Shoot-out at the OK Corral

[elwoodblues]

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Assume they take up positions an equal distance apart,

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Is this possible with 4 people? If they arrange themselves in a square:
a b
c d

A is closer to B than he is to D.

[TomCollins]

One has gravity boots, so they are forming a pyramid.

[TomCollins]

I'm assuming if two people die, then the loot is split 2 ways instead of 3. And if 3 people die, then one guy gets all the loot. As soon as one person dies, its over. So there is no real advantage in targeting anyone, since if you are going to shoot at A and hit or shoot at B and hit, the game is over, and you don't have to worry about being hit.

I can't see how there is a better strategy than shooting randomly.

I thought the problem was shoot until everyone is dead but 1. That is far more interesting unless I am missing something obvious.

[elwoodblues]

Excellent. I'll take the guy with gravity boots any day of the week.

[BettyBoopAA]

I can't see how there is a better strategy than shooting randomly.

put yourself in Abe's shoes, the worst scenario for him is for him to be shot by someone he could have shot first. With that line of thinking, he will choose to shoot at the best shooter, Bill

[TomCollins]

Start with a strategy of everyone shooting randomly besides A.

B kills A 60%*33% = 20%
B kills C 60%*33% = 20%
B kills D 60%*33% = 20%

C kills A 40%*33% = 13.3%
C kills B 40%*33% = 13.3%
C kills D 40%*33% = 13.3%

D kills A 20%*33% = 6.7%
D kills B 20%*33% = 6.7%
D kills C 20%*33% = 6.7%

In this case, A should shoot the opponent most likely to live, because in this case, more people end up dead. The opponent least likely to die is B, since he doesn't have B shooting at him. So A should always shoot B.

But now C figures this out, so he realizes if he doesn't shoot at B (since A should take care of him 80% of the time), hes better off shooting at just A or D.

This is actually becoming quite interesting. I will have to look at it later.

[TomCollins]

So A can do better than randomly shooting.

It looks like the maximum EV for every player can be reached when each of his opponents has an equal chance of dying. So this means the overall EV is when everyone dies an equal amount of time. This happens when each gunfighter decides to randomly shoot (but not equally random) at each other fighter. This comes out to roughly a 57% chance of survival for each fighter. However, the EV is about 24% of the loot. This is because occasionally every fighter will die.

Lesson learned: Share the damn money!

[HigherAce]

Well heres what im thinking. If someone shot at me I dont care how good they are I would be shooting at them next. Therefor I would think they all would shoot at Dan. Most probability of them ending the game there since there all good shots and even if they miss hes not gonna hit any of them. Think about it. If you were holding a gun agaisnt 3 people and you knew 1 out of you 3 was a horrible shot but you only needed one dead to win, what would give you best chances of ending in the money, so to speak. You fire at a good person and miss hes not gonna ignore a good shot and focus on a crappy one. Why find out which one is a little bit better outa the three in this situation instead of focusing on the one everyone knows is blind. Hence, Dan is dead. [img]/images/graemlins/smirk.gif[/img]

[TomCollins]

Remember all of our players are playing with perfect strategy, so they don't need to worry about pissing everyone off. They just want to maximize their profit. They maximize their profit when the most people die, so any case that provides the maximum EV of people dying, the better. I have not proved this, but I beleive your maximum EV happens when all three of your opponents have an equal chance of dying on the next round. Since everyone is attempting to reach this equilibrium, the optimum play happens when everyone has an equal chance of dying. A shoots B x% of the time, C y% of the time, and D z% of the time.

[BettyBoopAA]

"but I beleive your maximum EV happens when all three of your opponents have an equal chance of dying on the next round"
How is that possible, it's not. I believe this problem is a variation of the prisioners dilema. The best solution would appear for A, B and C to all shoot at D at the same time which would most likely eliminate D and there is 20% chance that one of the 3 would be killed by D.
The dilemna here makes everyone shoot at the same time and gets player A into thinking, that B is better off if he shoots A because there will be less people to split the money. If everyone shoots at someone different, we can expect 2 people to die on average(with a small chance that all are dead). Since no one trust anybody else, I don't see how A wouldn't shoot at B. B understands this and his best chance of survival is to shoot at A since his chance of killing A is higher than the other players chances and his best hope is that A misses him. B can't risk shooting at someone else in the case everyone misses, he doesn't want A to have another chance at him.
If you believe that then C shoots at D and D shoots at C.