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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".

We shall call the participents A, B, and C and the probabilities that they hit when they shoot a,b,c. 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:

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

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.