Solving the Mexican Standoff
The Mexican standoff, as epitomised by the finale of ‘The Good, the Bad and the Ugly’, is a game theory paradox which is widely regarded as unsolveable. In business terms, Mexican stand-off can be applied to a competitive tender where three bidders have just two rounds, in which one is eliminated in the first, in which to make their case. More familiarly, the semi-final of ‘The Weakest Link’ generally features a similar standoff, where, by playing non-altruistically, players who are not the best player can maximise their chances.
The situation is as follows. Three gunmen stand in a wide, flat area. Their guns are (in the classic Mexican standoff, but, we shall explore, not in the film version) each filled with six bullets. One is the fastest draw, another is the slowest draw, and by common agreement, the third is somewhere in between. Assuming that if two shoot at each other, the fastest will kill the other before he gets a shot off, and, assuming that it takes more time for the fastest to fire a second shot than it did for the slowest to fire a single shot, what are the optimal strategies for each shooter, and is there a guaranteed winner if all play rationally? Further, are there other strategies — for example, playing irrationally — which will benefit a player whose chances are otherwise poor?
Working from the first shot, it is impossible to solve the paradox. If the fastest fires, whichever person he shoots will die, and the other, provided he shoots the fastest, will live. But this gives both other players merely a fifty-percent chance of survival, with no confidence as to which player the fastest will choose to kill.
The key to the puzzle is to look at the second round. The result of the first round could be that only one player is left alive — for example, if the fastest shoots the second fastest and the slowest shoots the fastest — or it could be that two players are left alive — for example, if both the fastest and the slowest both shoot the middle player.
Looked at from the perspective of the slowest player, the second round is always a loss, because whoever else is left alive, the slowest player then loses. Therefore, the slowest player, if acting rationally, must take whatever course of action is necessary to ensure that there is only one round. But there is only one possible choice for this to be the case, and in which the slowest survives. That combination is when the fastest shoots the second fastest, thereby killing him before he gets off a shot, and the slowest shoots the fastest. In this arrangement, the choice of the second fastest is irrelevant. If the fastest shoots the slowest, then the slowest is dead anyway, and if the fastest and the second fastest shoot each other, the result is that the fastest and slowest go through to the second round, resulting in a certain loss for the slowest.
Therefore, a slowest player acting rationally must shoot the fastest player in the first round. The fastest may choose to shoot him, or may choose to shoot the second fastest, but the slowest player cannot influence this decision, and therefore must accept the notional 50% chance of survival, as against 0% in any other combination.
However, if the fastest player is able to reason this, then he knows that the slowest player will shoot him. Therefore, acting rationally, he must shoot the slowest player.
The second player can make his own decision without having to follow this reasoning. Simply, if he shoots the fastest player, then he faces the slowest in the second round. If he shoots the slowest player, he faces the fastest in the second round, which he will certainly lose. Therefore, the second player must shoot the fastest player, hoping that the fastest will shoot the slowest. As it happens — whether the second player knows this or not — the fastest must do this if he is acting rationally.
The result of this is that the second player always wins if the players are acting rationally, and if there are no other factors. The fastest player cannot ‘trick’ this, by, for example, acting irrationally and shooting the second player, since he is still shot by the slowest player acting rationally, and the slowest player never gains anything by deliberately acting irrationally (that is to say, by varying his game), since all other combinations inevitably lead to his death. The middle player has nothing to gain by acting irrationally — he is the winner if all players act rationally, and loses in any combination if the fastest chooses, irrationally, to shoot him, or if the slowest chooses, irrationally, to shoot him, since that then leaves him facing the fastest player in the second round.
Clearly this is not a good game for the fastest player to be entering. He would be better to try to configure some kind of two stage one on one knockout contest, which he will be certain to win in both rounds.
In the film ‘The Good, the Bad and the Ugly’, the character Blondie (aka, the man with no name) is the fastest player, and is the one who proposes the game. In an extraordinarily tense five minutes on screen, the fighters eye each other up, until the second fastest, Angel Eyes (‘the bad’) draws and, rationally, shoots Blondie. However, instead of shooting Tuco (‘the ugly’ — the slowest player), Blondie shoots Angel Eyes. It’s not clear whether Tuco would have acted rationally by shooting Blondie at this point, or would have remembered that Angel Eyes (‘the bad’) is not to be trusted, shooting him, because, in the event, it turns out that Blondie removed Tuco’s bullets the night before, thereby reducing the game to a one-on-one, which Angel Eyes should (acting rationally) not have accepted if he knew the real state of affairs.
However, there are other possible actions which the fastest player can take. In the Weakest Link version of the standoff, all three players get to vote together, with full knowledge of their strengths up to that point (if they have been paying attention), but the strongest player gets to tie-break. In this version, players vote off the weakest link. Acting altruistically, both the strongest and middle players will vote off ‘the weakest link’, who is the third player. But, if the players play rationally, the middle player will vote off the strongest, in the hope that the weakest player will also vote that way, giving a clear 2:1 majority with no tie-break, and leaving him to face the weakest rather than the strongest in the final round. The weakest player must vote for either of the other two. If he can count on the middle player voting rationally rather than altruistically, he will vote against the strongest player, as this will put him into the final, whereas if everyone votes altruistically, then he has lost anyway. There is a slight combination here, in that if the weakest player votes altruistically, and the middle player votes rationally, then the result is a three way tie-break, and the strongest player may be sufficiently displeased that he votes off the middle player rather than the weakest player. But, if the strongest really is the strongest (which is not as clear in the game as in the gunfight), then facing the strongest in the final will result in a loss, and the famous ‘you leave with nothing’, for the middle player which ever way he votes.
In both these games, counter-intuitively, the strongest player inevitably loses if the players play rationally. In the Weakest Link version of the game, the strongest player does not have the opportunity to remove the weakest-player’s bullets. However, in both versions, the strongest player can attempt to disguise himself as a player of a different strength, as can the other two.
In the gunfighting version of the game, the result is then as follows. If all players play rationally, based on the information available to them, then if the strongest player is able to pretend he is the second strongest, he wins, because the weakest and second strongest shoot each other, and he shoots the second strongest. In fact, he can shoot either, because he will still win in the second round. If disguising himself as second strongest is too difficult, he can disguise himself as the weakest — for example, by turning up with his hand in a bandage. In this case, the apparently second strongest shoots the real second strongest, who is shooting the true strongest disguised as the weakest, who is shooting him back. The apparent second strongest’s bullets only strike the apparent strongest after the real strongest’s bullets have already killed him. The strongest player then faces the weakest player in the second round, and wins.
The second strongest player gains nothing by disguising himself, if the other two are undisguised, because he wins anyway.
If the weakest disguises himself as the strongest, he loses, because the real strongest and the second strongest both shoot him. However, if he is able to talk up his position a little, to the extent that he now appears to be the second strongest, then the strongest and real second strongest will shoot each other, and he will shoot the real strongest, thereby winning the game, as there will be no more rounds.
If the players behave entirely irrationally, that is to say, if they have no accurate picture of who is the strongest, then the result is statistical rather than logical. Using the letters S, M, W for Strong, Middle and Weak, and > for a shot from left to right, and < for a shot from right to left, the combinations are as follows:
1) S>M, M>W, W>S = W wins, as there is no second round, and the bullets of S have killed M before he gets a shot off
2) S>M, M>W, W>M= S wins, as M kills W, leaving him to face S in the second round
3) S>M, M>S, W>S = W wins, as this is the same result as case 1
4) S>M, M>S, W>M= S wins, this results is a second round where S will defeat W
5) S>W, M>W, W>S= S wins on the second round
6) S>W, M>W, W>M=S wins on the second round
7) S>W, M>S, W>S=M wins, as in the ideal case where all act rationally
S>W, M>S, W>M=M wins, since W is dead before he fires
In the non-rational case, therefore, S wins 50% of the time, M wins 25% of the time, and W wins, surprisingly, 25% of the time.
Although S is clearly the most likely to benefit from this arrangement, it is the least advantageous to him except for the fully rational version with no disguise, where he inevitably loses.
If he can organise the game differently, into a knockout-type competition, then he will win each round against either opponent.
If he is able to disguise himself, then he wins, provided that the others do not disguise themselves.
In a fully random version, he wins 50% of the time.
In the fully rational version, he always loses, as does the weakest player.
An alternative scenario — which is to some extent what appears to be happening at the end of the Good, the Bad and the Ugly — is collusion between the strongest and weakest, who are aware that, playing fully rationally, they will both lose. However, the price of this, for the weakest player, must be that there is to be no second round, as he will inevitably lose this.
The Mexican Standoff, in this version, differs from rock-paper-scissors because, although the players can vary their strategies, they cannot vary the order of precedence of their skills.