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The solution given by the Car Talk boys has the right idea but they struggle a bit with the  mathematics.  This is a well known problem and occurs in most math puzzles books.  It is problem 20 in Mosteller's famous book  "Fifty Challenging Probability Problems".
The solution given by the Car Talk boys has the right idea but they struggle a bit with the  mathematics.  This is a well known problem and occurs in most math puzzles books.  It is problem 20 in Mosteller's famous book  "Fifty Challenging Probability Problems".


We shall call the participents A, B, and C and the probabilities that they hit when they shoot a,b,c.
A, B, and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 1/3, B's is 2/3, and C never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unhit.  What should A’s strategy be.
Then in the Car Talk problem a = 1/3, b = 2/3 and c = 1


Consider first a simpler duel where there are only two shooters M and N. Assume that M hits with probability m and misses with probability N with probability n. Assume that M starts first. Let q = (1-m)(1-n). Then q is the probability that in a cycle they both miss.  Then if M starts this two person duel the probability that he wins is:
A naturally is not feeling cheery about this enterprise. Having the first shot he sees that, if he hits B, C will then surely hit him, and so he is not going to shoot at B. If he shoots at C and misses him, then C clearly shoots the more dangerous B first, and A gets one shot at C with probability 1/3 of succeeding. If he misses this time the less said the better. On the other hand, suppose that  A hits C. Then B and A shoot alternately until one hits.  A’s chance of winning is


<center><math> m + qm + q^2m + q^3m + ...= m(1 + q + q^2 + ...) = m/(1-q).</math></center>
(1/3)(1/3) + (1/3)^2(2/3)*(1/3) + (1/3)^3(2/3)^2(1/3) + …


A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits B, C will then surely hit him, and so he is not going to shoot at B. If he shoots at C and misses him, then C clearly shoots the more dangerous B first, and A gets one shot at C with probability 1/3 of succeeding. If he misses this time, the less said the better. On the other hand, suppose A hits C. Then B and A shoot alternately until one hits. A's chance of winning is (.5)(.3) + (.5)^2(.7)(.3) + (.5)^3(.7)^2(.3) + ... . Each term corresponds to a sequence of misses by both C and A ending with a final hit by A. Summing the geometric series we get ... 3/13 < 3/10. Thus hitting B and finishing off with C has less probability of winning for A than just missing the first shot. So A fires his first shot into the ground and then tries to hit B with his next shot. C is out of luck.
Each term corresponds to a sequence of misses by both B and A ending with a final hit by A.
 
Summing the geometric series we get
 
<math>(1/3)(1/3) {1+ (1/3)(2/3) + [(1/3)(2/3)]^2 + …}<\math>
 
{math>= \frac{(1/3)(1/3)}{(1-(1/3)(2/3)} = \frac{1}{9}= \frac {7/9}{2/9} =  frac{2\9 < frac{1}{3}\math></math>

Revision as of 20:09, 4 March 2007

The solution given by the Car Talk boys has the right idea but they struggle a bit with the mathematics. This is a well known problem and occurs in most math puzzles books. It is problem 20 in Mosteller's famous book "Fifty Challenging Probability Problems".

A, B, and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 1/3, B's is 2/3, and C never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unhit. What should A’s strategy be.

A naturally is not feeling cheery about this enterprise. Having the first shot he sees that, if he hits B, C will then surely hit him, and so he is not going to shoot at B. If he shoots at C and misses him, then C clearly shoots the more dangerous B first, and A gets one shot at C with probability 1/3 of succeeding. If he misses this time the less said the better. On the other hand, suppose that A hits C. Then B and A shoot alternately until one hits. A’s chance of winning is

(1/3)(1/3) + (1/3)^2(2/3)*(1/3) + (1/3)^3(2/3)^2(1/3) + …

Each term corresponds to a sequence of misses by both B and A ending with a final hit by A.

Summing the geometric series we get

<math>(1/3)(1/3) {1+ (1/3)(2/3) + [(1/3)(2/3)]^2 + …}<\math>

{math>= \frac{(1/3)(1/3)}{(1-(1/3)(2/3)} = \frac{1}{9}= \frac {7/9}{2/9} = frac{2\9 < frac{1}{3}\math></math>

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