Thank you for the questions.
1)
The key here is that there is a notion of "information set" in an
extensive form game tree. This is a set of nodes that look alike, at
the time it is a person's move. Information sets in an extensive form
game tree are denoted by dotted lines connecting the nodes in the
information set.
Unfortunately it is really hard to draw game trees in ascii. Thus, I
hope that you will bear with me if I simply describe the tree.
The tree has 13 nodes and three levels. The moves for player 1 are
a,b,c and d. The moves for player two are A, B, C, and D.
The nodes of the tree are labelled 0 through 12.
The root of the tree is labelled 0.
The player 1 moves from the root are as follows:
There is an edge labelled a from 0 to 1.
There is an edge labelled b from 0 to 2.
There is an edge labelled c from 0 to 3.
There is an edge lablled d from 0 to 4.
There is a dotted line connecting nodes 3 and 4: these nodes are in
the same information set.
The player 2 moves are as follows:
There is an edge labelled A from 3 to 5
There is an edge labelled B from 3 to 6
There is an edge labelled C from 3 to 7
There is an edge labelled D from 3 to 8
There is an edge labelled A from 4 to 9
There is an edge labelled B from 4 to 10
There is an edge labelled C from 4 to 11
There is an edge labelled D from 4 to 12
To find the payoffs at the terminal nodes, we just check the
corresponding values in the matrx, where the rows of the matrix are
labelled a through d and the columns are labelled A through D.
Node 0 is a root node and has no payoff.
Node 1 has payoff (2,2)
Node 2 has payoff (2,2)
Nodes 3 and 4 are interior nodes and have no payoff
Node 5 is reached when player 1 chooses c and player 2 chooses A, so
its payoff is 3,3
Node 6 is reached when player 1 chooses c and player 2 chooses B, so
its payoff is 3,3
Node 7 is reached when player 1 chooses c and player 2 chooses C, so
its payoff is 2,0.
Node 8 is reached when player 1 chooses c and player 2 chooses D, so
its payoff is 0,0.
Similarly, the payoffs for nodes 9, 10, 11, and 12 are respectively:
9: (3,3)
10: (3,3)
11: (0,0)
12: (3,3)
One way to think about how this game is played is as follows.
Player 1 writes down a letter a-d on a piece of paper. If the letter
is a or b, he deals the paper face up and each player gets 2 dollars.
Otherwise, he deals the paper face down. Now player 2 has to write a
letter from A through D on a piece of paper. After player 2 shows the
paper to player 1, player 1 turns over his paper, and both players
collect the payoff given by the row/column in the payoff matrix
associated with their plays.
2)
Surprisingly, the answer to this question is yes: there is a subgame
perfect Nash equilibrium.
Consider the following strategy profile for player 1:
i. On the first round, play B. On the second round, if in the
previous round (B,R) was played, then play T. Otherwise play M.
For player 2, use the following strategy profile:
ii. On the first round, play R. On the second round, if in the
previous round (B,R) was played, then play L. Otherwise play C.
Let us call this strategy profiles s1 and s2 respectively, so that the
full strategy is s=(s1,s2).
Note that the payoff using this strategy is (4,4)+(3,1)=(7,5), since
B,R is played in the first round and (T,L) in the second.
Furthermore, note that (T,L) and (M,C) are the only pure strategy Nash
equilibria for a single round.
We need to prove two things:
A. (s1,s2) is a Nash equilibrium.
B. (s1,s2) is subgame perfect.
To show A, we need to show that if player 1 unilaterally changes his
strategy his payoff will not increase: he cannot make more than 7.
Whatever player 1 plays on the first round, player 2 will reply L or
C. Since (T,L) and (M,C) are Nash equilibria for one round, player 2
cannot do better on round 2 by varying his play unilaterally.
What if player 1 varies on round 1? If he plays T, then (T,R) will be
played in round 1, earning him a payoff of 5. In round 2, player 2
will play C, and so player 1 cannot make more than 1 in round 1. So
his maximum payoff if he varies is 6, which is worse than if he
stayed.
So player 1 cannot improve by varying.
Similarly, player 2 cannot improve by varying. As above, player 2
cannot vary in round 2. If player 2 chooses another move in round 1,
his max payoff in round 1 will be 2, and since player 1 will
subsequently play M, his max payoff in round 2 is also 2, so his total
max payoff is only 4.
Hence, (s1,s2) is indeed a Nash equilibrium.
To show B, we need to show that whatever happens in the first round,
the strategy (s1,s2), applied only to the second round, is a Nash
equilibrium. This is clear because (T,L) and (M,C) are each Nash
equilibria, and whatever happens in the first round, one of those two
will be played in the second round.
Search Strategy
---------------
I continued to use the text:
Game Theory Evolving, by Herbert Gintis. Princeton University Press,
Princeton, 2000.
I also looked at a number of sites on subgame perfect equilibrium,
none of which were very helpful. One reasonable one was the course
notes: http://www.nada.kth.se/~rjo/course/game/chap6/game_theory_ch6.ppt
; however, this is in powerpoint format and requires a plugin from
microsoft to read; I can forward the URL of the plug-in if you like,
but presumably you already have access to good definitions of these
terms.
I found both of these problems deceptively challenging. For problem 1,
I kept trying to find a game tree with full information and without
simultaneous moves, that is with no nontrivial "information sets". It
turns out that I doubt this is even possible - so I certainly learned
a lot better what an "information set" was.
For problem 2 I had to play around with a lot of different strategies.
I finally found a strategy I thought would work because it was a Nash
equilibrium, but it turned out not to be subgame-perfect. The key
thing was realizing that the moves a player made in the second round
depend, not only on his opponent's move in the first round, but on his
*own* move; otherwise, the subgame perfect definition will not hold. |