Sandbox
The solution given by the Car Talk boys has the right idea but they struggle a bit with the mathematics. Also the problem is not well defined. However this is a well known problem. It is problem 20 in Mosteller's famous book "Fifty Challenging Probability Problems". Here is his description of the duel as applied 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 suggest that A should make is first shot to the sky and otherwise shoot at the most skilled of the duelers still alive. Of course suggesting that A shoot at the sky is counterintuitve this is what makes the problem interesting. Neither Mosteller or the Car Talk shen defining the game say that shooting at the sky is an option but Mosteller 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 wins. Under his strategy the possible paths the duel are shown in the tree diagram below.
The nodes give the remaining duelers. The bar indicates the dueler who shoots next. The branches have the probabilities for the possible outcomes of a shot. All but one of these probabilies are obvious from skill of the dueler. The one that is not obvious when A has only B to shoot at. In this case we have a two person duel and we assume that A shoots first and the the duelers alternate shots until one is killed.
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 and B both miss their first shots (probability 2/3*1/3 = 2/9) for A still has probability 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 determine from the tree each players chance of winning the duel. Note that there is only one branch resulting in C winning and the probability that this branch occurs 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.
From this we see that A has the highest probability of winning and C the lowest probability. Thus we have survival of the unfitest.