Actually, I've figured out most of these. If someone could just tell me P(ABC -> BC) and P(ABC -> C) that would be really helpful! Thanks!
Three tanks fight in a three-way duel. Tank A has probability 1/2 of destroying the tank at which it fires, tank B has probability 1/3 of destroying the tank at which it fires, and tank C has probability 1/6 of destroying the tank at which it fires. The tanks fire together and each tank fires at the strongest opponent not yet destroyed. Form a Markov chain by taking as states the subsets of the set of tanks (i.e. take as states ABC, AC, BC, A, B, C, and none, indicating the tanks that could survive starting in state ABC. AB can be omitted because this state cannot be reached from ABC).
I'll denote the probability of tank A destroying B by P(AdB) and the probability of not destroying by P(nAdB).
P(ABC -> BC) = P(nAdB) * (P(BdA)*P(CdA) + P(BdA)*P(nCdA) + P(nBdA)*P(Cda)) =
1/2 * (1/3 * 1/6 + 1/3 * 5/6 + 2/3 * 1/6) = 1/2 * 8/18 = 8/36 = 2/9
P(ABC -> C) = P(AdB) * (the same as above) = 2/9