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