Three-Way Gunfight Riddle
Posted by Rob Herman at August 12th, 2007
Three gunfighters, named A, B, and C, are having a rather contrived gunfight. Here are the rules:
- The fighters shoot in turn: first A, then B, then C, then (if more than one is still alive) back to A, and so on.
- On a turn, a fighter gets to shoot only once, at one opponent.
- A hits the target 1/3 of the time. B hits 2/3 of the time. C always hits.
- Fighters are allowed to shoot at the ground or otherwise intentionally not hit any target.
- If all fighters intentionally miss in a row, the last one to do so is eliminated. *
What is A’s best strategy?
Extra credit: Generalize for different accuracies.
*This rule is not usually given as a part of this riddle, but it is necessary to avoid degeneracy while working out the solution, and (as you will see if you try to generalize) is particularly important if all of the fighters have accuracy over 50%. I chose it arbitrarily; it could be replaced with any rule that keeps everyone from intentionally missing forever.
The basic goal is to get one of the other two guys to make the first kill. In the given instance this works out. If A&B miss intentionally, C is forced to shoot, and will shoot the more accurate opponent, B. Thus B must shoot to kill, and will shoot at C, giving him the highest chance to avoid death before he gets another shot. An intentional miss by A gets him the first shot at the survivor of the others, and a full 1/3 win chance.
This solution works for all cases where A is the least accurate gunman.
For A being the most accurate, he must fire at the next-most-accurate and hope to be missed in return (any intentional missing will result in his being shot at by either the man he would have left standing anyway or the more accurate other opponent at least once, and thus will have him more or similarly likely to die)
For accuracies B > A > C, A must shoot at B, as B will shoot at A.
For C > A > B, similarly, A must shoot at C, as B will intentionally miss to draw a shot at A from C.
Summary:
Low man fires into the dirt
Middle man and high man fight to the death.
For bonus points, what is the probability that A will win the shootout?
The fight devolves to two consecutive duels. Win chance for the first shooter in each duel, calling his accuracy X and his opponent’s accuracy Y, is Sum[k=0:inf]{X * (1-X)^k * (1-Y)^k}
which reduces via infinite geometric series formula to X/[1-(1-X)(1-Y)] = X/(X + Y-XY)
Thus, A’s win chance is:
for A least accurate: B/(B+C-BC) * A/(A+B-AB) (1-B/(B+C-BC)) * A/(A+C-AC)
for A > B > C: A/(A+B-AB) * (1-A/(A+C-AC))
for A > C > B: A/(A+C-AC) * (1-A/(A+B-AB))
In the given example, A least accurate. I’ll walk through the logic that correlates to the mathematics. B duels C, then A duels the winner, so his win chance is his win chance in a duel against B * B’s win chance plus his win chance in a duel against C * C’s win chance (weighted average win chance against each opponent weighted based on the chance of facing said opponent).
Duel B vs C, B wins on 2/3 * 1/(1-(2/3 * 0) = 2/3. The 0 there being C’s miss chance, so total B wins 2/3, C wins 1/3.
Duel A vs B, A wins on 1/3 * 1/(1-(1/3 * 2/3) = 1/3 * 9/7 = 3/7
Duel A vs C, A wins on 1/3
Total A win chance = 2/3*3/7 1/3*1/3 = 2/7 + 1/9 = 25/63
Fixed your plus signs, Alatar. Sorry about the ongoing suckitude of the comment box.
Exactly right! *golf clap*
Who doesn’t love an infinite geometric series of gunfights?
Shinjo doesn’t. Frankly, she’s ashamed of you, Fu Leng.
And Lady Doji prefers iaijutsu duels to gunfights, but in this day and age she’ll take what she can get.
I should retheme this riddle around samurai.
eh, I’m not picky. I’ll take an infinite series of sword-fights, regardless of whether they involve samurai or pirates!