Backwards induction
The concept of backwards induction corresponds to the assumption that it is common
knowledge that each player will act rationally at each node where he moves — even if
his rationality would imply that such a node will not be reached.
1
Mechanically, it is
computed as follows. Consider a finite horizon perfect information game. Consider any
node that comes just before terminal nodes, that is, after each move stemming from this
node, the game ends. If the player who moves at this node acts rationally, he will choose
the best move for himself. Hence, we select one of the moves that give this player the
highest payoff. Assigning the payoff vector associated with this move to the node at
hand, we delete all the moves stemming from this node so that we have a shorter game,
where our node is a terminal node. Repeat this procedure until we reach the origin.
∗
These notes do not include all the topics that will be covered in the class. See the slides and
supplementary notes for a more complete picture.
1
More precisely: at each node i the player is certain that all the players will act rationally at all
nodes j that follow node i; and at each node i the player is certain that at each node j that follows
node i the player who moves at j will be certain that all the players will act rationally at all nodes k
that follow node j,...ad infinitum.
1
Example Consider the following wellknown game, called as the centipedes game. This
game illustrates the situation where it is mutually beneficial for all players to stay in
a relationship, while a player would like to exit the relationship, if she knows that the
other player will exit in the next day.
• 1
• 2
• 1
A
D
(1,1)
a
d
(0,4)
α
δ
(3,3)
(2,5)
In the third day, player 1 moves, choosing between going across (α) or down (δ). If
he goes across, he would get 2; if he goes down, he will get 3. Hence, we reckon that he
will go down. Therefore, we reduce the game as follows:
• 1
• 2
A
D
(1,1)
a
d
(0,4)
(3,3)
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In the second day, player 2 moves, choosing between going across (a) or down (d). If
she goes across, she will get 3; if she goes down, she will get 4. Hence, we reckon that
she will go down. Therefore, we reduce the game further as follows:
2
• 1
A
D
(1,1)
(0,4)
Now, player 1 gets 0 if he goes across (A), and gets 1 if he goes down (D). Therefore,
he goes down. The equilibrium that we have constructed is as follows:
• 1
• 2
• 1
A
D
(1,1)
a
d
(0,4)
α
δ
(3,3)
(2,5)
That is, at each node, the player who is to move goes down, exiting the relationship.
Let’s go over the assumptions that we have made in constructing our equilibrium.
We assumed that player 1 will act rationally at the last date, when we reckoned that he
goes down. When we reckoned that player 2 goes down in the second day, we assumed
that player 2 assumes that player 1 will act rationally on the third day, and also assumed
that she is rational, too. On the first day, player 1 anticipates all these. That is, he is
assumed to know that player 2 is rational, and that she will keep believing that player
1 will act rationally on the third day.
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 Spring '12
 Sjostrom
 Game Theory, player, Nash, Subgame perfect equilibrium, backwards induction

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