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Sir Lancelot and Sir Galahad are doing a shoot out, in which they try to shoot each other while shooting in the same time at each other. The probability of Sir Lancelot to hit Sir Galahad is 0.5 and the probability of Sir Galahad to hit Sir Lancelot is 0.25. All the shots are independent.

A. What is the probability that the shoot out will end in the n's round ?
B. If it is known that after k rounds the shoot out did not end, what is the chance that is will end within two rounds ?
C. What is the chance of Sir Lancelot to win ?
D. What is the chance of Sir Galahad to win ?

I solved A and B correctly. The answer for A is $[(3/8)^n-1]\cdot 5/8$ and for B is $(3/8)\cdot (5/8)$. In B I used the memory loss characteristic of the Geometric random variable.

I find it hard to solve C and D. Will appreciate your help. Thank you very much in advance.

user162354
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1 Answers1

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They both sound like losers to me.

By L wins, I assume what is meant is that L survives and G does not. Let $p$ be the probability L wins.

We condition on what happens in the first round. L wins if either (i) he hits G, and G misses or if (ii) they both miss, but ultimately L wins. We therefore obtain $$p=(0.5)(0.75)+(0.5)(0.75)p,$$ since in effect if they both miss the party begins all over again. Solve this linear equation for $p$.

The Galahad wins probability is calculated in the same way.

Another way: L wins in the following pairwise disjoint ways:

(i) L makes his first shot, and G misses, probability $(0.5)(0.75)$.

(ii) They both miss their first shots, but L nails the second shot, and G misses, probability $(0.5)(0.75)(0.5)(0.75)$.

(iii) They both miss their first two shots, but then L hits on the third, and G misses, probability $[(0.5)(0.75)]^2(0.5)(0.75)$.

(iv) And so on.

So the probability L wins is $$(0.5)(0.75)\left[1+r+r^2+r^3+\cdots \right],$$ where $r=(0.5)(0.75)$. If we sum the infinite geometric series, we get $\dfrac{(0.5)(0.75)}{1-(0.5)(0.75)}$.

André Nicolas
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  • Can you actually do that ? I thought that the catch is to somehow find the probability of Lancelot winning a round before Galahad does. – user162354 Jul 31 '14 at 20:09
  • You can do it the way you suggest also. It involves summing an infinite geometric series. But the way I did it is easier and introduces a useful idea. I can add the geometric series way, if no one does in the next half hour or so. – André Nicolas Jul 31 '14 at 20:12
  • That would be great, thanks. Don't get me wrong, I find your idea brilliant, I am quite surprised, but I would like to see also how to do it in the conventional way :-) – user162354 Jul 31 '14 at 20:16
  • By the way, under my interpretation of "within two rounds" the answer to B) is not right. The probability it ends in one round is $5/8$, the probability it ends in two is $(3/8)(5/8)$. Add. – André Nicolas Jul 31 '14 at 21:14
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    Thanks a lot, very helpful – user162354 Jul 31 '14 at 21:16
  • You are welcome. – André Nicolas Jul 31 '14 at 21:21