Guillermo Sabbioni
Game Theory (ECO4400)
Book: Games of Strategy, by Dixit & Skeath
Chapter 17: Bargaining
Bargaining is present in real life, all the time: countries, boss and employee, husband and
wife, parents and kids, etc. You can devise strategies to get a better result in bargaining
situations
All bargaining situations have two things in common:
 The total payoff that the parties can create and enjoy as a result of an agreement is
greater than the sum of the individual payoffs they can achieve separately.
Without this “surplus”, negotiation would be pointless.
 Following from the first one: bargaining is not a zero-sum game. When a surplus
exists, the negotiation is about how to divide it. It is not zero-sum because if they
fail to get to an agreement, the surplus vanishes and nobody gets anything.
No bargaining theory until recently. Even NE concept was not enough. Suppose two
people have to divide $1. Make a game where both (simultaneously) announce how much
they want (x and y). If x+y<=1 each gets what he announced. If x+y>1 they get nothing.
Then, any pair x,y such that x+y=1 is a NE (nobody can do better by deviating): multiple
NE.
1. Nash’s cooperative solution
This is Nash’s cooperative-game approach to bargaining.
a. Numerical example
Entrepreneur A produces a microchip that he can sell to any manufacturer for $900.
Entrepreneur B has a software package he can retail for $100. They realize that together
they can make a computer worth $3000. Therefore, together thy can create $2000 in
surplus. How to divide that extra value? Split the difference? The revenue or the profit?
If they split the $2000 profit equally, A gets $1900 and B gets $1100 of the revenue.
However, A argues that instead that they should have the same 66.66% profit with
respect to revenue: A should get $2700 (he contributed $900) and B should get $300 (he
contributed $100).
Assume instead of fighting they go to an arbitrator who decides a 4:1 division of profit in
favor of A (so he gets 80% of the surplus). Then A should get x-900 in profit, B should
get y-100 in profit, where the 4:1 division implies x-900=4(y-100) and x+y=3000.
Substitute x=3000-y in the first equation: 3000-y-900=4y-400 and obtain y=500 and
x=2500. This leaves A with a profit of 2500-900=1600 and B with a profit of 500100=400, which accomplishes the 4:1 division.
b. General theory
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Guillermo Sabbioni
Let total value be called v (only if they agree to a solution). If no agreement is reached, A
gets a and B gets b. These are called their BATNAs (best alternative to a negotiated
agreement). We need a+b<v for the negotiation to make sense, so there is a positive
surplus from the negotiation: v-a-b>0. Consider this rule: each player should be given
his BATNA plus a share of the surplus: A gets a fraction h of the surplus and B gets a
fraction k, such that h+k=1. Then, total revenue for each player (x for A and y for b)
should be as follows:
x=a+h(v-a-b)
y=b+k(v-a-b)
Rearranging them the Nash formulas are obtained:
x-a=h(v-a-b)
y-b=k(v-a-b)
Dividing the first equality by the second one allows to express the ratio h:k:
h/k=(x-a)/(y-b)
Solving for y as a function of x:
y=b+(k/h)*(x-a)
To use the whole surplus, we also must have x+y=v.
See fig 17.1 for the representation of the Nash’s cooperative solution. The BATNA
point is P, with coordinates (a,b), where no negotiation arises. Given that point P and
shares h and k, the line y=b+(k/h)*(x-a) can be graphed. At the intersection with the line
x+y=v we obtain the solution x and y at point Q (Nash solution).
*Note: h and k are not the solution to the problem as stated here. They are exogenously
given!! (not clear in the book, at least for me). The book says that this “vague” Nash
solution has its merits, because it can be used to encapsulate the results of many different
theories. The Nash formulas are then a description of the outcome of a bargaining process
yet not fully specified. All we can say is that h and k might describe the bargaining power
of both parties.
The thing is that Nash approach was different. He assumed that cooperative action is
possible (unlike in all the former chapters, where individuals only consider their own
payoffs when deciding their actions). Assuming players can delegate the implementation
of the agreement to a neutral third party, cooperative action might be possible. And that is
the way Nash modeled bargaining: as a cooperative game.
Nash formulated a set of principles for bargaining and proved that they implied a unique
outcome:
 The outcome should be invariant if the way in which payoffs are measured
changes linearly (no change in terms of risk-aversion attitude).
 The outcome should be efficient (no available mutual gain should go
unexploited). In the former example, the solution should have x+y=v (not less
than v). But not necessarily linear: see fig 17.2, for example, where y=f(x) is the
general form for the former linear expression x+y=v.
 If the set of possibilities is reduced by removing irrelevant ones, the outcome
should not be affected. Closely related to the IIA from Arrow.
Nash proved that the cooperative outcome that satisfied these 3 assumptions could be
characterized by this maximization:
Choose x and y to maximize ( x  a) h ( y  b) k subject to y=f(x).
Again, the values of h and k (which add up to 1) cannot be determined by the model.
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Guillermo Sabbioni
If you apply ln to the objective function to be maximized, you get:
h ln( x  a)  k ln( y  b)  C , such that you can graph the curves c1, c2… in fig 17.2.
The solution is again at Q, where the lines are tangent. This is the geometrical
representation.
Mathematically: Let X=x-a and Y=y-b. Also, from efficiency, replace Y=S-X, where S=va-b. The objective function is now: X h ( S  X ) k . Maximize with respect to the unknown
X and set that equal to 0:
hX h1 (S  X ) k  X h k (S  X ) k 1  0
Solving:
hX h1 (S  X ) k  X h k (S  X ) k 1
hX h X 1 (S  X ) k  X h k (S  X ) k (S  X ) 1
hX 1  k (S  X ) 1
h / X  k /( S  X )
X Y
x a y b
 so

, which is the original Nash formula we had before. Nash’s three
h k
h
k
conditions lead to the formula originally stated as a simple way of splitting the bargaining
surplus. Yet, the problem persists: a player who can do better by strategizing on his own
may reject the principles of the Nash solution. Then, challenging question: can Nash’s
cooperative solution arise as the NE of a noncooperative game played by the bargainers?
This can be done (wait until section 5).
2. Variable-threat bargaining
Suppose players can manipulate their BATNAS in a first stage (a and b are not fixed
anymore). In the second stage, we still have the Nash cooperative outcome. That
manipulation is called variable-threat bargaining.
See fig 17.3. If player A can increase his BATNA (move from P to P1) the new outcome
at Q’ gives him a better payoff (example: having already a good job offer when you go to
a new job interview).
Also, reducing the other player’s BATNA can lead to the same outcome (moving to P2,
for example). Example: “make yourself indispensable” for your employer, such that not
giving you a raise and loosing you as an employee (not getting to an agreement) leaves
him with a bad BATNA.
Finally, take a look at P3. This is the backstop point when your rival’s BATNA is reduced
more than yours.
Read yourselves the example about the baseball player’s strike in the 80’s.
3. Alternating-offers model 1: total value decays
Back to the more realistic non-cooperative game theory. Looking for a process that may
produce an equilibrium in a bargaining game. Possibility: alternating offers. One
proposes, the other accepts or not. If the other does not accept, he makes another
(alternative) offer. The game finishes when somebody accepts.
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Guillermo Sabbioni
This is a sequential game: we should find a rollback equilibrium. Therefore, start at the
end nodes. Problem: is there an end? Why would the players want to reach an agreement?
Two answers:
 The total value of the agreement decays as the agreement is delayed (this section).
 Time has value; impatience (section 5).
Example of a shrinking pie (see PowerPoint figure): a fan arrives at a football game
without a ticket. He is willing to pay at most $25 to see each quarter of the game. A guy
in the street sells his ticket. If the fan does not like the price, he goes to a bar and watches
the first quarter. At the end, he makes an offer to the potential seller (still on the street). If
the seller does not agree, the fan watches the second quarter on the bar. And so on… the
total value of the ticket declines as quarters go by: $100 at the beginning, then $75, then
$50, and finally $25.
Rollback analysis: at the end of the third quarter, any offer should be fine for the seller
(otherwise he is stuck with the ticket). Then, the fan should get it almost for free. At the
end of the second quarter, the seller could ask for at most $25 (the value of the third
quarter to the fan). At the end of the first quarter, the fan knows that after the second
quarter the seller can expect to get only $25, so that should be enough to get the ticket. At
the beginning of the game, the seller can ask for $50 and it should be accepted by the fan.
That includes the value of the first quarter and the $25 for which the fan could get the
other 3 quarters.
Trick: only one person makes an offer at each point in time.
Formally: A makes an offer to split v. If B refuses the available value declines to v-x1. B
now offers. If A refuses the value is now v-x1-x2. After 10 rounds (5 offers by A and 5
offers by B) the game finishes and the value is zero: v-x1-x2-x3…-x10=0. Rollback: one
round before the game is over only x10 is the available value. B could offer “almost
nothing” to A and get all the value. Therefore, the outcome in round 10 is (rounding):
“x10 for B, 0 for A”. In round 9, A (the one proposing) could at most obtain x9 and leave
x10 for B. In round 8, B (the one proposing) could also capture x8 (he has to leave x9 for
A)… finally: A can offer (x2+x4+x6+x8+x10) to B and keep (x1+x3+x5+x7+x9) for himself
in the first round. B should accept.
Trick: If B refuses to A’s first offer, the total value declines by x1. Then, x1 should go to
A in equilibrium. You get whatever is lost if the rival does not accept your offer.
If BATNAs are positive, the final offer at the end should include the BATNA (it was
assumed 0 in the example).
See that the strategies are complete plans of action: we know what the players would do
if the next stage was reached (even when in equilibrium they are not).
Notice: final outcome is fairer than “sudden decay” (ultimatum game: if you don’t
accept, the game is over and you get nothing).
4. Experimental evidence
Read yourselves.
Experiments of the ultimatum game. Lots of pairs. There is a proposer and a chooser.
There are almost no 99:1 proposals to split a dollar. Many are 50:50. Also, if a 75:25
proposal is made, it is usually rejected.
Problems:
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Guillermo Sabbioni
 Payoffs are not just the money obtained (fairness, pride, culture, etc).
 Maybe it is a repeated interaction among the players.
Experiments of the dictator game. The B player has no decision at all. Still, most of A
players keep 30 cents for the B players.
5. Alternating-offers model 2: impatience
Suppose total value does not decay, but people prefer to agree today rather than
tomorrow. Assume r=5% (you can consider that the interest rate or you can also call it
“impatience” parameter).
The thing is: you are indifferent between 95 dollars now or 100 dollars in the next period.
Reasons for introducing r: you could invest money today, or there might be a possibility
that the game finishes before getting to the next round.
Suppose two bargaining players with BATNAs equal to 0. The have to split $1. If an
offer is rejected, they face each other again the next period and again try to split $1. The
alternate the roles of proposer and chooser.
Let x be the amount obtained by the proposer in equilibrium (A or B, it does not matter).
If player A starts, he must offer at least 0.95x to B, because B would get x in the next
period if they don’t agree now. Offering 0.95x to B, player A is left with 1-0.95x. What is
left for A is exactly what the proposer gets in equilibrium: x.
Then: 1-0.95x=x. Solving: 1=1.95x and x=1/1.95=0.512
The proposer gets 0.512 (more than half) and the chooser 0.488.
Now suppose different patience: A is more impatient (indifferent between 90 cents now
and 1 dollar in the next period). Let x be the amount A gets in equilibrium and y be the
amount going to B.
A must give B at least 0.95y. Otherwise, B rejects the offer. Then, x=1-0.95y. If B
proposes, he must offer at least 0.90x to A. Then, y=1-0.90x. System of 2x2 equations.
Substitute the second in the first: x=1-0.95(1-0.90x). Simplify it:
x=1-0.95+0.90*0.95x and x(1-0.855)=0.05 so x=0.05/0.145=0.345.
Substituting back:
y=1-0.90x=1-0.90*0.345=0.690.
They don’t add up to 1 because those are the amounts A and B would offer.
If A moves first, he offers 1-0.345=0.655 to B and he keeps 0.345. B would accept.
If B moves first, he offers 1-0.690=0.31 to A and he keeps 0.690. A would accept.
Still: both players prefer to be proposers rather than choosers. Also, notice that being
first, A gets less money now (0.345 vs 0.512) than when both were equally impatient.
Being impatient hurts (even if you are proposer).
General algebra:
Suppose A is indifferent between $1 now and $(1+r) next period. Let a=1/(1+r).
Suppose B is indifferent between $1 now and $(1+s) next period. Let b=1/(1+s).
Suppose A gets x when he proposes and suppose B gets y if he is the proposer.
If A proposes, he must offer B at least what B can get next period: y/(1+s)=by. Then A
gets the rest: x=1-by. If B proposes, he should offer A at least x/(1+r)=ax. Then B gets
the rest: y=1-ax. Solve the system:
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Guillermo Sabbioni
x=1-by=1-b(1-ax)=1-b+abx
x(1-ab)=1-b so we have x=(1-b)/(1-ab).
We also have y=(1-a)/(1-ab).
Recall that a=1/(1+r) and b=1/(1+s). Then (this algebra is not in the book):
1 s 1
s
(1  r )(1  s)
1 s
=x
=x
1
1
(1  r )(1  s )  1
(1  s) (1  r )(1  s)  1
1
(1  r ) (1  s)
(1  r )(1  s )
s  rs
s  rs
s(1  r )
=x
=x
x
r  s  rs
1  r  s  rs  1
(1  r )(1  s)  1
1 b
x
=x
1  ab
1
1
1 s
r  rs
1 a
we have y 
r  s  rs
1  ab
The sum x+y is larger than 1 (like before):
s  rs
r  rs
r  s  2rs
x y 


1
r  s  rs r  s  rs
r  s  rs
Since r and s are usually very small numbers, their product rs is negligible and can be
ignored for the following approximate solution:
s
r
x
and y 
rs
rs
Since now the sum x+y adds up to 1, x and y are now the shares going to each player.
And the ratio x/y yields x/y=s/r. So the share is inversely proportional to the rate of
impatience. For A, for example, a larger r means a smaller x (keeping s constant).
Formally (without ignoring rs), we see that a larger r reduces x (this is not in the book
either):
Doing the same substitution with y 
dx d  s  rs  s (r  s  rs )  ( s  rs )(1  s) sr  ss  rss  ( s  rs  ss  rss )



dr dr  r  s  rs 
(r  s  rs ) 2
(r  s  rs ) 2
dx sr  ss  rss  s  rs  ss  rss
s


0
2
dr
(r  s  rs )
(r  s  rs ) 2
Recall the Nash cooperative approach: the shares were proportional to the bargaining
strengths (given from the outside). Now, this result could be interpreted as the rollback
equilibrium of a noncooperative bargaining (sequential) game where the impatience
determines the strength (inverse relationship).
6. Manipulating information
So far: common knowledge of the rival’s BATNA and impatience. What if imperfect
info? A good BATNA or low rate: you want to signal that. A low BATNA or high
discount factor: you want to hide it from the rival. Signaling and screening are relevant in
any bargaining situation. Read yourselves the example about buying a house.
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Guillermo Sabbioni
7. Bargaining with many parties and issues
Read yourselves.
SUGGESTED EXERCISES: 6.
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