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Homework 10
1. For this question, use the extensive form game shown above. Find the behavioral strategy of player 1 that is equivalent to her mixed strategy in which she plays
(B, r) with probability 0.4, (B, l) with probability 0.1, and (A, l) with probability 0.5.
Show formally that the mixed strategy and the behavioral strategy that you found
are equivalent, i.e. that they induce the same probability distribution on Z for any
general strategy of player 2.
Solution: The behavioral strategy must specify two distributions ((p, 1−p), (q, 1−
q)), one for each information set of Player 1.
(1 − q) =
p=
probability of reaching the first information set and choosing B with s
probability of reaching the first information set with s
probability of reaching the second information set choosing l with s
probability of reaching the second information set with s
=
0.5
0.5
= 0.4 + 0.1 = 0.5.
= 1.
We now need to show that this behavioral strategy is indeed equivalent. For this,
number the nodes at z1 , z2 , z3 , z4 , z5 , z6 . Consider any general strategy of the other
player. Since this is a game of perfect recall, a general strategy is equivalent to a
behavioral strategy (r1 , r2 , 1 − r1 − r2 ). If Player 1 plays the given mixed strategy
against this behavioral strategy, his payoff is
u(a1 , r)p(a1 ) + u(a2 , r)p(a2 ) + u(a3 , r)p(a3 ) + u(a4 , r)p(a4 ),
where a1 , a2 , a3 , a4 denote Player 1’s pure strategies. With the given probabilities, we
get u(a1 , r).4 + u(a2 , r).1 + u(a3 , r).5 + u(a4 , r)0 = u(z6 ).5 + u(a3 , r).5 = u(z6 ).5 +
(r1 ∗ u(z1 ) + r2 ∗ u(z2 ) + r3 ∗ u(z4 )).5.
Now consider Player 1 playing against (r1 , r2 , 1 − r1 − r2 ) with the behavioral
strategy we derived. His utility is u(z1 ) ∗ .5 ∗ r1 + u(z2 ) ∗ .5 ∗ r2 ∗ 1 + u(z3 ) ∗ .5 ∗
r2 ∗ 0 + u(z4 ) ∗ .5 ∗ (1 − r1 − r2 ) ∗ 1 + u(z5 ) ∗ .5 ∗ (1 − r1 − r2 ) ∗ 0 + u(z6 ).5 =
u(z1 ) ∗ .5 ∗ r1 + u(z2 ) ∗ .5 ∗ r2 + u(z4 ) ∗ .5 ∗ (1 − r1 − r2 ) + u(z6 ) ∗ .5, which is exactly
the same.
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2. For this question, use the extensive form game shown above. Show that the
highest expected payoff player 1 can guarantee with a behavioral strategy is 1/4,
while there is a mixed strategy that guarantees him an expected payoff of 1/2.
Solution: Consider a behavioral strategy ((p1 , 1 − p1 ), (p2 , 1 − p2 )) that Player
1 plays against (q, 1 − q), a general strategy of the other player. Player 1’s payoff is
p1 p2 q + (1 − p1 )(1 − p2 )(1 − q). To make it irrelevant what Player 2 plays, it needs to
p2
p1
= 1−p
. It follows that p1 = p2 .
be the case that p1 p2 = (1 − p1 )(1 − p2 ). Thus, 1−p
1
2
(If p1 > p2 , then 1 − p1 < 1 − p2 , and p1 /(1 − p1 ) > p2 /(1 − p2 ).) Hence player 1’s
payoff is p21 q + (1 − p1 )2 (1 − q). Again, to make it irrelevant what Player 2 plays, it
needs to be the case that p21 = (1 − p1 )2 . It follows that p1 = 1/2. Hence, the player’s
payoff from the behavioral strategy is 1/4 ∗ q + 1/4 ∗ (1 − q) = 1/4.
If Player 1 plays a mixed strategy, his payoff is s(l, l)q + s(r, r)(1 − q). To make
it irrelevant what Player 2 plays, player 1 should set s(l, l) = s(r, r) = 1/2, and
probabilities on all other pure strategies equal to 0. His payoff then is 1/2.
3. Formulate the following parlor game as an extensive game with imperfect information. First player 1 receives a card that is either H or L with equal probabilities.
Player 2 does not see the card. Player 1 may announce that her card is L, in which
case she must pay $1 to player 2, or may claim that her card is H, in which case player
2 may choose to concede or to insist on seeing player 1’s card. If player 2 concedes
then he must pay $1 to player 1. If he insists on seeing player 1’s card then player 1
must pay him $4 if her card is L and he must pay her $4 if her card is H. Find the
Nash equilibria of this game.
Solution: The parlor game is pictured below:
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The strategic form is:
The first two pure strategies are strictly dominated. Finding the mixed strategy
Nash Equilibrium is left as an easy exercise.
4. Prove formally that every game satisfying perfect recall is linear.
Solution: Assume an EFG has perfect recall for all i. Suppose by contradiction
there exist x1 ∈ u, x2 ∈ u, x1 6= x2 s.t. x1 , x2 ∈ P (z) for some z ∈ Z.
Assume without loss of generality that x2 % x1 . Let a be the action that leads
from x2 to x1 . Since x1 and x2 are in the same information set, by perfect recall,
there exists another node x3 such that a leads from x3 to x2 .
We can hence construct a chain
xn →a xn−1 →a ... →a x3 →a x2 →a x1 .
Since the game is finite, eventually the x’s have to start repeating. But then we
will have xn % xk and xk % xn for some xn 6= xk , which is a contradiction with the
order being antisymmetric.
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