There is a duel between 3 participants. (yeah, dumb)
A, B and C will shoot at each other until only one survives.

A is the worst shot, hitting target 1 time in 3 on average.
B is better, hitting target 1 time out of 2.
C always hits the target.

To make this fairer:
A will shoot first,
then B will shoot (if alive),
then C (if alive),
and another round until only one is alive.

Question: What should A do for his first shot ?

shoot the ground

A purposefully misses right? Or doesn't fire at all.

If nobody is dead when it comes to B's turn then B ought to shoot at C.

1/2 the time B kills C meaning it is A vs B with A to shoot first, giving A an expected win rate of 1/3 + 1/9 + 1/27...+...

The other 1/2 the time B misses and C instantly shoots B dead, giving A a 1/3 chance of killing C before being instantly shot dead.


If A shot and killed B then that's clearly the stupidest move because C instantly kills A.

If A shot and killed C then it would be A vs B with B to go first, giving A an expected win rate of 1/6 + 1/18 ...+...

So missing the first shot is the best strategy.

Ok, you guys are good. Credits to Simon Singh btw.
Now for nerds:

In this case, with obvious notations, we had pA=1/3, pB=1/2, pC=1.
How do you alter your strategy when you change pA, pB and pC,
with 0< pA ≤ pB ≤ pC ≤ 1 ?

But if A shoots at the ground, then there's only a 1/3 chance he'll hit it! He should shoot B to maximize his chances of hitting the ground.

I just read a question extremely similar to this one in a book titled Peerless Probabiliy Problems and Other Puzzles and thought it was a pretty cool question.

from Fermat's Enigma correct?

For mainstream audience, yes.

Original comes from C.Kinnaird (1946), Encyclopedia of Puzzles and Pastimes

A game-theory paper about truels (3-way duels):
truels and variations

This is a very good problem. I mean the problem of writing down the inequality for when A should shoot at the ground

Its also a good problem because its perfect for people who are new or inexperienced at approaching problems like this or thinking about things in this way. It is both counterintuitive and easily-grasped once its understood (in a way that the Monty Hall problem or the famous broken vase problem arent) so just about everyone can go through the "Well, I guess shoot B....no way thats stupid you cant shoot nobody...hmmm.....cool" progression.

I haven't read the paper in the above link yet, but here's my analysis.

If all players are good shots there would seem to be peace through mutually assured destruction. For example pA = .98, pB = .99, pC = 1. So A shoots at the ground and hopes one of B or C kills the other, and then A gets the first shot against the survivor which is 98% to hit. But B knows this, and he knows that C knows this also. Unless B and C have a death wish, they should both shoot into the ground also, and then everyone can go home. Not much good to be the survivor when A is waiting to pick you off.

But how good a shot does A need to be to keep the peace? 80%? 50%? For most people even a 1% chance of getting shot is too high, but then again most people don't sign up for a truel. On the other hand maybe these guys hate each other so much that a 98% chance of death is justified by a 2% chance of watching the other two guys die. From here on I will assume that their mutual hatred is such that life is not worth living unless the other two are dead. This avoids having to think about metagame issues.

So A has the option of shooting at the ground only because he's the worst shot and therefore B and C won't target him first. But B and C don't have that luxury so they must target each other when they get the chance, otherwise they become the target. A is the only one who has a strategy choice to make. If 0 < pA < pB < pC ≤ 1 (changing the original problem statement such that A is definitely a worse shot than B, and B is definitely a worse shot than C), A would never aim at B since if he hits, C gets the first shot at A. A is always better off hitting C and then B gets the first shot at A. So the strategy choice is whether A should shoot at the ground or shoot at C. If he doesn't hit, B then shoots at C, and if B misses then C shoots at B.

If A elects to shoot at the ground, he will eventually get the first shot at the survivor between B and C. But if A kills C, this puts A in a duel with B where B gets the first shot. The immediate question is whether that would ever be preferable. Consider pA = 1/n, pB = 1/n-1, pC = 1 for some very large value of n. If A shoots at the ground it is nearly certain that the sequence will be B misses C, C kills B, A misses C, C kills A. Intuitively A's chance of winning the truel is close to zero. So if instead of A shooting at the ground he aims at C and hits (very unlikely he would hit but the question is whether he should make the attempt), A has greatly improved his situation. Now he's in a duel with B, B gets the first shot but that doesn't matter much because B will almost certainly miss. They will probably take turns missing for a long time but eventually one of them hits. Intuitively A's chance of winning the truel after he gets lucky and hits C is almost 50%. Which shows that there are cases where A should target C. I'll post more in a bit.

You can solve it exactly. Just write out the expression for A living if he shoots at the ground or if he shoots at C. It looks complicated at first but you can simplify it with some tricks.

A hint if anybody is still working on this, a situation in which A won't shoot at the ground is if C is such a good shot compared to A and B that you would rather get heads up with B with B shooting first than get heads up with C in which you (as A) shoot first.

What's the broken vase problem?

I bought this on Amazon after reading this post and got it today. Awesome book, thanks!

I forget the names involved (One is Nash, though, for sure) but its a comical game-theory scenario originally posted in like OOT or something. Two guys are in the airport, both with identical vases bought at some store, and broken by the baggage handlers. They are in separate rooms being questioned by airport management, who are trying to determine how much to reimburse them. The manager is a sadistic bastard, so he says to write down a bid, and whoever bids lowest gets his bid plus 5 bucks (or whatever). They are both game-theory experts and completely rational. What do they write down?

They apparently write down $0 if they are not allowed to go negative, but it sparked a HILARIOUS 1000+ post thread. You could probably search for it in the archives.