Survival game: Consider $3$ players, $A, B$ and $ C$, taking turns shooting at each other. Any player can shoot at only one opponent at a time (and each of them has to make a shot whenever it is his/her turn).
Each shot of $A$ is successful with probability $1/3$, each shot of $B$ is successful with probability $1$, and each shot of C is successful with probability $1/2$ (with all the outcomes being independent).
$A$ goes first, then $B$, then $C$, then $A$, and so on, until one of them dies. Then, the remaining two will be shooting at each other, so that nobody ever makes two shots in a row: e.g. if $A$ gets $C$ shot, then $B$ goes next.
The game continues until only one player is left. Assume that every player is trying to find a strategy that maximises his/her probability of survival. Assume also that every player acts optimally and knows that the other players will act optimally too.
Who should player A shoot at first? What is the probability of survival of $ A$ (assuming he/she acts optimally)?
*Hint. A strategy is a sequence of decisions on who to shoot at at any given turn, given who is still left in the game. It is clear that, after $B$ shoots for the first time, there will be at most two players left, and, hence, for the remaining players, there will be no need to make any choices. Therefore, it is convenient to solve the problem recursively, starting from the decision of $B$, and assuming that all players are alive by the time $B$ shoots (otherwise, again, there are no decisions for $B$ to make).
It is clear that, given a choice between $A$ and $C$, $B$ will shoot at $C$, because playing against $A$ only will give $B$ a higher probability of survival than playing against $C$ only (i.e. $2/3$ vs. $1/2$). Knowing this, A needs to choose whether it is optimal to shoot at $B$ or at $C$. Considering the possible outcomes produced by each of the two choices, you will notice that, in one case, the survival probability can be computed by hand, and, in the other case, it can be reduced to the computation of an exit probability of a simple Markov chain.
if $A$ shoots $C$: $P$($A$ hits $C$ *$B$ misses $ A$ * $A$ hits $B$) + $P$ ($A$ misses $C$ * $B$ hits $ C$ * $A$ hits $B ) = 0 +1/3*1*1/3 = 1/9$
if $A$ shoots $B$: $P$($A$ hits $B$ *$C$ misses $ A$ * $A$ hits $C$) + $P$ ($A$ misses $B$ * $B$ hits $ C$ * $A$ hits $B ) = 1/3*1/2*1/3 +2/3*1*1/3 = 5/18$
So A should shoot B, is it correct?