Let's start out by considering two-player games. Suppose we have two players X and Y, whose probabilities of hitting on a given shot are $x$ and $y$ respectively, both greater than zero.
For this two-player game, let $p$ denote "the probability that player X will eventually win, given that it is currently his turn" and let $q$ denote "the probability that X will eventually win, given that it is currently Y's turn". (In both cases, assuming that all players are playing optimally to maximise their chance of winning.)
On X's turn, he has two options:
- Shoot at Y, with probability $x$ of hitting (in which case he automatically wins) and $1-x$ of missing (in which case it becomes Y's turn).
- Pass, and let Y take a turn.
If X shoots, his probability of winning is $x + (1-x)q$. (Probability $x$ of winning on this shot, plus probability that the turn passes to Y, but X eventually wins.)
If X passes, his probability of winning is $q$.
On Y's turn, he has the options of shooting at X (win probability: $y + (1-y)(1-p)$) or passing (win probability: $1-p$).
Hence, on his turn Y can achieve a win probability of at least $y$ by shooting at X. Hence, $q <= (1-y) < 1$.
It follows that $x > qx$ and hence $x + (1-x)q > q$. Therefore, in the two-player game, X should always shoot at Y rather than passing, and by a symmetrical argument Y should always shoot at X rather than passing. (Not a very surprising finding!)
Since shooting is always the optimal strategy for the two-player games, we know:
$p = x + (1-x)q$
$q = 1 - (y + (1-y)(1-p)) = p(1-y)$
Solving these simultaneously gives:
$p = x/(x+y-xy)$
$q = x(1-y)/(x+y-xy)$
We can apply this formula to each of the possible two-player scenarios:
- If A and B are playing, A's probability of victory works out to 3/7 if it's his shot next, and 1/7 otherwise.
- If A and C are playing, A's probability of victory is 1/3 if it's his shot next, and 0 otherwise.
- If B and C are playing, B's probability of victory is 2/3 if it's his shot next, and 0 otherwise.
For three-player games, you can apply a similar approach: for each possible scenario (A to shoot, B to shoot, C to shoot) and for each possible choice (shoot at next player, shoot at previous player, pass) you can specify the ultimate probability of victory as functions of the probabilities for other scenarios.
In the general case, finding optimal strategies can get very messy, because of circular dependencies between these probabilities. In some cases, you might even end up with deadlock. For example, if every player has a 99% chance of hitting, then you end up with a Mexican standoff where each player's optimal strategy is to pass until somebody else shoots first; in that scenario there is no well-defined probability of victory because the game goes on forever.
However, for this particular scenario there is an optimal strategy for each player. We can find it as follows:
Start by considering the case where both A and B have missed or passed, and C gets a turn. C has three options:
- Shoot A, reducing the problem to the two-player BC game with B next to shoot. We already know that C's chance of winning from here is 1/3.
- Shoot B, reducing the problem to the two-player AC game with A next to shoot. We already know that C's chance of winning from here is 2/3.
- Pass. Chance of winning: not yet known, depends on who the other players target.
From this, we know that if it gets to C's turn, then he has a strategy that gives at least a 2/3 probability of winning. Hence, if C gets a turn, the other players have no more than a 1/3 probability of winning. We also know that shooting at A is not his best option while B is in the game.
Now rewind to the case where A has passed or missed, and it's B's turn. One possible strategy B could follow is: "shoot at C until C is down, and then shoot at A".
If he follows this strategy, he has a 2/3 chance of hitting C on his next/first attempt; if he does hit, then we know he has a 4/7 chance of winning the ensuing "A vs B, A shoots first" game. Multiplying these gives at least an 8/21 chance of victory from this strategy. (Possibly more, because we haven't looked at cases where B misses C on his first attempt and ends up winning anyway.)
If B does not shoot at C, then C definitely gets a turn, and we know that this gives B no more than a 1/3 chance of winning. 1/3 is less than 8/21, so B's optimal strategy must be to shoot at C (and then A, once C is out of the picture).
So we know that as long as all three are in the game, B will be shooting at C, and C will either pass or shoot at B. At this point, nobody is shooting at A.
Now, let's consider A's options during the three-player phase of the game:
- Pass, until one of the other players is eliminated. When that happens, it will be A's turn. A's probability of victory will be either 1/3 (if C remains) or 3/7 (if B remains).
- Shoot at B. If A misses (2/3 chance), the consequences are the same as for passing, but if A does take out B (1/3 chance), then A will automatically lose as it's C's turn next. So this is clearly worse than passing.
- Shoot at C. If A misses (2/3 chance), the consequences are the same as for passing. If B does take out C (1/3 chance), then we go to the "A vs. B" game with B's turn next. We know that A's chances of victory from there are only 1/7, which is worse than either of the possibilities from passing. So this option is also worse than passing.
Hence, A's best option is to keep passing until one of the other players is eliminated. It then follows that C's optimal strategy is to target B first (because we already know targeting A is suboptimal, and if A and C both pass, B is just going to keep on shooting at C until he hits, which would give C no chance to win).
Hence, the possible outcomes with everybody playing optimally are:
A passes, B hits C (2/3 chance), A then has 3/7 chance of defeating B.
A passes, B misses C (1/3 chance), C hits B, A then has 1/3 chance of defeating C.
A's total chance of victory is 2/3 * 3/7 + 1/3 * 1/3 = 25/63.