For PROF XAVIER ONLY
Handout 11
ECO 444
Konrad Grabiszewski
Tullock Contest
There are N players competing for one prize. Player i values the prize at vi > 0. Players compete
in efforts and these efforts determine the probability of winning the contest. Effort, denoted by
si, is costly. Let Ci(si) denote the cost function. We will assume a linear cost function; i.e.,
Ci(si) = cisi. There could be only one winner; those who do not win get utility zero.
Normal-form representation.
• N is the set of players;
• Si = [0,∞) is the set of player i’s strategies; si is player i’s level of effort;
• ui is the utility function of player i defined as
ui(s1, ..., sN ) = vi si
s1 + ... + sN −Ci(si) (1)
1 Two identical players with common value
Assume that N = {Ann, Bob} and vA = vB = v > 0. Players being identical means that they
have the same cost function. In particular, we take Ci(si) = si. Player i’s utility function becomes
ui(sA, sB) = v si
sA+sB −si.
1.1 Best-response correspondences
FOD and SOD.
∂ui ∂si
= sj
(sA + sB)2 v − 1 (2)
∂2ui ∂s2i
< 0 (3)
1
Note that ∂ 2ui ∂s2i
< 0 for each si. Hence, FOC determines the maximum. Note that the best-response
correspondences of Ann and Bob satisfy the following.
sB (s̃A + sB)2
v = 1 (4)
sA (sA + s̃B)2
v = 1 (5)
1.2 Equilibrium strategies
From (4) and (5), we deduce the following.
s∗B (s∗A + s
∗ B)
2 v = 1 (6)
s∗A (s∗A + s
∗ B)
2 v = 1 (7)
Hence, it must be true that s∗A = s ∗ B. This implies the following.
s∗A = s ∗ B =
v
4 (8)
At equilibrium, the probability of Ann (or Bob) winning is 0.5, and each player’s utility is v 4 .
1.3 Comparative statics
In our model, there is only one exogenous parameter v. Equilibrium efforts of Ann and Bob increase
in v. The same is true with their equilibrium utilities. However, the equilibrium probability of
Ann (or Bob) winning does not depend on v because it is 0.5 no matter what v is.
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2 Two identical players with different values
Assume that N = {Ann, Bob} and vA > vB > 0. Players are identical. Player i’s utility function
becomes ui(sA, sB) = vi si
sA+sB −si.
2.1 Best-response correspondences
FOD and SOD.
∂ui ∂si
= sj
(sA + sB)2 vi − 1 (9)
∂2ui ∂s2i
< 0 (10)
Note that ∂ 2ui ∂s2i
< 0 for each si. Hence, FOC determines the maximum. Note that the best-response
correspondences of Ann and Bob satisfy the following.
sB (s̃A + sB)2
vA = 1 (11)
sA (sA + s̃B)2
vB = 1 (12)
2.2 Equilibrium strategies
From (11) and (12), we deduce the following.
s∗B (s∗A + s
∗ B)
2 vA = 1 (13)
s∗A (s∗A + s
∗ B)
2 vB = 1 (14)
Hence, it must be true that s∗AvB = s ∗ BvA which implies that s
∗ A =
vA vB s∗B and allows us to derive
the equilibrium values. Let’s do it step-by-step for Bob. In (13), we replace s∗A by vA vB s∗B.
s∗B( vA vB s∗B + s
∗ B
)2 vA = 1 (15)
3
s∗BvA =
( vA vB
+ 1
)2 (s∗B)
2 (16)
vA = (vA + vB)
2
v2B s∗B (17)
s∗B = vAv
2 B
(vA + vB)2 (18)
We obtain equilibrium efforts.
s∗A = v2AvB
(vA + vB)2 (19)
s∗B = vAv
2 B
(vA + vB)2 (20)
Equilibrium sum of efforts.
s∗A + s ∗ B =
v2AvB + vAv 2 B
(vA + vB)2 =
vAvB vA + vB
(21)
At equilibrium, the probability of Ann winning is vA vA+vB
. The probability of Bob winning is vB vA+vB
.
Recall that we assume that vA > vB > 0. Hence, at equilibrium, Ann’s effort is higher than Bob’s
effort and, at the same time, his probability of winning is higher than her probability of winning.
(How do you interpret these results?)
2.3 Comparative statics
In our model, there are three exogenous parameters, v, vA, and vB. Remember that we assumed
that vA > vB > 0. Ann’s equilibrium effort increases in both vA and vB. Bob’s equilibrium effort
increases in vB and decreases in vA. Ann’s equilibrium probability of winning increases in vA and
decreases in vB. Bob’s equilibrium probability of winning increases in vB and decreases in vA.
(How do you interpret these results?)
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