Mexican Standoff
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Three gunfighters. They all have their hands on their guns and are ready to draw. All three know that gunfighter A is the fastest draw but the least accurate shot. All three know that gunfighter B is the slowest draw but the most accurate shot. All three know that gunfighter C has an average draw (slower than A but faster than B) and an average shot (more accurate than A but less accurate than B).
One bullet in the chamber, one shot only.
Who would you rather be in this scenario? Who would you gun for? Who do you thinks wins most often in a million run simulation and who aims for whom?
***EDIT***
Ok, we've got a lot of splitting hairs going on here. So, to make things very clear:
A, B and C all draw in less than one second. There is no way for anoyone to have a clear advantage beyond a fraction of a second - not nearly enough time for either of the other two gunslingers to make a decision AFTER anyone else has moved. Gunfighter A is MARGINALLY faster than the other two and is MARGINALLY less accurate than the other two.
The 3 gunslingers are standing in an open field with an unobstructed view in any direction (other than the two opponents). Each one is standing precisely 20 feet from either of the other two, creating positions equivalent to the three corners of an equilateral triangle. Each one draws with his left hand. The sky is perfectly clear and the sun is directly overhead. Each one has exactly 20/20 vision. They have all had a hearty breakfast and brushed their teeth.
Seeing that, C will consider B to be the sole remaining threat, and aim at B. B, seeing that, will aim at C in turn. Neither would have any reason to aim at A.
The question as you've put it, though, needs clarification. What is each gunfighter's relative chance of hitting his opponent? Who "wins" if there are two survivors? Three? Is it more important for gunfighter X to survive, or to kill his opponents?
Quote: mkl654321A should fire his gun into the air!
Seeing that, C will consider B to be the sole remaining threat, and aim at B. B, seeing that, will aim at C in turn. Neither would have any reason to aim at A.
The question as you've put it, though, needs clarification. What is each gunfighter's relative chance of hitting his opponent? Who "wins" if there are two survivors? Three? Is it more important for gunfighter X to survive, or to kill his opponents?
The question is as clear as it can be. You have a moment to decide what to do. By the time "A" fires his gun into the air he might also have a bullet in his heart. I'm asking YOU to determine the relative chance of any possible outcomes and figure it out yourself. You have 3 seconds and the other two guys are about to draw...
Speaking of puzzles, I encourage you to keep them coming.
Quote: TheNightflyIdea taken from all of those great spaghetti westerns.
Three gunfighters. They all have their hands on their guns and are ready to draw. All three know that gunfighter A is the fastest draw but the least accurate shot. All three know that gunfighter B is the slowest draw but the most accurate shot. All three know that gunfighter C has an average draw (slower than A but faster than B) and an average shot (more accurate than A but less accurate than B).
One bullet in the chamber, one shot only.
Who would you rather be in this scenario? Who would you gun for? Who do you thinks wins most often in a million run simulation and who aims for whom?
You need to provide a diagram as to who is standing where.
Quote: chookYou need to provide a diagram as to who is standing where.
I assumed it was a triangle. That is all I think that needs to be said.
Actually, the wise gunman will prepare himself to possibly take a slug but will make his first (and in this case only) shot count: his greatest threat must be taken out. An average draw can still be nervous at that fight, so the greater threat is the one who is a better shot. Its the bullet that does the damage, not the sight of someone standing their with smoke coming out of his gun. Everyone there knows he may have to take a slug and it won't be fun. Chances are A, B and C have all been in gunfights before. So whoever marks a target knows its one slug. Shoot at the most dangerous shooter and throw your empty pistol at the one still standing while you try to step into the sun.
Which would I prefer to be? The set's armorer. He gets well paid to make sure those are blanks.
A will probably shoot at which ever he thinks is the calmest and smartest because that is his threat.
B will shoot at C because he will think A will shoot at B. If A's shot goes wide of the mark, A is not a threat but there is no time to then aim at C. So B will start off aiming at C. B may or may not live to see how his shot goes.
C wants to kill B. If A aims at C and his shot does not miss, well C will have a slug in him and probably not do well with his shot at A so he might as well go for the surer kill, B.
Quote: TheNightflyThe question is as clear as it can be. You have a moment to decide what to do. By the time "A" fires his gun into the air he might also have a bullet in his heart.
Well, no, if A is the quickest to shoot, he's going to get his shot off before the others draw. The answer to the question really does depend on whether the players can alter their choices after seeing A's action (and C's action) in turn. If A shoots into the ground/air and C *sees this and alters his decision* then C shoots at B before B shoots at C. If C's decision is set in stone before A acts, it's an entirely different problem.
....He does this all the time. The question is PERFECTLY stated. KenQuote: mkl654321A should fire his gun into the air!
Seeing that, C will consider B to be the sole remaining threat, and aim at B. B, seeing that, will aim at C in turn. Neither would have any reason to aim at A.
The question as you've put it, though, needs clarification. What is each gunfighter's relative chance of hitting his opponent? Who "wins" if there are two survivors? Three? Is it more important for gunfighter X to survive, or to kill his opponents?
Quote: WizardIf I were to ask the question, I would say that three gunmen are all quick-thinking perfect logicians. Would that satisfy you?
I assumed that. :) The edit made it clear that the decision has to be made prior to the draw and not after any shooter has fired. Given that, I'm still thinking about it...
Quote: MathExtremistWell, no, if A is the quickest to shoot, he's going to get his shot off before the others draw. The answer to the question really does depend on whether the players can alter their choices after seeing A's action (and C's action) in turn. If A shoots into the ground/air and C *sees this and alters his decision* then C shoots at B before B shoots at C. If C's decision is set in stone before A acts, it's an entirely different problem.
This is why I originally said the problem needed clarification. Since if the gunmen have time to alter their chosen target after A shoots, each player's decision is trivial, let's consider what would happen if they did NOT have time to switch targets:
All other things being equal, A would have the best chance of survival, because he might take out one of the others before they have a chance to get a shot off at him. So the question is, which opponent constitutes the greatest threat? It would be C, because even if it turns out that A is B's chosen target, C may choose B as his target and kill B before B ever gets his shot off at A. Therefore, A will target C.
B will reason as follows: logically, A will target C (as above). Therefore, I am home free if C targets A in return, and if C targets ME, he'll get his shot off before I can anyway, presuming that A misses him. So I can do nothing to affect the outcome; by the time I ever get my shot off, it'll all be over one way or another anyway. Therefore, B will not even draw his gun, in the hope that A and C will fire at each other.
C will reason as follows: I expect to be targeted by A. If B targets A, then I should target B; even if I miss B, b could still kill A (I am assuming that the gunmen would consider only one of their two opponents getting killed to still be a positive outcome). Also, however, if B targets ME, I should also target him, because I may kill him before he gets his shot off at me. Either way, I should target B.
So A should target C, C should target B, and B has no optimal strategy; he might as well save a bullet. Either that, or he can wait until the other two have fired and then take a "free shot" at the one he considers the ugliest.
The answer to this problem would be radically different if the standoff were iterated, i.e., if they played another round if there were two, or three, survivors. It would be an interesting problem if everybody had TWO bullets.
What matters here is the relative speeds of the gunmen, not their accuracy.
Quote: mkl654321This is why I originally said the problem needed clarification. Since if the gunmen have time to alter their chosen target after A shoots, each player's decision is trivial, let's consider what would happen if they did NOT have time to switch targets:
All other things being equal, A would have the best chance of survival, because he might take out one of the others before they have a chance to get a shot off at him. So the question is, which opponent constitutes the greatest threat? It would be C, because even if it turns out that A is B's chosen target, C may choose B as his target and kill B before B ever gets his shot off at A. Therefore, A will target C.
B will reason as follows: logically, A will target C (as above). Therefore, I am home free if C targets A in return, and if C targets ME, he'll get his shot off before I can anyway, presuming that A misses him. So I can do nothing to affect the outcome; by the time I ever get my shot off, it'll all be over one way or another anyway. Therefore, B will not even draw his gun, in the hope that A and C will fire at each other.
C will reason as follows: I expect to be targeted by A. If B targets A, then I should target B; even if I miss B, b could still kill A (I am assuming that the gunmen would consider only one of their two opponents getting killed to still be a positive outcome). Also, however, if B targets ME, I should also target him, because I may kill him before he gets his shot off at me. Either way, I should target B.
So A should target C, C should target B, and B has no optimal strategy; he might as well save a bullet. Either that, or he can wait until the other two have fired and then take a "free shot" at the one he considers the ugliest.
But if C's decision does not depend on where A is shooting, and if A is the perfect logician, as Wizard suggested, then A should inevitably realize, that his best shot is to aim B, not C, isn't it?
Quote: weaselmanBut if C's decision does not depend on where A is shooting, and if A is the perfect logician, as Wizard suggested, then A should inevitably realize, that his best shot is to aim B, not C, isn't it?
No, because from A's standpoint, the threat of B may be neutralized by C's targeting and shooting B before B ever gets a shot off (presumably, at A).
Quote: mkl654321No, because from A's standpoint, the threat of B may be neutralized by C's targeting and shooting B before B ever gets a shot off (presumably, at A).
It may be neutralized or it may not, if C misses. C on the other hand poses no threat to A, because he is aiming at B.
There is no point for A to target C - if he shoots him, he may be shot by B, and if he misses he has wasted his bullet. If A targets B on the other hand, it makes his chances better, because even if C misses, B could still be neutralized by A.
Three perfect logicians are in a truel (three-way duel). Each has one bullet in his gun. All are perfect shots. Logician A is the fastest, then B, and C is the slowest. All three know this. Deliberately missing is allowed. Who will live and who will die?
Quote: WizardI'd like to make a future "ask the wizard" question out of this. Does anybody object to the following wording? Do you think I should omit the hint that deliberately missing is allowed? Note that I swapped the speeds of B and C, for clarity.
Three perfect logicians are in a truel (three-way duel). Each has one bullet in his gun. All are perfect shots. Logician A is the fastest, then B, and C is the slowest. All three know this. Deliberately missing is allowed. Who will live and who will die?
C will only survive if A targets B. It would not be necessary for B to target A in return, because B would not survive to get the shot off even if he targeted C. So C may as well assume that A has targeted B, therefore he should aim at A.
B will only survive if A targets C. For the same reason as above, he also might as well aim at A.
A perceives this, and knows he cannot survive if he is targeted by both B and C. Therefore, he must make it clear to his opponents that he will either not fire, or deliberately miss. By doing this, he brings the player that he otherwise would have killed back into the mix; C serves as a deterrent to B's targeting A.
So: A will fire his gun into the air. B will then kill C.
In your scenario, not one of the perfect logicians would shoot. They would all simply raise their hands in the air. Each one would know that as soon as he's fired his gun he'd be dead.
Both B and C know that whomever A aims at will be dead as A will outdraw either one. Let's say A shoots B. C is then standing there with the only bullet that now matters (as B is dead and his bullet is useless) and he calmly shoots A. It's the same if A shoots C; B will then turn and shoot A. Therefore, A knows he cannot shoot either one or he will die. The outcome of this reasoning is that A puts his hands in the air. He's not worried about either B or C shooting him in this defenceless pose for if B shoots him then B would be shot by C and vice versa. As the other two are perfect logicians, neither one would do this. If A raises his hands in the air, neither B nor C will shoot each other as it wouldn't take a perfect logician to realize that once his bullet has been spent that he'd be shot by A.
Keep in mind that A is a perfect logician and not stupid enough to fire a bullet into the air. If he does this, B shoots C, runs over and picks up C's gun and shoots A... unless A can run faster than B which hasn't been stated.
If A does not shoot, then B will not shoot either. He knows that whomever he shoots will be dead and he will then be killed by the third gunfighter. He puts his hands in the air. The same goes for C.
This is why I worded the original question the way I did. It does not assume that any of the gunfighters are perfect logicians and it rightly presumes that in a real gunfight, no one would have the time to notice that someone pulling their gun would be firing to miss. I simply wanted to open a discussion and see the different lines of reasoning. There is no right answer as you would never know for certain what either of the other two gunfighters would be thinking.
C is perfect logician, and knows that he does not have a chance, unless A shoots B, so he should be aiming A.
A is a perfect logician too, and knows that he is dead as soon as he shoots B. So, his best choice is to target C, but since B is shooting C anyway, he might as well not shoot at all, save the bullet, and then shoot B, after C is gone. However, B is no idiot either, and will not shoot until he hears a shot from A. C does not have that luxury, but it does not make any more sense for him to shoot first, because as soon as he does, he'll be shot by B.
So, the answer is that nobody is going to shoot anyone, the will be just standing there forever with their hands on the guns.
I think, that the "perfect logician" assumption is fine, but the "perfect shot" one brings the duel to stand-still. A perfect logician would never participate in a duel knowing that his opponents are perfect shots.
This clip from the show talks about Game Theory, and specifically, a "Truel", or 3-way duel. It may not be exactly the situation being discussed, but it is similar enough to possibly shed some insight, and humourous besides.
Quote: DweenThis clip from the show talks about Game Theory, and specifically, a "Truel", or 3-way duel. It may not be exactly the situation being discussed, but it is similar enough to possibly shed some insight, and humourous besides.
That is a different puzzle. Again, there is a truel between A, B, and C. A has a 10% chance to hit his target, B 60%, and C 90%. A gets to go first. Who should he aim at? I assume B goes next, then C. Each person has one bullet.
But of course if you knew you were going to wind up in that situation, that's plan B.
Quote: WizardThat is a different puzzle. Again, there is a truel between A, B, and C. A has a 10% chance to hit his target, B 60%, and C 90%. A gets to go first. Who should he aim at? I assume B goes next, then C. Each person has one bullet.
That's also a different puzzle than the previous "perfect shot" scenario where all of them had 100% chances. If they're still perfect logicians and the shooters have to aim at *someone*, then based on those chances above A should play Russian roulette. He has a 90% chance of surviving, which is better than if A shoots toward anyone else. Knowing this, the other two will shoot at each other. This assumes the goal is to live, vs. to be the last standing. A's chances of being the last standing are 0 if he aims at himself...
Edit: now form the mental image of a 3-way gunfight where all the contestants draw at the same time, and where two draw on each other and the third puts the gun to his head.
Quote: MathExtremistThat's also a different puzzle than the previous "perfect shot" scenario where all of them had 100% chances. If they're still perfect logicians and the shooters have to aim at *someone*, then based on those chances above A should play Russian roulette. He has a 90% chance of surviving, which is better than if A shoots toward anyone else.
If the other two are aiming at each other, then A's chances to survive are 100% if he shoots any one of them instead of himself. If any of them is aiming at A, then his chances will still be improved by shooting at the guy aiming at him.
So, despite how intriguing it sounds, I don't think it is actually in A's best interest to play Russian roulette in this situation.
Quote: TheNightfly...A, B and C all draw in less than one second. There is no way for anoyone to have a clear advantage beyond a fraction of a second - not nearly enough time for either of the other two gunslingers to make a decision AFTER anyone else has moved. Gunfighter A is MARGINALLY faster than the other two and is MARGINALLY less accurate than the other two....
...in a real gunfight, no one would have the time to notice that someone pulling their gun would be firing to miss. ...
Let's assume that the gunfighters want to kill someone and don't want to be killed themselves, and that when the shooting starts, noone has the time to see who is shooting at whom. Also, assume that if a bullet hits someone, it kills him instantly before he can fire his gun.
B knows that he's the slowest to shoot his gun, so he knows his probability of not dying does not depend on who he tries to kill; so he's equally likely to try to kill A or C.
Shooter A must ask himself who is the bigger threat to him, C or B; it's B because B is more accurate, and A and C are equally likely to be B's target. (But, more importantly, see next line.)
Shooter C knows he's slower than A, so C's probability of dying could not be reduced by his trying to kill A, so C would try to kill B. (Shooter A would understand this reasoning, so would not plan on trying to shoot C.)
So A and C each plan on trying to kill B, and B picks A or C at random to try to kill.
A and C have equal survival probabilities. And their probabilities are higher than B's unless A's and C's shooting accuracies were considerably lower than B's.
Since gunman A knows he's fastest and gunman B is most accurate, but slowest; gunman A shoots at gunman B
Since gunman B knows he's the slowest, he cannot beat either A or C to the draw. Therefore, he shoots at gunman A, since A is the most likely to have missed with his quicker shot.
Gunman C could shoot at either A or B, but since neither of them aimed at him, he survives.
Quote: weaselmanIf the other two are aiming at each other, then A's chances to survive are 100% if he shoots any one of them instead of himself. If any of them is aiming at A, then his chances will still be improved by shooting at the guy aiming at him.
So, despite how intriguing it sounds, I don't think it is actually in A's best interest to play Russian roulette in this situation.
Let's assume that if it gets to C's turn and all three are still alive, then C will choose randomly who to shoot at. So consider the possibility that A shoots at B or C, and misses, and then B shoots at C, and misses. There is a chance that C will choose to shoot at A. So the odds of death are more than 0% if you aim at B or C. So I pose the question, what A's probability of surival under these options:
1. Deliberately misses.
2. Aims at himself.
3. Aims at B.
4. Aims at C.
If A aims at himself, his odds of success are still 10%. I admit that seems silly, especially if he can deliberately miss with 100% chance, but just take that on faith.
Quote: weaselmanIf the other two are aiming at each other, then A's chances to survive are 100% if he shoots any one of them instead of himself. If any of them is aiming at A, then his chances will still be improved by shooting at the guy aiming at him.
So, despite how intriguing it sounds, I don't think it is actually in A's best interest to play Russian roulette in this situation.
Right, but they're perfect logicians so if they know A is going to aim at one of them, they might aim back (and they have a much higher chance of killing A than he has of killing himself. :)
Quote: MathExtremistRight, but they're perfect logicians so if they know A is going to aim at one of them, they might aim back (and they have a much higher chance of killing A than he has of killing himself. :)
Aiming A won't help them survive, because A will shoot first anyway, and, once he has taken a shot, he is no threat.
Besides, even if they are aiming at A, he is better off aiming back at one of them because it'll improve his chances by 10% as opposed to making them 10% worse if he tried to shoot himself instead.
- The three participants form a triangle.
- Each has one bullet only.
- A goes first, then B, and C.
- A's probability of hitting an intended target is 10%.
- B's probability of hitting an intended target is 60%.
- C's probability of hitting an intended target is 90%.
- There are no accidental shootings.
- Shooting in the air (deliberately missing) and shooting yourself are allowed, and are always successful.
- After each round, if there are two or three survivors, then each participant is given a new bullet.
- All three participants are perfect logicians.
Rule 9 is not stated on the show, but I think can be assumed. I also assume that if all a round results in no deaths that the same strategy is repeated the next round.
Here are my probabilities of survival according to each initial target. Anybody care to dispute this before I make an "ask the wizard" question out of it?
Air 13.887%
A 0.000%
B 12.560%
C 13.094%
Quote: WizardRule 9 is not stated on the show, but I think can be assumed. I also assume that if all a round results in no deaths that the same strategy is repeated the next round.
I'm glad that you clarified Rule 9, because if the players each had only one shot and one bullet, period, then their respective strategies would be different.
I also would like to know if enough time elapses between the three shots that B, then C, would be able to change their minds once they saw the results of prior shots. Or do they just pick a target and blaze away? The latter seems more likely, so I'll answer based on that.
A is in bad shape if either opponent remains standing after the first round. He therefore wants to give himself the best chance of eliminating both opponents. If he aims at C, B might be doing the same, and if so, C dies 64% of the time. Of the remaining 36% (when A and B both miss him), C will kill B (presuming that B is C's chosen target, which seems likely) 90% of the remainder. So 64% of the time the game is reduced to A vs. B, 32.4% of the time, to A vs. C, and 3.6% of the time, everybody misses, and we do it all over.
If A aims at B, he would actually worsen his chances, because he wouldn't WANT to hit him; that would ensure that his remaining opponent would be the deadly accurate C. If A misses B, then he will be facing B as his remaining opponent 60% of the time (ignoring the possibility of a triple miss for the moment). So this is clearly a worse outcome than the roughly 2/3 possibility that his remaining opponent will be B (by aiming at C).
The next question is, does A improve his outcome by aiming at C at all? The answer would be yes, because he gives himself the best chance that his remaining opponent will be B rather than C.
B should target C for similar reasons. He would much rather have his remaining opponent be A than C. C should target B for the same reason.
So I would expect A and B to both target C. The most likely outcome would be that A would miss C, B would kill C, and then it would be B vs. A, with a 10% chance of A winning a given round, B a 54% chance, and a 36% chance of a do-over. In a B misses, C kills B scenario, the chances would be A 10%, C 81%, do-over 9%.
In either case, A's chances are greater than zero, and B's chances are better than C's (because he shoots before C), so I don't see how your percentages were derived.
Quote: mkl654321I'm glad that you clarified Rule 9, because if the players each had only one shot and one bullet, period, then their respective strategies would be different.
I also would like to know if enough time elapses between the three shots that B, then C, would be able to change their minds once they saw the results of prior shots. Or do they just pick a target and blaze away? The latter seems more likely, so I'll answer based on that.
I should have wrote that after a round where there are two or more survivors that each participant is given another bullet. Additional rounds follow the same rule about going in order, skipping anybody who was already killed. Does that help? I'd like to give others a chance to get the exact probabilities before showing my math.
A, B and C should all repeatedly shoot in the air. That makes all of their survival rates 100%.
If they are perfect logicians, they should come to this conclusion. ;)
Quote: thecesspitA, B and C should all repeatedly shoot in the air. That makes all of their survival rates 100%.
I'm not sure if I should add another rule for this, but they desire to have only one survivor. On the QI show they use the premise that they are fighting over a woman. So it would be better to die than live without her.
Quote: WizardI'm not sure if I should add another rule for this, but they desire to have only one survivor. On the QI show they use the premise that they are fighting over a woman. So it would be better to die than live without her.
Which contradicts the premise that the three duellists are perfect logicans.
If they were French, they'd just all move into the same house, and A would have her on Mondays and Thursdays, B on Tuesdays and Fridays, C on Wednesdays and Saturdays, and Sunday would be her day of rest.
Quote: mkl654321Which contradicts the premise that the three duellists are perfect logicans.
If they were French, they'd just all move into the same house, and A would have her on Mondays and Thursdays, B on Tuesdays and Fridays, C on Wednesdays and Saturdays, and Sunday would be her day of rest.
Nice :-). I guess I'll have to add a rule that they are not French. How about if they were German?
Quote: WizardNice :-). I guess I'll have to add a rule that they are not French. How about if they were German?
Same time-based solution, just that they would all live in different houses.
Quote: WizardI'm not sure if I should add another rule for this, but they desire to have only one survivor. On the QI show they use the premise that they are fighting over a woman. So it would be better to die than live without her.
Ah the all wish to win rather than merely survive :)
http://www.youtube.com/watch?v=sXldafIl5DQ
Quote: Toes14Too much thinking for me - I pick Clint Eastwood, he always wins those Mexican Standoffs! Here's the finale of The Good, The Bad, & the Ugly from Youtube. It's one of my favorite movie scenes of all time. The background music really heightens the tension.
Here is an old thread I'd like to reawaken. I just watched The Good, The Bad, & the Ugly, in no small part due to my interest in the game theory behind duels and in this case a truel (three-way duel).
Let me set the scene. A lot of money is buried in a grave in a remote cemetery, with LOTS of graves. The "good," "bad," and the "ugly" are all there, but none of them trust each other. Only the Good knows the grave. However, he can't just dig it up, because then the bad and/or ugly will shoot him and take the money.
How to resolve this Mexican standoff? They have a truel. I assume the goal is that two are to be killed, so the only survivor gets the money. Before they start the Good allegedly writes down the name of the drive on a rock in the middle of the triangle.
Here is what I think happens:
1. Bad tries to shoot Good.
2. Before he gets the shot off Good shoot Bad, but doesn't kill him.
3. Wounded, Bad tries to shoot Good again, but Good kills him before he has the chance.
4. It seems that Ugly was ready to shoot Bad, but his gun had no bullets (Good took them out without his knowledge).
5. Good has a clear shot at Ugly, but chooses not to shoot him.
6. The Good reveals he never wrote anything on the rock.
I'm not sure what my question is, but I'd be interested in any commentary of the scene. The movie seems to show all three as being good shots, especially Ugly. I have a feelings there is a good puzzle somewhere in this scene, but I don't understand the dynamics of real life duels well enough to formulate the question.
So, I'd be interested in a deeper analysis of the scene. What do you guys say?
probably the best one. Eli Wallach made the movie great.
The ending was a typical spaghetti western Mexican Standoff.
A ploy by the writer to have the audience on the edge of
their seats at the very end. I thought the title gave it away. The
Good guy, Blondie. The Bad guy, Angel Eyes. The Ugly guy,
Tuco. Good guy had to win, that was the whole point. I don't
know what the puzzle would be. Give us a hint.
Quote: EvenBobI don't know what the puzzle would be. Give us a hint.
I'm not sure where to begin. How about, what are the basic rules of a dual? Assuming you were a good shot, what would be the reason not to shoot first?
Quote: WizardI'm not sure where to begin. How about, what are the basic rules of a dual?
I don't think there are any rules, thats why its a standoff.
Everything is equal, anything can happen, including walking
away. Thats where the suspense happens, the unknown
conclusion. A regular dual has lots of rules you have to abide,
not so a Mexican Standoff. Its every man for himself.
Quote: WizardWhy do they always just stand there staring each other down for a long time in the Westerns?
Because a Mexican Standoff is technically a stalemate, the
chance of you winning is equal to you losing. Just jumping
in could get you killed because you don't have the edge. I
think they stand there looking for an edge. I remember
seeing that movie in a drive in in 1966, and we thought the
ending was just terrific. All Clint Eastwood's westerns were
big hits. We would buy those thin cigars and in our minds
we were rough and tough.
Quote: EvenBobBecause a Mexican Standoff is technically a stalemate, the
chance of you winning is equal to you losing.
How do you know?
Quote: WizardHow do you know?
"A Mexican standoff is a slang term defined as a stalemate or impasse; a confrontation that neither side can foreseeably win."
Wikipedia
"Noun 1.Mexican standoff - a situation in which no one can emerge as a clear winner."
Free Dictionary
Casino gambling is the opposite of a Mexican Standoff.
