strategy question for david s (and others)

here's a very interesting (i hope) strategy problem for you


so there are these three guys called david, ray and mason (we'll call them the good, the bad and the ugly for the purposes of this question, but not necessarily in any order) who fall out after a big game of poker and decide to play a variation of russian roulette, wild west style, winner take all


each one has a revolver and two bullets


the deal is that they will shoot in order: ugly first; bad second; good third; ugly fourth; bad fifth; and good sixth - if they are hit they are dead and out of the game - any left alive after a maximum of two shots each share the pot - no teamwork is allowed - each plays best strategy at every shot


when it's their turn to shoot (if they are still alive) they can shoot at whoever they want


the bad news is that ugly only hits his target one-third of the time; bad hits his target half the time; and good has never been known to miss; - each one knows the others' skill level, which is why they they have agreed the order of fire - ugly; bad; good


ugly draws his revolver for the first shot


who does he aim at and why?


and can this strategy be used if you regard yourself as the worst player of three left in a no limit poker tournament?


(please don't look at other replies until after you have posted yours - thanks)


mike

Well this is a somewhat trivial game theory question. When I was first learning gmae theory we did a gmae called "The Good, the Bad, and the Ugly" that involved 3 players firing at the same time. It can be shown that both the ugly and bad must fire at the good (doing something else is dominated). Borrowing a line from David I will leave other to elaborate.


Randy Refeld

I'll give it a fast shot:


First you shoot good, then bad, then ugly.

Of course you don't shoot your self.


Is it a trick-question ? Maybe I should do some calculation.


* and can this strategy be used if you regard yourself as the worst player of three left in a no limit poker tournament? *


No ! I don't think we will be allowed to carry guns to the poker-room !

It seems like the ugly has to shoot the good, because if he kill the bad, the good will kill him.


Similarly, the bad will do the same, because if he kills the ugly, the good will kill him.


So the good only survives the first round to fire 1/3 of the time(2/3*1/2). He always kills the bad with his first shot.


After 1 round, the bad and ugly are alive 2/3 of the time. The other 1/3, the good and the ugly are alive.


So the Good wins 2/9 of the time overall.

The Bad wins 2/3*2/3*1/2 or 2/9 of the time, and splits the pot just as often with the ugly.

So the Bad survives 4/9 times, and wins the pot outright 2/9. The ugly wins the whole pot 1/3 of the time, and splits with the Bad 2/9.


Final Tally:

Good wins 2/9.

Bad wins 2/9, splits with Ugly 2/9.

Ugly wins 1/3, splits with Bad 2/9.


I don't know what this has to do with poker, except position is of paramount importance!


Dan Z.

If ugly shoots at good:


1/3 he kills good.

if he kills good:

-1/2 bad kills ugly

-1/2 bad misses

--1/3 ugly kills bad; ugly survives

--2/3 ugly misses

---1/2 bad kills ugly

---1/2 bad misses; bad and ugly survive


2/3 he doesnt kill good.

-1/2 then bad kills good,

--1/3 ugly kills bad; ugly survives

--2/3 ugly misses bad

---1/2 bad kills ugly

---1/2 bad misses ugly; bad and ugly survive

-1/2 bad misses good

--1 good kills bad

---1/3 ugly kills good; ugly survives

---2/3 ugly misses good,

----1 good kills ugly


Ugly shoots at bad:


1/3 Ugly kills bad

-1 good kills ugly


2/3 ugly misses bad

-1/2 bad kills good

--1/3 ugly kills bad; Ugly survive

--2/3 ugly misses bad

---1/2 bad kills ugly

---1/2 bad misses ugly; B and U survive

-1/2 bad misses good

--1 good kills bad

---1/3 ugly kills good; Ugly survives

---2/3 ugly misses good

----1 good kills ugly


Ugly misses on purpose:


1/2 bad kills good

--1/3 ugly kills bad; U survives

--2/3 ugly misses bad

---1/2 bad kills ugly

---1/2 bad misses ugly; B and U survive


1/2 bad misses good

-1 good kills bad

--1/3 ugly kills good; Ugly survives

--2/3 ugly misses good

---1 good kills ugly


EV for shooting at good:


1/3*1/2*1/3(1)+1/3*1/2*2/3*1/2(0.5)+2/3*1/2*1/3(1)+2/3*1/2*2/3*1/2(0.5)+2/3*1/2*1*1/3(1)= 0.36


EV for shooting at bad:


2/3*1/2*1/3*(1)+2/3*1/2*2/3*1/2*(0.5)+2/3*1/2*1*1/3*(1)=

0.28


EV for missing:


1/2*1/3*(1)+1/2*2/3*1/2*(0.5)+1/2*1*1/3*(1)=

0.42


UGLY should miss on purpose!!!

... maybe it is a trick-Q !?


* UGLY should miss on purpose!!! *


Can he do that ? Is that not cheating !?

Ugly aims at Good. He wants the 1/3 chance to get the best shot out of the game at once.


In my view, psychology is more important at no limit than any probablities. But if I regard myself as the worst player of the remaining three then attacking poker is best and you should attack the best player. Why wouldn't you?


Mr. Ugly could also just shoot himself and be done with it all. Of couse there is the 2/3 chance that he will miss and have to live for another round. Life has its misfortunes!

but position comment is good! ty

but lol to guns comment! ty

but ty for the humour!

not exactly cheating lol!

if ugly is allowed to draw first, (after realizing what he gott himself into), he should shoot the good first, and then before the bad draws his gun, shoot him too! /images/smile.gif

well, it wasn't quite as trivial a question as RandyRefeld thought!


but nf got it exactly right, above, so there is no point repeating his work - thank you nf


in my opinion, the strategy could be used in certain situations in tournaments


sometimes it could be better not to become involved with bigger stacks, but just to let them shoot it out between themselves


if one knocks the other out, leaving you headsup with a better player, you can then fire your last best shot at that single opponent and hope to hit him some of the time


ironically, mason and others are hinting at exactly this strategy in another forum at this very moment - see : http://www.twoplustwo.com/cgi-bin/newforums/holdem.pl?read=77083


thank you everyone who joined in

An easier way to see this is to work from last action. (Also true in poker, I think):


Good will take the risk-free last shot if last to act on the last round, so Bad will try to prevent that by firing at Good in round 2. This means Good will aim at Bad in rnd 1; which means Bad will aim at Good in rnd 1 too. In sum, 2 points: If ugly misses, he survives round 1. And, at the beginning of round 2, there will be someone dead.


So Ugly Will be headsup against a better gunman. Ugly will take 1 shot, then the other guy 1. All that changes if Ugly hits on round 1 is that his opponent will take 1 additional shot before Ugly. It's clearly worse.

Whoops! There's one unstated assumption here:


It's better for Ugly to be headsup with either bad or ugly, rather than headsups with Bad and giving an extra shot. This doesn't hold when Bad gets worse.


Meaning, hitting good right away outweighs getting shot at by bad an extra time. (This effect counts double, since Ugly is more likely to survive another shot against Bad, plus Bad is more likely to miss Good - making it less likely ugly would suvive a headsup confrontation anyway)

brilliant question, you can call it game theory or anything else, but what ever it is, this kind of thing sorts the men from the boys!