Sandbox
The solution given by the Car Talk boys has the right idea but they struggle a bit with the mathematics and in fact the problem is not really well defined. 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". Here is his description of the duel adapted 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?
Both Mosteller and Car Talk give arguments that for his first shot A should shoot into the air and in all other cases when dueler has to choose between two duelers he will choose fire at the most skillful of the two. The fact that for his first shot A should not shoot at anyone is considered counter-intuitive. Note that in his formulation of the problem Mosteller did not state that dueler could shoot into the ski. He Writes
In discussing this with Thomas Lehrer, I raised the question whether that (shooting at the ski) 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.
If we assume that the Duelers use the strategy suggested by Mosteller then we can compute the probability that each of the three survives. Under his strategy the possible paths the duel could follow are shown in the tree diagram below:
The nodes show the remaining duelers. The bar indicates the dueler who shoots next. The branches have the probabilities for the possible outcomes of a shot. All these probabilities are determined from the probabilities that the duelers shots will be successful. The one that is not is when C has been killed and A shoots at B. If A misses B will shoot at him and this will continue until either A or B is killed. That is we have a two person duel. Let p be the probability that A wins and q the probability that B wins this two person duel. Then A wins if he hits B with his first shot (probability 1/3). If A misses and then B must also miss (probability 2/3*1/3 = 2/9) for A to have a chance of winning. If this happens he has the same chance p of winning. Thus p = 1/3 + 2/9p. Solving for p we have p = 3/7 and so q = 4/7.
Now we can see from the tree each players chance of winning the duel. Note that there is only one branch resulting in C winning. The probability that this branch will be followed is 1/3*2/3 = .2/9=222. There is also only one branch that will result in B winning and this branch has probability 2/3*4/7 = 8/21 = .381. Thus the probability at A wins is 1-8/21 - 2/9 = 25/63 = .397.
A has the highest probability of winning and C the lowest probability. Thus we have survival of the unfitedness.