The game of Chicken models two drivers, both headed for a single-lane bridge from opposite directions. The first to swerve away yields the bridge to the other. If neither swerves, the result is a potentially fatal head-on collision. If one stays straight while the other swerves, then he wastes less time, and also gains face as a very macho fellow and all teenage girls immediately have the hots for him.
| Swerve | Straight | |
| Swerve | -1, -1 | -1, +2 |
| Straight | +2, -1 | -13, -13 |
| Chicken with numerical payoffs. | ||
The optimal strategy for each player (according to game theory) is to go straight with probability P and swerve with probability 1-P.
And keep your randomness secret – if the opponent knows your random choice then it effectively is a deterministic choice, not random at all, and the opponent then always wins the game! If you want to be deterministic, then use P=0 or P=1 for genuine determinism – don't go for fake determinism.Your strategy is then defined by a single number P. Say your opponent's strategy similarly is defined by a single number also (call it Q). Then your expected payoff is
An unstable optimum situation is P=Q=0.2. In this situation both players get expected payoff=-1. It is "optimum" in the sense that neither player can do better by unilaterally changing their strategy.
If you, instead of P=0.2, use P=0.1999 (slightly perturbed) then your opponent will exploit that error by making Q=1 (always going straight). You then will suffer more (i.e. get hugely worse expected payoff, namely -3.3988 instead of -1). So you are incentivized to keep P≥0.2.
The reason the original situation is "unstable" is that if one player slightly increases P i.e, instead of P=0.2, use P=0.2001 (now perturbed in the increasing direction) then your opponent can strategically exploit that by making Q=0, i.e. always swerving. But in the event your opponent always swerves, then you are being an idiot – you could have gained hugely more payoff by going straight (payoff=+2 rather than -0.3997).
The reason the original situation is "unstable" is that if one player slightly increases P above 0.2, the opponent is best off always swerving, at which point the first player is best off always going straight. So a tiny perturbation slams us into the state where we have one "macho guy" who always wins payoff=+2 and never swerves, versus one "wimp" who always swerves and always gets payoff=-1. There are two stable macho-V-wimp states – it's just a question of which one you want to keep being.
In chicken games played between political parties or candidates (which can arise in certain voting systems), most parties consider it unacceptable to always be the "wimp" and always lose the election to the "macho" party. But if both parties try to be macho, the result is disaster (i.e. they both lose).
As a contemporary example, I am writing this in mid-March 2008. Hillary Clinton and Barack Obama are competing for the Democratic US Presidential Nomination. Hillary chose to be "macho" by employing "negative ads" about her opponent Obama, and also to try to get unelected "superdelegates" to vote for, and install, her even if she loses the popular vote; and also to try to change the rules midgame about Florida and Michigan. These moves may result in her defeating Obama, but also risk disaster – voters will then prefer the Republican McCain to a "Democratic" party that behaves in this manner. If Obama also chooses to be "macho" then that danger is increased. However, there is, as you can imagine, considerable pressure on both to avoid being "wimps," e.g. television commentators every day are urging each to "show the American public that they can be strong, fight, and not just lie down and take it" (that is almost an exact quote).