Shoot-out at the OK Corral
 

Suppose four gun-fighters rob a bank, and decide to split the booty three ways.

The only problem is, they can't decide who gets left out. To settle the matter, Abe suggests they have a shoot-out.

Now these four gun-fighters are good friends, and they all happen to know Abe is the best shot. Bill and Corby are mediocre, and Dan is truly sucky.

Assume they take up positions an equal distance apart, none of them shoots any faster or slower than any of the others, and they all act rationally once the shooting starts.

If Abe can hit a target at that distance with 80% accuracy, Bill and Corby are 60% and 40% accurate respectively, and poor Dan misses 80% of the time, who is most likely to be eliminated? Who is most likely to survive?

Re: Shoot-out at the OK Corral
 

"...and they all act rationally once the shooting starts."

Danny-boy. Cuz if he's acting rationally he'll run like hell.

Guess
 

Assuming:
1) The booty is equally divided among the survivors.
2) The shooting stops as soon as one robber gets shot.
3) "act rationally" means use a Nash equilibrium strategy for the game where the only payoff is percentage of the booty.

Then I think there is only one Nash Equilibrium for the game. Each robber randomly chooses an opponent every round and shoots. Abe is the most likely to survive. Poor Dan is the most likely to die.

If we replace Assumption 2) with

2) The shooting stops when at least 3 robbers are dead.

Then, the problem becomes difficult.

Re: Shoot-out at the OK Corral
 

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Assume they take up positions an equal distance apart

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So I guess they're not all on the ground.

Re: Shoot-out at the OK Corral
 

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Assume they take up positions an equal distance apart

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So I guess they're not all on the ground.

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No. They're wearing gravity boots.

Re: Guess
 

[ QUOTE ]
Assuming:
1) The booty is equally divided among the survivors.
2) The shooting stops as soon as one robber gets shot.
3) "act rationally" means use a Nash equilibrium strategy for the game where the only payoff is percentage of the booty.

Then I think there is only one Nash Equilibrium for the game. Each robber randomly chooses an opponent every round and shoots. Abe is the most likely to survive. Poor Dan is the most likely to die.

If we replace Assumption 2) with

2) The shooting stops when at least 3 robbers are dead.

Then, the problem becomes difficult.

[/ QUOTE ]

I'm not sure what the "Nash Equilibrium Theory" is - I'll look it up in a moment.

But, isn't there a strategy each of the robbers could use that would maximize his chances of surviving (as opposed to shooting randomly)?

Re: Shoot-out at the OK Corral
 

i think that because they are good friends greed will not overcome and they will stop shooting after at least one is killed on any volley of shots

because Abe is the best shot everyone will try to take him out so on each volley that has to be fired he will have a 60% chance of being killed by Bill, and of the 40% of the time he's not killed by Bill he has a 40% x 40% = 16% chance of being killed by Corby, and a 24% x 20% = 4.8% chance of being killed by Dan = total 80.8% chance of being killed

Abe will be worried most at being killed by Bill so he will aim at Bill and will kill Bill 80% of the time

therefore Abe has the most chance of being killed first and will hang up his boots 80.8% of the time

Bill has almost as much a chance of being killed first at 80% of the time

Abe and Bill will both be killed 80% x 80.8% = 64.64% of the time

imo, good old Corby and Dan will survive 100% of the time

Re: Guess
 

Well, suppose you're Abe. Whoever you shoot at, if you hit them, it's over. So you don't gain anything by trying to take out one person over another. You can't try to take out Bill or Corby, thinking that after they're dead, you have a better chance of surviving. After they're dead, it's over.

Either someone's dead and it's over or everyone survives and the shooting continues. But if everyone survives, who cares who you shot at originally?

Re: Shoot-out at the OK Corral My Solution
 

"Assume they take up positions an equal distance apart, none of them shoots any faster or slower than any of the others, and they all act rationally once the shooting starts"

Each person wants to maximim their survival and is greedy so they want the most people dead. This means the correct game theory solution for all 4 is to shoot at once. Abe and Bill will both shoot at each other at the same time as they both want to take out the player that can eliminate them. Since Corby and Dan understand this, their best action is to shoot at each other which makes
Corby the least likely to die at 20%

Re: Shoot-out at the OK Corral
 

60 will have 80 then 40 and 20 then 20
80 will have 20,40,and 60 then 20,40 then 20
40 will have no one the first round then at worst 80 then 20
20 will have no one the first round no one the second round and then at worst 80 the third

Ah screw it, my head hurts