Centipede game in normal form

In the unique subgame perfect equilibrium, each player chooses to defect at every opportunity. This, of course, means defection at the first stage. In the Nash equilibria, however, the actions that would be taken after the initial choice opportunities even though they are never reached since the first player defects immediately may be cooperative.

Defection by the first player is the unique subgame perfect equilibrium and required by any Nash equilibrium , it can be established by backward induction. Suppose two players reach the final round of the game; the second player will do better by defecting and taking a slightly larger share of the pot.

Since we suppose the second player will defect, the first player does better by defecting in the second to last round, taking a slightly higher payoff than she would have received by allowing the second player to defect in the last round. But knowing this, the second player ought to defect in the third to last round, taking a slightly higher payoff than she would have received by allowing the first player to defect in the second to last round.

This reasoning proceeds backwards through the game tree until one concludes that the best action is for the first player to defect in the first round. The same reasoning can apply to any node in the game tree. In the example pictured above, this reasoning proceeds as follows. If we were to reach the last round of the game, Player 2 would do better by choosing d instead of r. However, given that 2 will choose d , 1 should choose D in the second to last round, receiving 3 instead of 2.

Given that 1 would choose D in the second to last round, 2 should choose d in the third to last round, receiving 2 instead of 1. But given this, Player 1 should choose D in the first round, receiving 1 instead of 0.

There are a large number of Nash equilibria in a centipede game, but in each, the first player defects on the first round and the second player defects in the next round frequently enough to dissuade the first player from passing. Being in a Nash equilibrium does not require that strategies be rational at every point in the game as in the subgame perfect equilibrium.

This means that strategies that are cooperative in the never-reached later rounds of the game could still be in a Nash equilibrium. In the example above, one Nash equilibrium is for both players to defect on each round even in the later rounds that are never reached.

Another Nash equilibrium is for player 1 to defect on the first round, but pass on the third round and for player 2 to defect at any opportunity. Several studies have demonstrated that the Nash equilibrium and likewise, subgame perfect equilibrium play is rarely observed. Instead, subjects regularly show partial cooperation, playing "R" or "r" for several moves before eventually choosing "D" or "d".

It is also rare for subjects to cooperate through the whole game. As in many other game theoretic experiments, scholars have investigated the effect of increasing the stakes.

As with other games, for instance the ultimatum game , as the stakes increase the play approaches but does not reach Nash equilibrium play. Since the empirical studies have produced results that are inconsistent with the traditional equilibrium analysis, several explanations of this behavior have been offered.

Rosenthal suggested that if one has reason to believe her opponent will deviate from Nash behavior, then it may be advantageous to not defect on the first round.

One reason to suppose that people may deviate from the equilibria behavior is if some are altruistic. The basic idea is that if you are playing against an altruist, that person will always cooperate, and hence, to maximize your payoff you should defect on the last round rather than the first.

If enough people are altruists, sacrificing the payoff of first-round defection is worth the price in order to determine whether or not your opponent is an altruist. Passing strictly decreases a player? If the opponent also passes, the two players are faced with the same choice situation with reversed roles and increased payoffs.

The game has a finite number of moves which is known in advance to both players. In the above diagram, a 1 at a black circle "decision node" denotes a decision opportunity for player 1.

A 2 at a decision node tells us that person 2 can make a decision here. The top number at the end of each vertical line is a payoff for player 1 and the bottom number is a payoff for player 2. Player 1 has the first move: if she chooses D, both players get 1; if she chooses A, the opportunity to make a decision passes to player 2.

Player 2 has the second move: if he chooses D, player 1 gets payoff of D and he gets 3; if he chooses A, the opportunity to make a decision passes to player 1. And so on to the end of the game tree. If both players always choose A, they both receive payoff of at the end of the game tree. We have just observed that both players receive payoff of if both players always choose A rather than D.

Note also that both players receive payoff of 1 if player 1 chooses D on his first move. What does game theory predict will happen? Game Theory predicts that Player 1 will choose D in his first move and thus both players will receive payoff of 1!

How can that be the case? Click here to learn how using backward induction to solve a game in extensive form leads to the prediction player 1 will choose D on his first move. Furthermore,since all Nash equilibria make the same outcomes prediction, any usual refinements of Nash equilibrium also make the same prediction.

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