Shoot-out at the OK Corral

[TomCollins]

You aren't accounting for the fact that when two people die, you are much better off. Suppose B and C always shoot D, and D shoots at C, since this is what your strategy is.

So in this strategy (without acconting for A's actions), D will live .6*.4 of the time, C will die .2 of the time.

Compare my strategy to yours.
You always shoot at D.
So you will have the following results:
All Live: 3.8%
C Dies Only: 1.0%
D Dies Only: 76.2%
Both Die: 19.0%

So by the time the game ends, the EV of the players are as follows:

A: 36.6% of booty
B: 36.6% of booty
C: 26.4% of booty
D: .3% of booty.

So lets just see if A can do any better.
Suppose he decides to try to shoot B instead.

My EVs end up as:
A: 51.4%
B: 6.1%
C: 34.8%
D: 7.7%

So A stands to take a HUGE improvement by shooting B instead of D. D will still die 76% of the time, instead of 80% of the time. But now B dies 80% of the time as well, instead of 0%.

Since B, C, D use optimal strategy, they will adjust to this by trying to shoot A more often as well.

The equilibrium is almost certianly when everyone is equally likely to die.

You also state that it is impossible to reach that.
It is not. You do something like A shoots B 23% of the time C 37% of the time and D the remainder. I don't have hte exact numbers calculated, but its very possible to reach that.

[BettyBoopAA]

Tom,
you missunderstood my solution. What the players should do is shoot at D, they will get approx 1/3 of the booty. However this is not what they will do.

A and B are both afraid that each other will not shoot at D and shoot at the other, thus lowering their EV.
Thus A and B aim at each other. Since C and D know this, they both aim at each other.
How do you shoot at someone 23% of the time if you have one shot? You have to choose someone.

[TomCollins]

You flip a 100-sided dice and if its 23 or lower, you shoot him. It is very simple.

[tek]

If they were smart, all three would shoot Dan (preferably in the back while he's walking out of the saloon to Main Street).

Since he is sucky, they wouldn't want him to get lucky and take out one of the other three fair to good shooters.

Nor would it be logical for the other three to risk taking each other out.

[BettyBoopAA]

You flip a 100-sided dice and if its 23 or lower, you shoot him. It is very simple.

Yes I understand but your solution involves trust, You are player A you roll your dice, you are not worried that player B's best solution is now to shoot at you since that would maximize his value.
A, B and C should all shoot at D, but by not attacking the weak combined with a lack of trust as each player is better of if they and they alone break from the ideal solution thus the 2 weekest shooters will have the best chance to survive.

[TomCollins]

My solution has nothing to do with trust. In fact, my solution is quite the opposite.

If all tried to gang up on D, then A would realize he would fare better by trying to shoot B or C instead. That way, D is most likely going to die, but now the possibility of 2 deaths or even 3 deaths increases tremendously.

As long as my assumption is true, that a player maximizes his EV when each of the other players is equally likely to die, there is a Nash Equilibrium when all 4 players are equally likely to die. No player can improve his chances beyond this.

It cannot be an ideal situation for an individual if he can break from a strategy and do better, as well as if another individual can break from the strategy and cause you to do worse.

[TomCollins]

In fact, since all of these people are perfectly logical, they would realize there is no way to win, and decide that it is better to split the money equally than to risk the small chance that everyone dies.

So my FINAL ANSWER is they both realize this and decide that a gunfight is a terrible idea with expert game theorists.

[BettyBoopAA]

"As long as my assumption is true, that a player maximizes his EV when each of the other players is equally likely to die, there is a Nash Equilibrium when all 4 players are equally likely to die. No player can improve his chances beyond this"

I don't think this is true at all in this problem. A, B and C are all better if if they shoot D, 95.2% of the time, they will kill D in the first round and they will die 1/3*20% (6.6% of the time)
Player A-C all faced with the question would you have 1/3X all of the time are 1/2X 50 percent of the time so your assumption is not correct.

The Trust part comes will A B and C trust each other to do the correct thing. It like the 2 prisoners who must remain silent and not admit the other one is guilty.

[TomCollins]

This is a fact. I have proved the math here. Whether or not you believe me is another story.

Suppose the following percentages of dying. A= 6.6%, B=6.6%, C=6.6%, D=1-(.2)(.4)(.6)= 95.2%.

With another scenerio, where A shoots B every time.
A = 6.6%, B = 76.52%, C = 6.6%, and D=76%.

In these two scenerios, A fares much better. The odds that B C and D die have risen tremendously (since it was impossible before). Also, the odds of 2 players dying happens nearly 3/5 of the time. A does fantastically better here. So why would A not take this chance?

Of course, if they could all agree to do this and trust each other, there would not be a free-for-all, and A, B, and C could put themselves in a situation where D cannot improve and, and all three do better than if they could not trust each other.

If there could be an alliance formed, there are 4 possible alliances. Each player prefers to be in a alliance to not being in one. Also, each player fares best when his "alliance" wins more often. Every player would then want to be in an alliance with A. A would prefer to be in one with B and C.

So this is where the prisoners dilema comes in, since each of the three A,B, and C are better off cooperating. But each is rewarded by deviating.

The key here is the words "and they all act rationally once the shooting starts.". Rationality in terms of the prisoner dilema results in deviating from the "optimum" outcome for the group. So I stand by my point that The gunfight is cancelled, and everyone splits it equally.

However, the smartest thing for A,B, and C to do collectively is to trade their shares of the loot with each other, so that no one can deviate from the plan. If A, B, and C each agree to have 1/2 of the other opponents share loot if all three live and all of it if one dies, and keep none of their own share, they could act completely rationally, and do better than 25%.

So my NEW solution is that A, B, and C agree to take 1/2 of each of the other two's share at the end of the gunfight if they all live. Otherwise, they get the other survivors loot. If they are the only survivor in the alliance, they get to keep their own money.

In this case, there is much less an incentive to kill someone in your alliance, since the only way you get money is if they survive, or you kill both of them.

[jimdmcevoy]

Well put Tom