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ECMC02H
Prof. G. Cleveland
Problem Set for Game Theory
1. The game of “chicken” is played by two macho teenagers (with more money and spare time
than brains) who speed towards each other on a single-lane road. The one who swerves to avoid
a collision is labelled a chicken, whereas the one who does not swerve is considered to be
fearless and admirable. If neither swerves, both will be hurt (perhaps die) but if both swerve,
both participants lose some face. Payoffs to the chicken game are shown below:
Person #2
Person
#1
Swerve
Do not swerve
Swerve
(1,1)
(0,10)
Do not
swerve
(10, 0)
(-100, -100)
(a) Does this game have a Nash equilibrium? What is it and why?
(b) Is a threat by either player not to swerve a credible threat? Why or why not?
(c) Would the ability of one player to firmly commit to a no-chicken strategy (for example, by
locking the steering wheel into the straight-ahead position and throwing away the key) be
desirable for that player?
2. Two firms are considering bringing out competing brands of cigarettes with a special
ingredient. Payoffs to the companies are given below:
Firm #2
Produce
Firm #1
Do not produce
Produce
(3, 3)
(5, 4)
Do not
produce
(4, 5)
(2, 2)
(a) Does this game have a Nash equilibrium? What is it and why?
(b) If one of these players could move first (in a sequential game), would there be any first-mover
advantages for either player? Explain
(c) Might Firm #2 find it in its interests to bribe Firm #1 to stay out of the market?
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3. The Wave Energy Technology company has a monopoly on the production of a particular
type of highly-desirable waterbed. Demand for these beds is relatively inelastic…at a price
of $1000 per bed, 25,000 will be sold while at a price of $600, 30,000 will be sold. The only
costs associated with waterbed production are the initial costs of building a plant. This
company has already invested in a plant capable of producing up to 25,000 beds and this is a
sunk cost (and therefore irrelevant for current pricing decisions).
(a) suppose a would-be entrant to this industry could always be assured of getting half the
market, but would have to invest $10 million in a plant to accomplish this. Construct a
sequential game showing the payoffs to this firm from the entry/no entry decision. What is
the Nash equilibrium and why?
(b) Suppose the Wave Energy Technology firm could invest $5 million more to enlarge its
existing plant so that it would produce 40,000 beds (another sunk cost). Would this strategy
be a profitable way to deter entry by its rival? Explain.
4. Assume now that mixed strategies are possible – not just pure strategies. Take a simple
“zero-sum” game. In zero-sum games, the payoffs to both players are assumed to sum to
zero, so that one player's gain is inevitably another player's loss. These games correspond to
most of the sports which we play and/or watch, since generally one side either wins or loses
(or there is a tie). Thus, if we designate the outcome as +1 for a win, 0 for a tie, and -1 for a
loss, you can see that two players must always have their results "sum to zero". These games
do not allow for collusion, since there is no possible agreement that will make both parties
better off.
a) Begin with a simple game of "matching" in which two players simultaneously put out either
one or two fingers. One of the players - Jon - has chosen "evens" in advance; the other player
- Sue - has chosen "odds." If the number of fingers put out by the two players matches, then
Jon wins; if they do not match, then Sue wins. Set out the payoff matrix for this game. Is
there a Nash equilibrium in pure strategies?
b) Now suppose that we allow each player to choose a mixed strategy. In a mixed strategy, the
player chooses to play randomly, with a probability assigned to each "pure" strategy.
Surprisingly, this is often the best alternative. Suppose that Jon plays the mixed strategy of
putting out one finger with probability PJ and two fingers with probability (1-PJ). Show that
if Sue can figure out this strategy and respond with the best possible strategy, that Jon's
maximin will occur at PJ = 0.5. Then show that if the positions are reversed and Sue chooses
PS (defined the same way for her as PJ is for Jon), that her best choice is PS = 0.5. Explain
the "common sense" of this result.
5.
Suppose that in a football game, the offense can choose one of two strategies - run or pass while the defense can choose one of two strategies - run defense or pass defense. The results
of the play are as follows: if the offense runs, it gains 3 yards against a run defense and 9
yards against a pass defense; if the offense passes, it gains 15 yards against a run defense and
0 yards against a pass defense. Show that a mixed strategy is best for both sides. Explain the
common sense of this result.
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Solutions
1.
2.
3.
4. The matching game is summarized in the zero sum payoff matrix below (where the payoffs to
Jon are shown first):
Sue
One
Finger
Jon
Two
Fingers
One
Finger
+1, -1
-1, +1
Two
Fingers
-1, +1
+1, -1
There is no Nash equilibrium; this accords with our intuition. Each player will refuse to continue
with a simple strategy if the other person is winning, so no simple strategy can be a stable result.
Suppose instead that Jon plays the mixed strategy of putting out one finger with probability PJ
and two fingers with probability (1-PJ). Then Sue's payoff will be
R1 = (-1)(PJ) + (+1)(1-PJ) = 1 - 2PJ if she puts out one finger, and
R2 = (+1)(PJ) + (-1)(1-PJ) = 2PJ - 1 if she puts out two fingers.
If PJ < 0.5, then Sue obtains a positive result by always putting out one finger;
if PJ > 0.5, then Sue obtains a positive result by always putting out two fingers. Only if Jon
chooses PJ = 0.5 will Sue end up with a zero result (which also means a zero result for Jon, since
this is a zero sum game).
The argument for Sue is identical, but it is worth going through it. Suppose that Sue plays the
mixed strategy of putting out one finger with probability PS and two fingers with probability (1PS). Then Jon's payoff will be
R1 = (+1)(PS) + (-1)(1-PS) = 2PS - 1 if he puts out one finger, and
R2 = (-1)(PS) + (+1)(1-PS) = 1 - 2PS if he puts out two fingers.
If PS < 0.5, then Jon obtains a positive result by always putting out two fingers;
if PS > 0.5, then Jon obtains a positive result by always putting out one finger. Only if Sue
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chooses PS = 0.5 will Jon end up with a zero result (which also means a zero result for Sue).
The common sense of this is obvious. In the game of matching, if you can pick up any pattern in
the other player's choices, then you can exploit that pattern. If you are playing "evens" and, at
any point, you figure out that the other player is "likely" to put out one finger (meaning that the
probability of putting out one finger is greater than 0.5), then of course you put out one finger so
as to win most of the time. The only way to avoid this is for each player to employ a completely
random strategy.
5. Consider first the decisions by the defense. If the defenders defend against the run, you always
pass; if the defenders defend against the pass, you always run. Of course, no football team would
behave so stupidly on defense (that is, no one would reveal exactly how they are going to play).
Suppose now that the defense chooses the run defense with probability PRD and thus chooses the
pass defense with probability 1 - PRD (of course, the two probabilities must add to one). Then,
if the offense runs, the expected gain is
GR = (+3)(PRD) + (+9)(1-PRD) = 9 - 6PRD
If the offense passes, the expected gain is
GP = (+15)(PRD) + (+0)(1-PRD) = 15PRD
The two results are equal (that is, GR=GP) if 9 - 6PRD = 15PRD or PRD = 3/7
When PRD = 3/7, GR = GP = 6.43
The defense will always choose PRD = 3/7 because if they do not, the offense can do better than
6.43 (assuming that the offense is only interested in the expected result - for football fans, you
know that there are short yardage situations and end of game situations that dictate different
strategies!). If the defense chooses PRD > 3/7, then the offense will always pass because
GP > GR, and this will hurt the defense because GP > 6.43 On the other hand, if the defense
chooses PRD < 3/7, then the offense will always run because GR > GP, and this will hurt the
defense because GR > 6.43. Thus the best that the defense can do is to choose PRD = 3/7.
Now consider the decisions by the offense. If the offense runs, you always defend against the
run; if the offense passes, you always defend against the pass. Again, no football team would
behave so stupidly on offense. Suppose now that the offense the run with probability PR and thus
chooses the pass with probability 1 - PR Then, if the defense chooses the run defense, the
expected gain is GRD = (+3)(PR) + (+15)(1-PRD) = 15 - 12PRD
If the defense chooses the pass defense, the expected gain is
GPD = (+9)(PR) + (+0)(1-PR) = 9PR
The two results are equal (that is, GRD=GPD) if 15 - 12PR = 9PR or PR = 5/7
When PR = 5/7, GRD = GPD = 6.43
The offense will always choose PR = 5/7 because if they do not, the defense can do better than
6.43 If the offense chooses PR > 5/7, then the defense will always defend against the run
because GRD < GPD, and this will hurt the offense because GRD < 6.43 On the other hand, if the
defense chooses PR < 5/7, then the defense will always defend against the pass because
GRD > GPD, and this will hurt the offense because GPD < 6.43. Thus the best that the offense can
do is to choose PR = 5/7.
The common sense of the mixed strategy is that it will generally hurt you in any sport to be too
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predictable!