game theory

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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