-p_A)^{n-1}(1-p_B)^{n-1}p_A + (1-p_B)^{n-1}(1-p_B)^{n-1}p_B,$$ where the first term is if A and B miss and then A hits and the second term, the same but instead B hits. Is there anything wrong with this answer? Would I perhaps need to add in the probability that they both hit each other? –  CAF Dec 2 '12 at 22:14
Also, what was wrong with my argument about saying that P(A and B hit| game ends on nth round) = 1/3? Thanks! –  CAF Dec 2 '12 at 22:17
The idea is good. I was hinting (for (d)) at thinking about it that way after making formal calculation. We do ned to take acount of the fact that the three alternatives you list are in general not equally likely. –  André Nicolas Dec 2 '12 at 22:28
oh, I see. So the 1/3 may not be valid because this assumes that each of the ways the game could end are equally likely. What about my answer to c) that I had 3 comments above? –  CAF Dec 2 '12 at 22:39

Let's start with part a) You are off to a good start but your summation is not quite right. It should be an inifinite sum equaling: $$ \mathbb{Pr}(A \ wins \ duel) = \sum_{n=1}^{\infty}[(1-p_A)(1-p_B)]^{n-1}(p_A)(1-p_B)$$ The form of this series should look familiar to you. The summands are as above because we are adding up the probability that $B$ is shot but $A$ remains unscathed after $n = 1,2,3,..$ draws of their firearms. In the round where $B$ is shot, $A$ must still be alive and the probability of this in any one draw is $(p_A)(1-p_B)$. See if you can reason out the other part of the summand (the one raised to $n-1$) and then calculate the value using the well known closed form solution.

Part b) will require only a slight modification to the sum above.

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