Sandbox

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 will present his proof addaped to the Car Talk problem.

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

<math>(1/3)(1/3) + (1/3)^2(2/3)(1/3) + (1/3)^3(2/3)^2(1/3) \cdots</math>

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 + \cdots] </math>

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

Thus hitting C and finishing off with B has less probability of winning for A than just misssing the first shot. So A fires his first shot into the ground and then tries to hit C with is net shot. C is out of luck.

Mosteller writes:

In discussing this with Thomas Lehrer, I raised the question whether that was an honorable solution under the code duello. Lehrer replied that the honor involved in three-cornered duels has never been established and so we are on safe ground to allow A a deliberate miss. </blockquate>