ECON5111:
Economics of Strategy
Sequential games – Threats, oligopoly,
bargaining and conflict
Sequential games with complete info
• So far: simultaneous-move games where all players act at once without observing
the actions of other players
• Good way to model interactions in which players literally decide simultaneously, or
decide at different times but without knowing what the other has done
• Many strategic interactions are dynamic: A moves, B sees what A has done and
then moves, etc.
• This section: sequential games of complete information – players know each
other’s payoffs perfectly
• Later: sequential games of incomplete information
Experiment: Ultimatum game
• One of you is Player 1 and the other is Player 2.
• Player 1 has before him a “pie” of size 20EC$. He decides whether to give a little
(5EC$) or a lot (15EC$) of the pie to Player 2.
• Player 2 observes what Player 1 has offered and has the choice to accept or reject.
• If Player 2 accepts, Player 2 gets what was offered and Player 1 gets 20 minus
what he offered.
• If Player 2 rejects, BOTH PLAYERS GET 0.
• Player 1s: write “give $5” or “give $15”
• Player 2s: write what you would do in response to BOTH possible actions by Player 1
• E.g. Accept $5, Reject $15
Experiment: Ultimatum game
PLEASE PAUSE THE
VIDEO AND
PARTICIPATE IN THE
EXPERIMENT
BEFORE
CONTINUING
Ultimatum game in extensive form (the game tree)
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)
Ultimatum game in extensive form (the game tree)
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)Player 1’s
actions
Ultimatum game in extensive form (the game tree)
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)Player 1’s
actions
Player 2’s
actions
Ultimatum game in extensive form (the game tree)
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)Player 1’s
actions
Player 2’s
actions
Payoffs (Player
1’s listed first,
Player 2’s listed
second)
Decision
nodes
Ultimatum game
• Representing in normal form (payoff matrix)?
• To do so, recall the definition of strategy: a description of what the player would do
at every possible point in the game at which that player could do something
• Player 1’s strategies: Give 15, Give 5
• Player 2’s strategies: Accept all offers, reject all offers, accept high offer and reject
low offer, accept low offer and reject high offer
The matrix
Player 2
Accept all A high, R low R high, A low Reject all
Player 1
$15 5, 15 5, 15 0, 0 0, 0
$5 15, 5 0, 0 15, 5 0, 0
Ultimatum game
• Notice that “reject all” is a strictly dominated strategy for Player B – the strategy
“accept all” always gives B higher payoff
• Notice that “reject high, accept low” and “reject low, accept high” are weakly
dominated strategies for B – the strategy “accept all” gives B payoffs that are
always either greater or equal
Introducing the game matrix
Player 2
Accept all A high, R low R high, A low Reject all
Player 1
$10 5, 15 5, 15 0, 0 0, 0
$5 15, 5 0, 0 15, 5 0, 0
Nash equilibria of the ultimatum game
• This game has three Nash equilibria in pure strategies:
1. Player 1 offers $5 and Player 2 accepts all offers
2. Player 1 offers $15 and Player 2 accepts high offers and rejects low offers
3. Player 1 offers $5 and Player 2 rejects high offers and accepts low offers
• Do any of these equilibria not make sense?
Nash equilibria of the ultimatum game
• This game has three Nash equilibria in pure strategies:
1. Player 1 offers $5 and Player 2 accepts all offers
2. Player 1 offers $10 and Player 2 accepts high offers and rejects low offers
3. Player 1 offers $5 and Player 2 rejects high offers and accepts low offers
• Do any of these equilibria not make sense?
• Equilibria 2 and 3 are inadmissible – Player 2 is playing weakly dominated strategies
• Can we eliminate them on even more intuitive grounds?
• Can we eliminate them more simply – without writing matrix?
• Process for finding only the “reasonable” NE in a sequential game: backward
induction
• Uses game tree instead of matrix (in more complex games, writing the matrix is not feasible)
• What are “reasonable” NE? More on this later
Backward induction in the ultimatum game
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)
At the last stage, what would player 2 do?
Backward induction in the ultimatum game
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)
At the last stage, what would player 2 do?
15 is better than 0, so
he’d accept a high
offer.
5 is better than 0, so
he’d accept a low
offer.
Backward induction in the ultimatum game
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)
In the first stage, what would Player 1 do, knowing
Player 2’s payoffs and so knowing how Player 2 would
act in the last stage?
Backward induction in the ultimatum game
Player 1 Player 2
Reject
Offer 5
Offer 15
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)
In the first stage, what would Player 1 do, knowing
Player 2’s payoffs and so knowing how Player 2 would
act in the last stage?
Since Player 2 would
always accept any offer,
Player 1 has the choice of
offering 5 and getting 15
or offering 15 and getting
5. 15>5, so Player 1
would offer 5.
Backward induction
1. At the last stage of the game, what would the player making a choice in that stage
do, if he ever arrives to that stage?
2. At the second-to-last stage of the game: knowing the answer to #1, what would
the player acting here do, if he ever arrives to that stage?
…….
3. At the first stage of the game, knowing how everyone would act in all subsequent
stages, what would the player who acts first do in this stage?
In sum: work backwards from the end of the game.
Subgame-perfect Nash equilibria
• Only one of the 3 Nash equilibria are found through backward induction:
Equilibrium 1, where Player 1 offers $5 and Player 2 accepts all offers
• This equilibrium refinement is called subgame perfection. Nash equilibria found
through backward induction are subgame perfect Nash equilibria (SPNE).
Credible and non-credible threats
• Nash equilibria of sequential games may feature non-credible threats; subgame
perfect Nash equilibria do not
• A non-credible threat is a strategy in which a player “threatens” to take an action
at a certain stage of the game, but if he ever did arrive to that stage, taking that
action would not be in his interest
• In equilibrium 2, Player 2’s strategy to reject low offers is not a credible threat – once a low
offer has been made, Player 2 is better off accepting it.
• The only reason Player 1 offers $15 in equilibrium 2 is that Player 1 believes Player 2’s non-
credible threat to reject $5.
• Subgame perfect Nash equilibria (SPNEs) are those in which players don’t make
non-credible threats, and other players don’t believe non-credible threats.
An important note
• Notice that the equilibrium is “Player 1 offers $5, Player $2 accepts all offers”, not
“Player 1 offers $5 and Player 2 accepts $5”.
• The second isn’t an accurate description of this or any equilibrium:
• We need to know what Player 2 would do if $15 was offered, even if that
doesn’t happen in equilibrium.
• Any Nash equilibrium must specify a strategy per player, and a strategy
includes what that player would choose at every stage of the game in which he
can act… even if he’ll never have to make certain choices in equilibrium.
Stackelberg competition
• Sequential version of Cournot competition seen in week 2
• One of the two firms, the leader, chooses q1. The other firm, the follower, observes
q1 and chooses his quantity, q2
• Both firms are choosing their quantities to maximize their respective profits,
keeping in mind that their profits depend on their quantity choice AND
competitor’s quantity choice
• Suppose inverse market demand is P = 100 – 2Q
• Suppose the leader’s total costs are 4q1 and the follower’s are 4q2.
• What quantities will they choose?
Stackelberg competition
• Use backward induction: start from the last stage of the game
• In the last stage of the game, the follower has observed the leader’s choice of q1. He then chooses
his quantity to maximize:
π2= revenue – cost
= [100 – 2(q1 + q2)]q2 – 4q2
= 100q2 – 2q1q2 – 2q2
2 – 4q2
Differentiating with respect to q2 and setting equal to 0 gives us
96 – 2q1 – 4q2 = 0
q2 = 24 – 0.5q1
This is the follower’s reaction function
Stackelberg competition
• Now we know what the follower would do in the last stage of the game. Anticipating that, what would the
leader do?
• In the first stage of the game, the leader chooses his quantity to maximize:
π1= revenue – cost
= [100 – 2(q1 + 24 – 0.5q1)]q1 – 4q1
= 100q1 – 2q1
2 – 48q1 + q1
2 – 4q1
= 48q1 – q1
2
Differentiating with respect to q1 and setting equal to 0 gives us:
48 – 2q1 = 0
q1 = 24
Substituting this into the follower’s reaction function gives us q2 = 24 – 0.5(24) = 12
The follower’s reaction function
Stackelberg competition
• The SPNE: Leader makes q1 = 24, Follower makes q2 = 24 – 0.5q1
• Leader’s profit = [100 – 2(24+12)]*24 – 4*24 = 576
• Follower’s profit = [100 – 2(24+12)]*12 – 4*12 = 288
• The leader has an advantage: he makes more profits than the follower and more profits than in the
simultaneous Cournot competition case (from lecture 2, these profits are 512)
• If one firm in a duopoly competing in quantity can act first in a visible way for its opponent, it
benefits from doing so
• Notice that after observing the follower’s choice of 12, the leader would like to retroactively lower
his own quantity. Specifically, if the follower is making 12,
Leader’s profit = [100 – 2(q1 + 12)]q1 – 4q1
Differentiating with respect to q1 and setting equal to 0 gives us
q1 = 18
Stackelberg competition
• If the leader sets q1 = 18 and the follower sets q2 = 12, the leader’s profits are
[100 – 2(18 + 12)]*18 – 4*18 = 648 > 576
• So why doesn’t the leader just make 18 in the first place?
Stackelberg competition
• If the leader sets q1 = 18 and the follower sets q2 = 12, the leader’s profits are
[100 – 2(18 + 12)]*18 – 4*18 = 648 > 576
• So why doesn’t the leader just make 18 in the first place?
• Because the follower wouldn’t respond to that by making 12! The follower
would then produce q2 = 24 – 0.5(18) = 15
• That would give the leader profits of
[100 – 2(15+18)]*18 – 4*18 = 540 < 576
Stackelberg competition
• Notice that after observing the follower’s choice of 12, the leader would like to
retroactively lower his own quantity.
• From lecture 2: one way to think of Nash equilibrium is that no player wants to
switch their strategy once the strategies of all others are revealed
• Q: So how can this be a Nash equilibrium (much less a subgame perfect one)?
Stackelberg competition
• Q: So how can this be a Nash equilibrium (much less a subgame perfect one)?
• A:
• Each player is indeed best responding to the other’s strategy
• The leader’s equilibrium strategy is q1 = 24
• The follower’s equilibrium strategy is q2 = 24 – 0.5q1, NOT q2 = 12
• Remember that a strategy specifies what a player would do at every point in the game at which
he may have to make a decision
• So a strategy for the follower in this game has to specify what q2 he would choose in response
to any possible q1
• The follower’s best response function is the equivalent of “accept all offers” for player 2 in the
ultimatum game
A game of entry deterrence
• Suppose the following sequential game:
• at the first stage an entrant firm can choose to enter the market or not
• at the second stage the incumbent can choose to “fight” or “accommodate”
• at the third stage, if the incumbent accommodates, the two firms play a Stackelberg game with
the incumbent as leader
• Inverse market demand is P = 100 – 2Q
• Both firms’ costs are 4q
• If the entrant does not enter, the incumbent is a monopolist and makes profits 1152 (calculated in
lecture 2), while the entrant makes profits of 100 (outside option – e.g. resale value of capital)
• If the incumbent fights, he makes a quantity that would induce the entrant to produce 0
A game of entry deterrence
• What would the incumbent have to produce to induce the entrant to produce 0?
• Entrant’s reaction function is q2 = 24 – 0.5q1= 0 => q1 = 48
• If the incumbent makes 48, the entrant makes 0 and therefore has profits of 0
• What is the incumbent’s profit if he makes 48 and the entrant makes 0?
[100 – 2(48+0)]*48 – 4*48 = 0
• Therefore, if the entrant enters and the incumbent fights, both make 0 output AND
therefore 0 profit.
• If the entrant enters and the incumbent accommodates, the incumbent makes 576
and the entrant makes 288 (from previous calculations in Stackelberg game)
The entry deterrence game in extensive form
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
0, 0
q1
q2
Payoff for
entrant?, Payoff
for incumbent?
The entry deterrence game in extensive form
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
288, 576
0, 0
Solving by backward induction
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
288, 576
0, 0
• Incumbent’s threat to
produce 48 is not credible.
In that stage of the game,
he would accommodate.
Solving by backward induction
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
288, 576
0, 0
Anticipating that the incumbent
would accommodate, the entrant
prefers to get 288 to getting his
outside option of 100, so he would
enter.
A game of entry deterrence
• We first used backward induction to find the outcome of the Stackelberg game the
two firms play if one enters and the other accommodates
• Again by backward induction, we then see whether the incumbent would choose
to fight or accommodate
• Finally, we see whether the entrant would choose to enter or not knowing how
the rest of the game would go if he entered and how it would go if he didn’t
• Unique SPNE: Incumbent accommodates and makes 24, entrant enters and makes
24 – 0.5q1.
• How do we find SPNE in games where backward induction isn’t possible?
The entry deterrence game, version 2
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
Entrant and incumbent play
Cournot game
0, 0
The entry deterrence game, version 2
• This new game has sequential components as well as simultaneous components
• Backward induction doesn’t help because there isn’t one player who moves last in the Cournot game
• Subgame perfect Nash equilibrium: A Nash equilibrium is subgame perfect if it is also a Nash
equilibrium of every subgame of the larger game
• Subgame (in simple terms): a part of the game that could exist as a game in its own right (even a
game as simple as one person making a binary choice counts as a game)
• A subgame that starts with a decision node must contain all the nodes that follow that decision node
• At any node in a subgame, players know for sure they are in THAT subgame (relevant later on when
we introduce incomplete information)
Subgames of the second entry deterrence game
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
Entrant and incumbent play
Cournot game
0, 0
NOT subgames!
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
Entrant and incumbent play
Cournot game
0, 0
The entry deterrence game, version 2
• The Nash equilibrium of the Cournot subgame is that they both produce 16 (and
both gets profits of 512)
The entry deterrence game, version 2
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
512, 512
0, 0
The entry deterrence game, version 2
• The Nash equilibrium of the Cournot subgame is that they both produce 16 (and
both gets profits of 512)
• The Nash equilibrium of the subgame where the incumbent chooses to fight or
accommodate and they play Cournot if he accommodates is that he accommodates
and both produce 16
The entry deterrence game, version 2
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
512, 512
0, 0
The entry deterrence game, version 2
• The Nash equilibrium of the Cournot subgame is that they both produce 16 (and
both gets profits of 512)
• The Nash equilibrium of the subgame where the incumbent chooses to fight or
accommodate and they play Cournot if he accommodates is that he accommodates
and both produce 16
• The SPNE of the larger game is that the entrant chooses to enter, the incumbent
chooses to accommodate, and they both choose quantity 16 in the Cournot game
The entry deterrence game, version 2
Incumbent
Entrant
Enter
Not enter
Accommodate
Fight
100, 1152
512, 512
0, 0
Subgames of the ultimatum game
Player 1 Player 2
Reject
Offer 5
Offer 10
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)
NOT subgames!
Player 1 Player 2
Reject
Offer 5
Offer 10
Accept
Accept
Reject
(5, 15)
(0, 0)
(15, 5)
(0, 0)
The importance of the structure of the game
• Changing a simultaneous move game into a sequential game may change the
outcome
• Changing the order in which decisions are made may also change the outcome
Legislature and CB
• Two players: Legislature and Central Bank
• Legislature can run a budget deficit or not
• Central Bank can set low interest rate or set high interest rate
• Legislature has a dominant strategy to run a budget deficit (provide public goods or tax breaks to key
interest groups → get reelected)
• All else equal, Legislature prefers low rates
• Central Bank prefers to keep interest rates low if the budget is balanced, but prefers to keep them
high if the budget is in deficit (counteracting the effects of expansionary fiscal policy with tight
monetary policy and vice versa)
• All else equal, CB prefers balanced budget
• Outcome in simultaneous move game?
Legislature and CB
Legislature
Deficit Balanced
Central
Bank
High 2, 2 3, 1
Low
1, 4 4, 3
Legislature and CB
Legislature
Deficit Balanced
Central
Bank
High 2, 2 3, 1
Low
1, 4 4, 3
• CB does not have a
dominant strategy:
what is best
depends on what
Legislature does
Legislature and CB
Legislature
Deficit Balanced
Central
Bank
High 2, 2 3, 1
Low
1, 4 4, 3
• Legislature has
dominant strategy:
deficit
• Outcome of
simultaneous game:
Deficit, High rates
• Not Pareto efficient:
would both be
better off if CB
player Low and L
played Balanced
CB moves first
LegislatureCentral Bank
Balanced
Low
High
Deficit
Deficit
Balanced
(2, 2)
(3, 1)
(1, 4)
(4, 3)
CB moves first
LegislatureCentral Bank
Balanced
Low
High
Deficit
Deficit
Balanced
(2, 2)
(3, 1)
(1, 4)
(4, 3)
CB moves first
LegislatureCentral Bank
Balanced
Low
High
Deficit
Deficit
Balanced
(2, 2)
(3, 1)
(1, 4)
(4, 3)
SPNE: Legislature runs
deficit no matter what
CB did. CB sets high
rate
Same (non Pareto
efficient) outcome as
in the simultaneous-
move game!
L moves first
Legislature Central Bank
Low
Balanced
Deficit
High
High
Low
(2, 2)
(4, 1)
(1, 3)
(3, 4)
L moves first
Legislature Central Bank
Low
Balanced
Deficit
High
High
Low
(2, 2)
(4, 1)
(1, 3)
(3, 4)
L moves first
Legislature Central Bank
Low
Balanced
Deficit
High
High
Low
(2, 2)
(4, 1)
(1, 3)
(3, 4)
SPNE: CB sets low
rates if budget is
balanced, high if
deficit. L runs balanced
budget
Different order of
moves → different
outcome!
CB and L: from simultaneous to sequential
• In the simultaneous game, the L had a dominant strategy to run deficit
• In the sequential game in which L moves first, deficit is no longer a dominant
strategy
• This is because in the sequential game, their choice can influence the choice of
the other player
• The outcome has both the L and the CB better off – (3, 4) vs (2, 2)
Changing the order of decisions: the trial
• There is a man accused of a capital offense: depending on the outcome of the trial
he will be acquitted, sentenced to life in prison or sentenced to death
• There are 3 possible ways in which he can be tried:
1. Status quo method: first determine innocence or guilt, then decide
punishment
2. Roman way: after evidence is presented, start from most severe punishment
(death) and decide if it should be imposed. If not, work down the list in order of
decreasing severity deciding if each subsequent punishment is the appropriate
one.
3. Mandatory sentencing: first determine the appropriate punishment for the
crime, then decide guilt or innocence
The judges
• Whichever trial method is used, there are 3 judges and decisions will be made by
majority vote between them
Judge A’s ranking Judge B’s ranking Judge C’s ranking
Best Death penalty Life imprisonment Acquittal
Middle Life imprisonment Acquittal Death penalty
Worst Acquittal Death penalty Life imprisonment
Status quo method
• Guilt or innocence is determined, then punishment is determined
• How will the judges decide on the question of guilt or innocence?
• By first looking at the last stage of the game (backward induction)
• In the last stage of the game (if it is reached) punishment is determined. A and C
prefer death penalty to life imprisonment, so death penalty will be the chosen
punishment.
• In light of this, how will the judges decide on guilt vs. innocence?
• B and C prefer acquittal to death penalty
• Outcome: acquittal
Roman method
• Go down the list starting from harsh punishments to determine what should be
imposed
• In the second stage of the game when life imprisonment is on the table, the choice
is effectively between life imprisonment and acquittal
• A and B prefer life imprisonment to acquittal, so that will be the outcome
• This means that when the death penalty is on the table, the choice is effectively
between death penalty and life imprisonment
• A and C prefer death penalty to life imprisonment
• Outcome: death penalty
Mandatory sentencing method
• First determine punishment, then determine guilt vs. innocence
• After punishment is determined, the choice in the second stage is between the
chosen punishment and acquittal
• Between death penalty and acquittal, B and C prefer acquittal – acquittal is chosen
• Between life imprisonment and acquittal, A and B prefer life imprisonment – life
imprisonment is chosen
• So in the first stage, the choice is between life imprisonment and acquittal
• Outcome: life imprisonment
More choice is always better?
• “Incredible threats” – player making the threat knows he wouldn’t carry if out if
that subgame was reached
• Knows other players know it too and thus don’t “believe him”
• The player making the threat may actually be worse off because he has a choice
• Could do better by tying his hands in advance of the strategic interaction
A leftist legislature and a rightist president
• 1980s, United States
• The (left-wing) Congress wants to a program in urban renewal (U)
•
• The (right-wing) president wants an antiballistic missile system (M)
• Both the Congress and the President would prefer to have both than to get neither
of the two
Outcome Congress’ ranking President’s ranking
Both U & M 3 3
U 4 1
M 1 4
Neither 2 2
Traditional veto
U
Veto
Veto
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE?
Traditional veto
U
Veto
Veto
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE?
Traditional veto
U
Veto
Veto
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE?
Traditional veto
U
Veto
Veto
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE: President signs U&M and M,
vetos U. Congress passes U&M.
• Outcome: Both urban renewal and
antiballistic missile system are
funded
• Payoffs: Congress and President
both get 3 (their second-best
option)
Line-item veto
U
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE?
Veto M
(1, 4)
(4, 1)
Line-item veto
U
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE?
Veto M
(1, 4)
(4, 1)
Line-item veto
U
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE?
Veto M
(1, 4)
(4, 1)
Line-item veto
U
(3, 3)
(2, 2)
(4, 1)
(2, 2)
(2, 2)
(2, 2)
(1, 4)
Congress
President
• SPNE: President vetos U, signs M,
and line-item vetos U from the
U&M proposal. Congress either
doesn’t pass anything or passes U.
• Outcome: nothing gets passed
• Payoffs: Congress and President
both get 2 (their third-best option)
Veto M
(1, 4)
(4, 1)
The power of backward induction
• Much like it is not always practical to represent a complex simultaneous game in a
payoff matrix, it is not always practical to depict a sequential game in a tree
• This may be because actions are continuous (like in Stackelberg game) or simply
because the game is very complex or long
• Next, we will see some games like this, and see how backward induction can
relatively quickly resolve them even without us needing to draw the tree
• Goal: application to protracted processes of negotiation, bargaining and conflict
Thought experiment: Capture the flags
• 2 teams, A and B, one spokesperson from each team
• There are 11 flags to be taken
• Each team can take either 1 or 2 flags per turn (0 or 3 not allowed). Team A starts
• Winning team is the one that takes the LAST FLAG (by itself or as one of 2 flags)
• Each member of winning team gets 10EC$
• 2 rounds in total
Thought experiment discussion
• Who wins?
• Backward induction – start from the end of the game
• Team that is left with 2 or 1 flags wins for sure
• How to make the other team leave you with 2 or 1 flags?
• Leave them with 3 flags on your previous turn
• If they take 1, they leave you with 2 and you win
• If they take 2, they leave you with 1 and you win
Thought experiment discussion
• Team A wants this to happen at the end
Team moving Flags left at end of
move
Flags taken during
move
A
B
A
B
A
B
A 0 1 if 1 left
2 if 2 left
Thought experiment discussion
• Team A wants this to happen at the end
• To make sure B leaves A with either 1 or 2
flags, B has to be left with 3
• If B takes 1, A then takes 2 and wins
• If B takes 2, A then takes 1 and wins
• If A leaves B with 3, A wins for sure
• How can A ensure it will leave B with 3?
Team moving Flags left at end of
move
Flags taken during
move
A
B
A
B
A 3
B 2 or
1
1 or
2
A 0 1 if 1 left
2 if 2 left
Thought experiment discussion
• If A leaves B with 3, A wins for sure
• How can A ensure it will leave B with 3?
• By leaving B with 6 at the previous turn
• When left with 6, B can take 1 or 2
• If B takes 1, leaves A with 5. A can take 2
and leave B with 3, ensuring win
• If B takes 2, leaves A with 4. A can take 1
and leave B with 3, ensuring win
• How can A ensure it will leave B with 6?
Team moving Flags left at end of
move
Flags taken during
move
A
B
A 6
B 5 or
4
1 or
2
A 3 1 if 4 left
2 if 5 left
B 2 or
1
1 or
2
A 0 1 if 1 left
2 if 2 left
Thought experiment discussion
• How can A ensure it will leave B with 6?
• By leaving B with 9 at the previous turn
• When left with 9, B can take 1 or 2
• If B takes 1, leaves A with 8. A can take 2
and leave B with 6. So A can leave B with
3. So A can win for sure
• If B takes 2, leaves A with 7. A can take 1
and leave B with 6. So A can leave B with
3. So A can win for sure
• How can A ensure it leaves B with 9?
Team moving Flags left at end of
move
Flags taken during
move
A 9
B 8 or
7
1 or
2
A 6 1 if 7 left
2 if 8 left
B 5 or
4
1 or
2
A 3 1 if 4 left
2 if 5 left
B 2 or
1
1 or
2
A 0 1 if 1 left
2 if 2 left
Thought experiment discussion
• How can A ensure it leaves B with 9?
• By taking 2 in its very first move
• Once A has taken 2 in the very first move, a win
by B is impossible as long as A follows the right
strategy (leaving B with 6, then 3)
• First-mover advantage
• If A takes 1 in its very first move, now B can
leave A with 9 flags, therefore 6 flags, therefore
3 flags… therefore B can win
• Wrong first move = no chance of recovery
if B plays smart
• Right first move = guaranteed win if A
plays smart
Team moving Flags left at end of
move
Flags taken during
move
A 9 2
B 8 or
7
1 or
2
A 6 1 if 7 left
2 if 8 left
B 5 or
4
1 or
2
A 3 1 if 4 left
2 if 5 left
B 2 or
1
1 or
2
A 0 1 if 1 left
2 if 2 left
The pirate game
• Players: 5 pirates, called A, B, C, D and E
• Are trying to decide how to split 100 gold coins
• The structure of the game:
1. A proposes a distribution.
2. The 5 pirates vote on the distribution (in case of a tie, the proposer A is the tie-breaker)
3. If proposal passes, coins are split and game ends
4. If proposal does not pass, A is thrown overboard and B proposes a distribution next.
5. Etc.
The pirate game
• Players: 5 pirates, called A, B, C, D and E
• Are trying to decide how to split 100 gold coins
• The structure of the game:
1. A proposes a distribution.
2. The 5 pirates vote on the distribution (in case of a tie, the proposer A has the casting vote)
3. If proposal passes, coins are split and game ends
4. If proposal does not pass, A is thrown overboard and B proposes a distribution next.
5. Etc.
• Each pirate:
• Wants to live
• Conditional on living, wants to maximise his number of gold coins
• Conditional on living and having the same number of gold coins, would prefer to throw another pirate
overboard
• What is the outcome of this game?
The pirate game: last stage
• In the last stage (if it is reached), every pirate except D and E has been thrown
overboard.
• D has to make a proposal, but has no incentive to offer anything to E: even if E votes
against, D votes for, and as proposer has the tie-breaking casting vote
• In the last stage, D offers 100 to himself and 0 to E. The proposal passes.
• What happens in the second-to-last stage?
The pirate game: second-to-last stage
• In the last stage, D offers 100 to himself and 0 to E. The proposal passes.
• What happens in the second-to-last stage?
• C proposes a split to D and E. C knows that D knows he can get 100 by throwing him
overboard and proposing at the next stage – no sense for C to try to win over D
• But C knows that if his proposal is rejected, E will get 0 in the next round: can offer E
1 coin to make E prefer to accept the proposal
• In the second-to-last stage, C offers 99 to himself, 0 to D and 1 to E. The proposal
passes.
• What happens in the third-to-last stage?
The pirate game: third-to-last stage
• In the last stage, D offers 100 to himself and 0 to E. The proposal passes.
• In the second-to-last stage, C offers 99 to himself, 0 to D and 1 to E. The proposal
passes.
• What happens in the third-to-last stage?
• B proposes a split to C, D and E. Needs 1 on his side for proposal to pass. No sense in
trying to get C. Can get E but only by offering him 2. Can get D by offering him just 1
• In the third-to-last-stage, B offers 99 to himself, 0 to C, 1 to D, 0 to E. The proposal
passes.
• What happens in the first stage?
The pirate game: the first stage
• In the last stage, D offers 100 to himself and 0 to E. The proposal passes.
• In the second-to-last stage, C offers 99 to himself, 0 to D and 1 to E. The proposal passes.
• In the third-to-last-stage, B offers 99 to himself, 0 to C, 1 to D, 0 to E. The proposal passes.
• What happens in the first stage?
• A proposes a split to B, C, D and E. Needs 2 others on his side. No sense in appealing to B. Can get D
but only by offering him 2. Can get C and E by offering them 1.
• In the first stage, A proposes 98 for himself, 0 for B, 1 for C, 0 for D, 1 for E. The
proposal passes.
Lessons from the pirate game
• No proposing pirate ever makes an offer that will not pass (the consequences of a
“no” vote are dire)
• As a result, the game ends with A’s offer: minimal time wasted negotiating
• But what makes a given offer “sure to pass” is how it compares to the alternative,
which is what would happen in the next stage if it were to fail!
• The “off equilibrium path” parts of the game – those never reached in equilibrium
- are often crucial to determining what the equilibrium is
Bargaining
• 2 players, Player A and Player B, bargain over a “pie” of size 1
• Players take turns making offers and counteroffers
• If an offer is accepted, the game ends. If an offer is rejected, the “rejecter” makes a
counter-offer
• Players may be impatient: they discount future periods by a factor d, where ½ < d < 1
• Game may go on for 1, 2, infinite stages
• If an agreement isn’t reached at the end of the last stage of the game, both get 0
Bargaining: 1-stage
• Simplest/shortest game: 2 players, Player A and Player B
• They are bargaining over a “pie” of size 1
• First stage: Player A makes an offer (x, 1-x), where x is the share of the pie he gets
and 1-x is the share of the pie he offers Player B
• Player B either accepts or rejects. If he accepts, he gets 1-x and Player A gets x. If he
rejects, both get 0.
1-stage bargaining in extensive form
Player A
Player B
x, 1-x
Accept
Reject
x, 1-x
0, 0
Simple bargaining
• In the last stage of the game, what would Player B do?
• Player B would accept any offer greater than or equal to 0.
• Given Player B’s strategy, Player A would offer the smallest possible amount, 0, and
Player B would accept.
• SPNE: Player A offers 0, Player B accepts all offers.
Bargaining: 2-stage
• Stage 1: Player A offers and Player B can accept or reject. If B accepts, game ends
and they split the pie according to A’s proposal.
• Stage 2: If B rejects, B makes an offer. If A accepts, game ends and they split the pie
according to A’s proposal. Last stage – if A rejects, both get 0.
• Players are impatient – dx today is as good as x in the next period
• d is a discount factor, ½ < d < 1
Bargaining: 2-stage
• What would happen in the last stage where A must accept or reject B’s offer?
• A would accept an offer of 0, which would give B the full pie – 1.
• What would happen in the first stage where A makes an offer?
• Note that B can always reject A’s offer and get 1 in the next period.
• A must offer something big enough to prevent B from rejecting.
• A would offer d. d today is as good as 1 tomorrow for B, so B would accept.
• Outcome: A offers d in the first stage, B accepts. A gets 1-d, B gets d.
Bargaining: 3-stage
• Stage 1: A offers and B can accept or reject. If he accepts, game ends and they split
the pie according to A’s proposal.
• Stage 2: If B rejects, B makes an offer. If A accepts, they split the pie according to B’s
proposal.
• Stage 3: A makes an offer. If B accepts, they split the pie according to A’s proposal.
Last stage – if B rejects, both get 0.
• Players are impatient – dx today is as good as x in the next period
• d is a discount factor, ½ < d < 1
Bargaining: 3-stage
• What would happen in the 3rd stage of the game, where A makes an offer?
• A would offer 0 and B would accept, giving A the full pie – 1.
Bargaining: 3-stage
• What would happen in the 3rd stage of the game, where A makes an offer?
• A would offer 0 and B would accept, giving A the full pie – 1.
• What would happen in the 2nd stage of the game, where B makes an offer?
• Note that A can always reject and get 1 in the 3rd stage of the game.
• B must make A an offer big enough to prevent him from rejecting.
• B would offer A d – it’s as good for A to get d today as to get 1 tomorrow. B would have 1-d left
over for himself.
Bargaining: 3-stage
• What would happen in the 1st stage of the game, where A makes an offer?
• Note that B can always reject and get 1-d in the 2nd stage of the game.
• A must make B an offer big enough to prevent him from rejecting.
• A would offer B d(1-d) – it’s as good for B to get d(1-d) today as to get 1-d tomorrow. A would
have 1-d(1-d) left over for himself.
• B would accept!
Bargaining: Many stages
Number of stages A gets B gets
1 1 0
2 1-d d
3 1-d(1-d) = 1-d+d2 d(1-d) = d-d2
4 1-d(1-d(1-d)) = 1-d+d2-d3 d(1-d(1-d)) = d-d2+d3
5 1-d+d2-d3+d4 d-d2+d3-d4
6 1-d+d2-d3+d4-d5 d-d2+d3-d4+d5
In each game, A makes the first offer. The outcome is that the first offer is
accepted. These are the distributions of the pie in games of 1-6 stages:
A little trick with geometric series
In the 6-stage game, A gets 1-d+d2-d3+d4-d5. Call this x6.
dx6 = d(1-d+d
2-d3+d4-d5) = d-d2+d3-d4+d5-d6
x6 + dx6 = (1+d)x6 = 1-d+d
2-d3+d4-d5
+ d-d2+d3-d4+d5-d6
Every term but the 1 and –d6 cancel out, giving us (1+d)x6 = 1- d
6
Therefore x6, the payoff A gets in a 6-stage game, is (1-d
6)/(1+d)
B gets 1-x6, or 1 -(1-d
6)/(1+d) = (1+d-1+d6)/(1+d) = (d+d6)/(1+d)
Where n is an even number (A starts but B finishes):
In an n-round game, A gets (1-dn)/(1+d)
In an n-round game, B gets (d+dn)/(1+d)
The case of infinite stages
In an n-round game, A gets (1-dn)/(1+d)
In an n-round game, B gets (d+dn)/(1+d)
Since d∞ = 0,
In an infinite-stage game, Player 1 gets 1/(1+d)
In an infinite-stage game, Player 2 gets d/(1+d)
The unique SPNE: Whenever a player offers, he proposes 1/(1+d) for himself and the
rest for the other player. Players accept any offer giving them at least d/(1+d) and
rejects all lesser offers.
Rubinstein bargaining
• This is a very simplified version of the bargaining model solved by Ariel Rubinstein in a 1982 paper
• Crucial results:
• Agreement is reached in the very first round, even in an infinite-stage game!
• If d<1, then 1/(1+d) > d/(1+d). If players are even a little bit impatient, there is a first-mover
advantage – the player who proposes first gets a higher payoff in equilibrium.
• If d=1 (infinite patience or very fast counter-offers), 1/(1+d) = d/(1+d) = ½. The players split the
pie equally in the first round.
• (Not in our model) If players have different discount factors, patience pays: equilibrium payoff
increases in your d
• Again, what happens off the equilibrium path shapes the equilibrium!
Fighting over an unsplittable pie
• Agreement reached in first round, and if players are sufficiently patient, pie is split
equally
• Why do we observe unequal distributions?
• Why do we observe drawn-out conflicts?
Wars of attrition
• 2 players fighting over a pie that cannot be split: either A gets it, or B does
• In each period, each player can Fight or Quit
• Quitting gives you 0 of the pie, no matter what the other player does
• If you Fight while the other player Quits, you get the whole pie, of value 1
• If you both Fight, you both incur a cost of c that period (c<1)
• The game ends when one player (or both) quit
• Both players are trying to “outlast” each other, incurring costs (through fighting) in
the hope the other guy will quit first and they will be the last one standing
Wars of attrition: examples
• WWI trench warfare
• Competition between firms
• Aggressive (and costly) campaigns meant to bring about a dominant market position
• Labour union strikes
• Boycotts and embargoes
• Interpersonal relationships
• Each flatmate waiting for the others to do the cleaning
• Each person silently waiting for the other to end a fight by saying “sorry, you were right”
• Our simple example: 2 periods, no discounting
War of Attrition: last stage
B
Fight Quit
A
Fight
-c, -c 1, 0
Quit
0, 1 0, 0
War of Attrition: last stage
B
Fight Quit
A
Fight
-c, -c 1, 0
Quit
0, 1 0, 0
• No reason to fight if you
think the other guy is
fighting: game ends after
this stage, anyway!
• Costs incurred from fighting
in previous periods are sunk
costs: already spent, do not
influence players’ decisions
today
• In the last stage, each player
wants to mismatch what the
other does
• Two NE in this subgame: (A
Fight, B Quit), (A Quit, B
Fight)
War of Attrition: first stage
B
Fight Quit
A
Fight
1, 0
Quit
0, 1 0, 0
-c+[payoff from stage 2],
-c+[payoff from stage 2]
• In the first stage, there may
be reason to fight even if
the other guy is fighting
because of the chance to
stay in the game and get the
pie tomorrow!
• What are the payoffs from
stage 2?
• Depends on which
equilibrium is played in
stage 2!
War of Attrition: first stage if A wins in second stage
B
Fight Quit
A
Fight
1, 0
Quit
0, 1 0, 0
-c+[1], -c+[0]
• If A wins in second stage:
• A gets 1 in second stage
• B gets 0 in second stage
• -c+1 > 0 since 1>c, so A
fights in the first stage
• In the first stage, A fights
and B quits
• If both players know A wins
tomorrow, he wins today
War of Attrition: first stage if B wins in second stage
B
Fight Quit
A
Fight
1, 0
Quit
0, 1 0, 0
-c+[0], -c+[1]
• If B wins in second stage:
• A gets 0 in second stage
• B gets 1 in second stage
• -c+1 > 0 since 1>c, so B
fights in the first stage
• In the first stage, B fights
and A quits
• If both players know B wins
tomorrow, he wins today
Wars of attrition
• Two SPNE of the 2-stage game
• A fights today and fights tomorrow, B quits today and quits tomorrow
• A quits today and quits tomorrow, B fights today and fights tomorrow
• Game ends in first stage – no delay!
• So why do we observe protracted wars of attrition?
• Equilibrium in mixed strategies?
War of Attrition: last stage
B
Fight Quit
A
Fight
-c, -c 1, 0
Quit
0, 1 0, 0
• If B fights with
probability p and quits
with probability 1-p:
• A’s payoff of Fight =
− + 1(1 − )
• A’s payoff of Quit =
0 + 0(1 − )
• A is indifferent if:
− + 1 − = 0
=
1
1 +
War of Attrition: last stage
B
Fight Quit
A
Fight
-c, -c 1, 0
Quit
0, 1 0, 0
• If B fights with probability
p and quits with
probability 1-p:
• A is indifferent if:
=
1
1 +
• Game is symmetric, so B is
indifferent if A fights with
probability q:
=
1
1 +
War of Attrition: first stage
B
Fight Quit
A
Fight
1, 0
Quit
0, 1 0, 0
• In last stage, both players
fight with probability
1
1+
• That means that in second
stage, they each get 0 on
expectation
• So the payoffs from stage
2 are 0 for both players in
stage 1!
-c+[payoff from stage 2],
-c+[payoff from stage 2]
War of Attrition: first stage
B
Fight Quit
A
Fight
1, 0
Quit
0, 1 0, 0
• In last stage, both players
fight with probability
1
1+
• That means that in second
stage, they each get 0 on
expectation
• So the payoffs from stage
2 are 0 for both players in
stage 1!
-c+[0], -c+[0]
War of Attrition: first stage
B
Fight Quit
A
Fight
-c, -c 1, 0
Quit
0, 1 0, 0
• If B fights with
probability p and quits
with probability 1-p:
• A’s payoff of Fight =
− + 1(1 − )
• A’s payoff of Quit =
0 + 0(1 − )
• A is indifferent if:
− + 1 − = 0
=
1
1 +
War of Attrition: first stage
B
Fight Quit
A
Fight
-c, -c 1, 0
Quit
0, 1 0, 0
• If B fights with probability
p and quits with
probability 1-p:
• A is indifferent if:
=
1
1 +
• Game is symmetric, so B is
indifferent if A fights with
probability q:
=
1
1 +
War of Attrition: first stage
B
Fight Quit
A
Fight
-c, -c 1, 0
Quit
0, 1 0, 0
• In the mixed strategy
equilibrium, each player
fights in every period with
probability
1
1+
• Also an equilibrium in
games with 3, 4, …., n
stages
• Reasoning: if you
know that in the nth
stage you will both
play those mixes, you
both play those mixes
in the n-1th stage…
and the n-2th stage….
War of Attrition and prolonged conflict over resources
• Probability of a 1-period
fight =
1
1+
×
1
1+
=
1
1+
2
• Probability of a 2-period
fight =
1
1+
2
×
1
1+
2
=
1
1+
4
• Probability of an n-period
fight =
1
1+
2
Low cost of
fighting: c = 0.25
High cost of
fighting: c = 0.9
Probability of an n-period fight