Three gunslingers are going to fight in a gun duel (or truel if you prefer). The rules are such that they will be arranged into a triangle in some order and one of the cowboys is picked randomly to go first. They can aim wherever they like and from then they each take turns to take their shots going round and round the triangle until two cowboys are dead.

Alice is a crackshot and never misses, Bob can hit a target 80% of the time and Carol can only hit half of the time.

What should each of them do and what are their chances of survival? We will assume that they are all logical and will play to the best of their ability.

One way into the problem is to think about what would happen if they could all hit 100% of the time. If any of them shoot and kill one of their opponents then the remaining rival will kill the first. Playing logically all three should just repeatedly shoot into the air which means the game never resolves.

Applying that logic to our main game, the worst shot Carol should waste all of her shots until there is only one person left. Neither Alice nor Bob are going to waste their shots on Carol, so she can afford to wait.

Alice and Bob should both try to take out the other before turning on Carol since they are the main competition for each other. Alice’s probabilities are the easiest to work out because they don't go infinite. 1/2 the time Alice will get to go before Bob in the playing order and so she can take him out. Then Carol gets one attempt to take Alice out, but the other half of the time she'll miss and so Alice will survive.

The other half of the time Bob gets to shoot first and so Alice will have to survive both a shot from Bob and then one from Carol. Put together we get Alice's total chances of survival are 1/2*1/2+1/2*1/5*1/2=3/10. Immediately this should be slightly surprising because it is less than a third, so the best shooting ability leads to a below par chance of survival mostly because it paints a large target onto you.

The other cases are going to get a bit infinite. For Bob to survive he needs to first shoot Alice which happens 2/5ths of the time, then survive one shot from Carol taking us to 1/5th of the time and combined with his probability to actually hit Carol makes a success of 4/25 straight away. However if he misses his chance to take out Carol with his first attempt (1/5) and Carol also misses (1/2) then he gets another go. This can go on forever and so we get a geometric series. 4/25(1+1/10+1/100+1/1000+...)=4/25 * (1-1/10) = 4/25 ÷ 9/10 = 8/45.

Since we have the probabilities of success for two of the gunslingers we can subtract them from 1 to find Carol’s, which ends up as 47/90. Comparing fractions doesn't make it easy to see these, so as a summary we have Alice 30%, Bob 17.8% and Carol 52.2%. The worst aimer ends up with the best chance of success.

There are many variations on this problem with larger Mexican Standoff set ups. As an entry point this Wikipedia article goes a little deeper.

Paper Folding Puzzle

Paper Folding Puzzle

The Problem from Good Will Hunting

The Problem from Good Will Hunting