n?u=RePEc:bgu:wpaper:1611&r=hpe

TWO-STAGE ELIMINATION

CONTESTS WITH OPTIMAL

HEAD STARTS

Noam Cohen, Guy Maor and

Aner Sela

Discussion Paper No. 16-11

November 2016

Monaster Center for

Economic Research

Ben-Gurion University of the Negev

P.O. Box 653

Beer Sheva, Israel

Fax: 972-8-6472941

Tel: 972-8-6472286

Two-Stage Elimination Contests with Optimal Head Starts

Cohen Noam, Maor Guy and Sela Aner

October 6, 2016

Abstract

We study two-stage elimination Tullock contests. In the …rst stage all the players compete against

each other of which some advance to the second stage while the others are eliminated.

The …nalists

compete against each other in the second stage and one of them wins the prize.

To maximize the

expected total e¤ort the designer can give a head start to the winner of the …rst stage when he competes

against the other …nalists in the second stage. We show that the optimal head start, independent of the

number of …nalists, always increases the players’expected total e¤ort. We also show how the number of

players and …nalists a¤ect the value of the optimal head start.

Keywords: Multi-stage contests, Tullock contests, head starts.

JEL classi…cation: C70, D44, L12, O32

Cohen Maor and Sela: Department of Economics, Ben-Gurion University of the Negev, Beer–Sheva 84105, Israel.

1

1 Introduction

In elimination contests in each stage some of the contestants are eliminated while the others advance to the

next stage until the …nal stage in which some of the players (usually one of them) win prizes. In elimination

tournaments in sport, players or teams usually play pair-wise matches and the winner advances to the next

stage while the loser is eliminated from the competition. Examples include the ATP tennis tournaments

and the professional playo¤s in US-Basketball or World Cup playo¤s.

In this paper we study two-stage

elimination Tullock contests (Tullock 1980) 1 where in the …rst stage all the n players compete against each

other and k; k n; players (…nalists) proceed to the second (…nal) stage. The contestants exert their e¤orts

once in the …rst stage and the …nalists are determined sequentially. The …rst winner is determined by the

probability success function considering the e¤orts of all the players. The second winner is determined by

the probability success function considering the e¤orts of all the contestants excluding the e¤ort of the …rst

winner. This sequential process continues until all the …nalists are determined. The winners in the second

stage win the prizes. 2

To illustrate, such a two-stage elimination contest can be a preliminary contest in

which all the contestants are interviewed for a job and a …nal stage in which a few …nalists are interviewed

and tested. 3

Fu and Lu (2012) studied a multi-stage sequential elimination Tullock contest, and showed that the

optimal contest eliminates one contestant at each stage until the …nal stage. Then, the winner of the …nal

takes the entire prize sum. However, their result to eliminate only one player in the …rst stage does not

hold when the number of stages is limited, especially when there are only two stages. In such a case, the

contest designer can decide about the number of players who advance from the …rst stage to the second one

and the incentives he can give them. Such an incentive could be a head start, an exogenously determined

mechanism that increases the winning probabilities of some contestants. By using a head start, the contest

1 A number of studies provided axiomatic justi…cation for the Tullock contest (see, for example, Skaperdas (1996) and Clark

and Riis (1998)). Baye and Hoppe (2003) have identi…ed conditions under which a variety of rent-seeking contests, innovation

tournaments, and patent-race games are strategically equivalent to the Tullock contest.

2 Clark and Riis (1996) considered the same method to select the winners in one-stage Tullock contests.

3 Amegashie et al. (2007) studied such a two-stage elimination all-pay auction with four heterogeneous players and budget

constraints. He experimentally showed that players use all of their available budgets in the …rst stage.

2

designer can a¤ect the e¤ort levels of the contestants. 4

This was shown by Konrad (2002) in a one-stage

two-player all-pay auction under complete information, Kirkegaard (2012) in an asymmetric all-pay auction

under incomplete information and Segev and Sela (2014) in a sequential all-pay auction under incomplete

information. Franke et al. (2011) studied one-stage Tullock contests with heterogeneous contestants where

the designer who wishes to maximize the contestants’total e¤ort may allocate asymmetric success functions

(which are equivalent to di¤erent head starts) for the contestants. They showed that the optimal contest

rule is biased in favor of weaker contestants, i.e., stronger contestants are relatively discouraged while weaker

contestants are encouraged both to participate and to exert higher e¤orts. 5

In other words, head starts

should be given to the weak players. While Franke et al. found the optimal head starts in a one-stage

Tullock contest which are given according to the players’types, we …nd the optimal head start in a two-stage

Tullock contest which is given to the winner of the …rst stage according to his performance in that stage.

For simplicity, we begin with the analysis of the most common case in which there are only two …nalists

in the second (…nal) stage. We demonstrate that the entire prize sum should be awarded to the winner of

the …nal stage only. Later, we generalize the model for any number of …nalists, and we assume that the

entire prize sum is awarded to the winner in the …nal stage. Then, we show that awarding a head start to

the winner in the …rst stage, independent of the number of …nalists, always increases the players’total e¤ort.

This result implies that the elimination two-stage contest with a head start dominates the one-stage contest

with respect to the contestants’total e¤ort.

We also explicitly calculate the optimal head start and show that its value decreases in the number of

players. The optimal head start is not monotonic in the number of …nalists. We illustrate that it decreases

in the number of …nalists, obtains its minimal value and then increases again. In particular, the optimal

head start has the same value if the number of …nalists is either two or equal to the number of players.

Last, we calculate the optimal head start together with the optimal number of …nalists. We …rst show

that if the number of players is three then the optimal number of …nalists is three as well. We then show that

4 See Tsoulouhas et al. (2007) for asymmetric rules in labor tournaments, and Epstein et al. (2011) for asymmetric rules in

public procurement and lobbying contests.

5 Nti (2004) determined the optimal contest success function in the one-stage Tullock contest with two players, while Dasgupta

and Nti (1998) did this for n homogeneous players.

3

the optimal number of …nalists increases in the value of the head start. By our analysis, we can calculate

the combination of the optimal head start and the optimal number of …nalists for any number of players.

For instance, given the optimal head start, if the number of players is three the optimal number of …nalists

is three as well (all the players advance to the …nal stage), but if the number of players is 100, the optimal

number of …nalists is 27. Our calculations show that, given the optimal head start, when the number of

players is relatively large, a change in the number of players yields a very small change in the optimal number

of …nalists.

1.1 Related literature

In this paper we assume that the designer can determine the players’contest success functions. However,

in the literature on elimination tournaments it is usually assumed that the structure of the tournament is

given and the contest designer can determine the players’allocation in the tournament. Rosen (1986), for

example, studied an elimination tournament with homogeneous players where the probability of winning a

match is a stochastic function of the players’e¤orts. Gradstein and Konrad (1999) studied a rent-seeking

contest à la Tullock (with homogenous players), and Groh et al. (2012) studied an elimination tournament

with four asymmetric players where players are matched in the all-pay auction in each of the stages. Groh et

al. found optimal seedings for di¤erent criteria and, in particular, they showed that total expected e¤ort in

the elimination tournament where the two strongest players meet in the …nal with a probability of one equals

the total e¤ort in the all-pay auction where all players compete simultaneously. This result is contrasted

with the main …nding of Gradstein and Konrad who found that simultaneous contests are strictly superior

if the contest’s rules are discriminatory enough.

Similarly to previous works we show that it is not pro…table for the contest designer to split the prize

in the …nal stage. For example, Moldovanu and Sela (2001) showed that in one-stage all-pay contests under

incomplete information when cost functions are linear or concave in e¤ort, it is optimal to allocate the

entire prize sum to a single …rst prize, but when cost functions are convex, several positive prizes may be

optimal. Later (2006) these authors studied two-stage all-pay contests with multiple prizes under incomplete

information and showed that for a contest designer who maximizes the expected total e¤ort, if the cost

4

functions are linear in e¤ort, it is optimal to allocate the entire prize sum to a single …rst prize. In Tullock

contests, Schweinzer and Segev (2009) demonstrated that the optimal prize structure of symmetric n-player

Tullock tournaments assigns the entire prize sum to the winner, provided that a symmetric pure strategy

equilibrium exists.

In elimination contests there are several methods di¤erent from ours for choosing the …nalists who advance

to the …nal stage. Amegashie (1999) and Moldovanu and Sela (2006) split the contestants among several

sub-contests in the …rst stage and then the winners compete against each other in the …nal stage while the

other contestants are eliminated. Amegashie showed that the comparison between one-stage and two-stage

Tullock contests is ambiguous and depends on the parameters of the Tullock success functions at each stage.

In addition, he found that the players’expected e¤ort in the two-stage elimination Tullock contest and the

standard one stage Tullock contest is the same. Moldovanu and Sela also compared between the multi-stage

and the one-stage winner-take-all contests from the point of view of a designer who maximizes the expected

total e¤ort. For the case of a linear cost of e¤ort, they showed that if the designer maximizes the expected

total e¤ort, the optimal architecture is a single grand static contest.

We theoretically show that the two-stage elimination Tullock contest with the optimal head start yields

a higher total e¤ort than the one-stage Tullock contest. Similar to our …ndings, Sheremeta’s (2010 a and b)

experimental studies showed that the two-stage Tullock contest generates higher aggregate e¤ort than the

corresponding one-stage contest.

The rest of the paper is organized as follows. In Section 2 we analyze the subgame perfect equilibrium

of the two-stage elimination contest with two …nalists and in Section 3 we analyze the general model with

any number of …nalists. In Section 4 we analyze the two-stage elimination contest with optimal head starts.

2 The model with two …nalists

We …rst study an elimination two-stage Tullock contest with n 3 players. In the …rst stage, all the n

players simultaneously exert their e¤orts and only two of them (so called …nalists) advance to the second

(…nal) stage, while all the other players are eliminated. If player i; i = 1; 2; :::; n exerts an e¤ort x i in the

5

…rst stage then his probability to be the …rst to advance to the …nal stage is

and his probability to be the second to advance to the …nal stage is

nX

k=1

k6=i

x i

nX

j=1

x k

nX

x j

x i

nX

x j x j

j=1 j=1

j6=k

Thus, player i’s probability to advance to the …nal stage is

x i

+ nX

x j

j=1

nX

k=1

k6=i

x k

nX

x i

x j

j=1 j=1

j6=k

(1)

nX

x j

We assume that the …rst prize for the winner in the …nal stage is > 0:5 while the second prize for the

loser is 1

: A player has an incentive to be the …rst to advance to the …nal stage if he receives a head

start that improves his probability to win the …rst prize in the …nal stage. Formally, if player i is the …rst to

advance to the …nal stage and player j is the second, then if they exert e¤orts of y i ; y j respectively, player i’s

y i

probability to win the …rst prize is

y i+y j

and 1

y i

y i+y j

to win the second prize, where the head start 1

is a commonly known constant determined by the contest designer. In the following we analyze the subgame

perfect equilibrium of the elimination two-stage contest with two …nalists. We begin with the analysis of the

second stage and go backwards to the …rst one.

2.1 The second stage

The maximization problem in the second stage of the …nalist who won in the …rst stage is

max

y 1

y 2

y 1

+ (1 )

y 1

y 1 + y 2 y 1 + y 2

where y 2 is the other …nalist’s e¤ort whose maximization problem is

max

y 2

y 2

y 1

+ (1 )

y 2

y 1 + y 2 y 1 + y 2

The F.O.C. satisfy

y 2 (2 1)

(y 1 + y 2 ) 2 = y 1(2 1)

(y 1 + y 2 ) 2 = 1

6

Thus, the equilibrium e¤orts in the second stage are

y 1 = y 2 =

(2 1)

(1 + ) 2 (2)

Since > 0:5 the …nalists’e¤orts in the second stage satisfy y 1 = y 2 > 0: The expected payo¤ of the winner

in the …rst stage is

u 1 = 2 2 + 2 + 1

(1 + ) 2 (3)

and the expected payo¤ of the other …nalist is

u 2 = 2 2 + 2 + + 2

(1 + ) 2 (4)

It can be easily shown that if is su¢ ciently high (i.e., close enough to 1) then u 1 ; u 2 > 0: We can now

proceed to the analysis of the equilibrium in the …rst stage.

2.2 The …rst stage

By (1), (3) and (4), the maximization problem of player i; i = 1; 2; :::; n in the …rst stage is

2 2 + 2 + 1 x i

max

x i (1 + ) 2 nX

j=1

+ 2 2 + 2 + + 2

(1 + ) 2 n

X

k=1

k6=i

x j

x k

nX

x i

nX

x j x j

j=1 j=1

j6=k

x i

7

The F.O.C. is

= 1

X n

x j

2 j=1

2 + 2 + 1 j6=i

(1 + ) 2 nX

( x j ) 2

j=1

2 2 + 2 + + 2

(1 + ) 2 n X

k=1

k6=i

2 2 + 2 + + 2

(1 + ) 2 n X

k=1

j6=i

(

x k

nX

x j ) 2

j=1

x k

nX

j=1

x j

x i

+ nX

x j

j=1

j6=k

X

n x j

j=1

j6=k;i

(

nX

x j ) 2

j=1

j6=k

By symmetry, we obtain

Thus, we have

2 2 + 2 + 1

(1 + ) 2 n 1

n 2 x + 2 2 + 2 + + 2

(1 + ) 2 n 2 3n + 1

n 2 (n 1)x = 1

Proposition 1 In the subgame perfect equilibrium, the strategy of every player in the …rst stage of the

elimination two-stage contest with two …nalists is

x = 2 (n 2 + n 3n + 1) + (4n 2 4n 2 + 10n 10n 4 + 4) + (n 2 2n n + 1)

(1 + ) 2 n 2 (n 1)

while the …nalists’strategies in the second stage are given by (2).

The players’total e¤ort in both stages is then given by

T E = nx + 2y = (5)

2 (n 2 + n 3n + 1) + (4n 2 4n 2 + 10n 10n 4 + 4) + (n 2 2n n + 1)

(1 + ) 2 n(n 1)

2(2 1)

+

(1 + ) 2

Note that

dT E

d

= n(2 1) + (6n 4)

(1 + ) 2 n(n 1)

8

y

0.80

0.75

0.70

0.65

0.60

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

x

Figure 1: The expected total e¤ort as a function of for n=3.

Thus, if > 1 we obtain that dT E

d

> 0, namely, it is optimal to award the entire prize sum (which is equal

to 1) to the winner in the second stage. From this point on we will assume that only one prize is awarded,

i.e., = 1. Then, we can see that when n approaches 1 we obtain

lim T E n!1 =

(1 + )2

(1 + ) 2 = 1

The optimal head start that yields the highest expected total e¤ort is given by

d

d T E() = 2

n ( + 1) 3 (2n + n 1) = 0

(n 1)

This implies that

= 2n 1

n 1

(6)

Since d

dn = 1

(n 1) 2 < 0, the maximal value of the optimal head start is obtained for n = 3 ( = 2:5)

and is decreasing in the number of players n. When n approaches in…nity it converges to 2. Figure 1

presents the expected total e¤ort as a function of the head start when the number of players is n = 3.

[Figure 1 to be here].

We can see that when n = 3, the highest total e¤ort is obtained for = 2:5 where the players’total

e¤ort increases for every 1 2:5 and decreases for all > 2:5: If we compare the players’performances

9

in our model and the one-stage Tullock model we obtain that

Proposition 2 For any number of players n 3, the expected total e¤ort in the two-stage elimination

contest either with or without the optimal head start is larger than in the one-stage contest.

Proof. The total e¤ort in the standard (one-stage) Tullock contest with n symmetric players and a prize

equal to 1 is

T E s = n 1

n

By (5), the di¤erence between the total e¤ort in the two-stage elimination Tullock contest with the optimal

head start = 2:5 and the one-stage contest is

T E( = 2:5) =

1

3n 2 (n 1) > 0

2n

Furthermore, by (5), the di¤erence between the total e¤ort in the two-stage elimination Tullock contest

without any head start and the one-stage contest is

T E( = 1) =

1

4n 2 (n 2) > 0

4n

Figure 2 presents the total e¤ort in the two-stage elimination contest and in the standard (one-stage) one.

[Figure 2 to be here].

In the next section we generalize the model to any number of …nalists k; 2 k n.

3 The model with k 2 …nalists

We now study a two-stage Tullock contest when in the …rst stage all the n players simultaneously exert their

e¤orts and k; 2 k n of them advance to the second stage. Following the result in the previous section

according to which the entire prize sum should be allocated to the winner in the second stage, we assume

that the k …nalists compete against each other to win a prize equal to 1, where the winner in the …rst stage

has a head start advantage that increases his winning probability compared to his rivals. By symmetry, we

denote the e¤ort of every contestant j; j 6= i by x j = x. Then, if player i exerts an e¤ort x i his probability

to win in the …rst stage is

x i

(n 1)x + x i

10

y

1.0

0.9

0.8

0.7

0.6

0.5

5 10 15 20 25 30 35 40 45 50

x

Figure 2: The total e¤ort as a function of the number of players in the one-stage contest (red curve) and in

the two-stage contest with the optimal head start (black curve) and without a head start (the green curve).

The probability of player i to be the second to advance to the …nal is

x(n 1)

(n 1)x + x i

x i

(n 2)x + x i

Similarly, the probability of player i to be the s-th, s = 3; 4; :::; k to advance to the …nal is

x(n 1)

(n 1)x + x i

s 1 (n 1)!

= x

(n s)!

x(n 2)

::::

(n 2)x + x i

sY 1

x i + (n j + 1)x x i

x i + (n

j=2

x(n (s 1))

(n s + 1)x + x i

x i

(n s)x + x i

s)x

In sum, the probability of player i to advance to the …nal stage is given by

2

x i

kX

s 1 (n 1)!

sY 1

x i

+ 4x

(n 1)x + x i (n s)! x i + (n j + 1)x x i + (n

s=2

j=2

3

5 (7)

s)x

Each player has an incentive to be the …rst to advance to the …nal stage since then he receives a head start

that improves his probability to win in the …nal. Formally, let y j = y; 1 j k and j 6= i, then if player i

is the …rst to advance to the …nal stage among k …nalists and he exerts an e¤ort y i his probability to win is

y i

y i + (k 1)y

11

where 1 is a commonly known head start determined by the contest designer and y = y j ; j 6= i is the

symmetric e¤ort of all the …nalists except player i. We begin with the analysis of the second stage and go

backwards to the …rst one.

3.1 The second stage

The maximization problem in the second stage of the …nalist (denoted as player 1) who won in the …rst stage

is

max

y 1

y 1

y 1 + (k 1)y

y 1

where y is the symmetric e¤ort of all the other …nalists. The maximization problem of a …nalist j; j 6= 1;

who did not win in the …rst stage is

max

y j

y j

y 1 + y j + (k 2)y

y j

where y 1 is the e¤ort of the winner in the …rst stage, and y is the symmetric e¤ort of all the other …nalists.

Since y j = y; the F.O.C. yield

Thus, the equilibrium e¤orts in the second stage are

(k 1)y

(y 1 + (k 1)y) 2 = y 1 + (k 2)y

(y 1 + (k 1)y) 2 = 1

y 1 = (k 1)2 (k 1)(k 2)

((k 1) + 1) 2 (8)

y = y j =

Then, the total e¤ort in the second stage is

(k 1)

; j = 2; :::; k

((k 1) + 1)

2

T E 2 = (k 1)y + y 1

=

(k 1)(2(k 1) (k 2))

((k 1) + 1) 2

The expected payo¤ of the winner in the …rst stage is

u 1 =

and the expected payo¤s of the other …nalists are

((k 1) (k 2))2

((k 1) + 1) 2 (9)

12

1

u j =

; j = 2; :::; k (10)

((k 1) + 1)

2

We can now proceed to the analysis of the equilibrium in the …rst stage.

3.2 The …rst stage

By (7), (9) and (10), the maximization problem of player i; i = 1; ::::; n in the …rst stage is

((k 1) (k 2)) 2 x i

Max xi

((k 1) + 1) 2 x i + (n 1)x

2

1

kX

s 1 (n 1)!

sY 1

x i

+

4x

((k 1) + 1) 2 (n s)! x i + (n j + 1)x x i + (n

s=2

j=2

3

5 x i

s)x

where x is the symmetric e¤ort of all the other players. By the symmetry, the F.O.C. yields

=

Thus, we obtain that

((k 1) k + 2)) 2 n 1

((k 1) + 1) 2 n 2 x + 1

kX

((k 1) + 1) 2 x

s=2

"

1

kX

((k 1) + 1) 2

s=2

s 1 (n 1)!

x

(n s)!

sX

i=2

1

(n i + 2)x

((k 1) k + 2))2

((k 1) + 1) 2 n 1

n 2 x + 1

((k 1) + 1) 2 kX

s=2

s 1 (n 1)!

(n s)!

(n s + 1)!

n!x s 1 n s

(n s + 1)!

n!x s 1 1

(n s + 1)

n s

n(n s + 1)x

sX

i=2

(n s + 1) 2 x

#

1

n(n i + 2)x

!

= 1

Proposition 3 In the subgame perfect equilibrium the strategy of every player in the …rst stage of the

elimination two-stage contest with k 2 …nalists is

x =

((k 1) k + 2))2 n 1 1

kX

((k 1) + 1) 2 n 2 +

((k 1) + 1) 2

while the …nalists’strategies in the second stage are given by (8).

s=2

n s

n(n s + 1)

sX

i=2

!

1

n(n i + 2)

Then, the total e¤ort in the …rst stage is

T E 1 = nx =

and the total e¤ort in both stages is

((k 1) k + 2))2 n 1 1

kX

((k 1) + 1) 2 +

n ((k 1) + 1) 2

s=2

n s

(n s + 1)

sX

i=2

!

1

(n i + 2)

T E = T E 1 + T E 2 = (11)

!

((k 1) k + 2)) 2 n 1 1

kX n s

sX 1

((k 1) + 1) 2 +

n ((k 1) + 1) 2 (n s + 1) (n i + 2)

+

(k 1)(2(k 1) k + 2)

((k 1) + 1) 2 13

s=2

i=2

4 Results

We have assumed that the winner in the …rst stage receives a head start > 1 in the second stage. The

following result shows that awarding a head start in the …rst stage is an e¢ cient tool for maximizing the

players’total e¤ort.

Proposition 4 In the two-stage elimination contest awarding a head start > 1 to the winner in the …rst

stage, independent of the number of …nalists, always increases the players’total e¤ort.

Proof. The derivative of (11) yields

dT E

d

=

2(k 1) 1)(n 1)(k 2 (1 k))

3

((k

(k + 1) n

!

kX n s

sX 1

+

+ ( 1)(k 1) 2 )

(n s + 1) (n i + 2)

s=2

i=2

(12)

When approaches one we obtain that

dT E

lim

!1 d = 2k 1 (k 1)(n 1)

k 3 n

kX

s=2

n s

(n s + 1)

sX

i=2

!!

1

(n i + 2)

Since

kX n s

(n s + 1)

s=2

(k 1)n

2

n 1

We obtain,

dT E

lim

!1 d

1)2

2(k

k 3 ( n 1

n

n 2

n 1 ) 0

Therefore, the head start > 1, independent of the number of the …nalists k; increases the players’total

e¤ort.

Now, given that a head start is an e¢ cient tool for increasing the players’total e¤ort, it is worthwhile to

…nd out what the optimal head start is and examine how it changes as a function of the number of …nalists.

Proposition 5 In the two-stage elimination contest with n players and k …nalists, the optimal head start

for the winner in the …rst stage is

k+1

1 X

=

(k 1) 2 (n k i + 2

n i + 2

i=2

1) + 1 (13)

14

Proof. The F.O.C. of the maximization of the total e¤ort given by (11) is

2(k 1)

(k + 1) 3 ((k 1)(n 1)(k 2 (1 k))

n

!

kX n s

sX 1

+ (

(n s + 1) (n i + 2) ) + ( 1)(k 1) 2 ) = 0

s=2

By some calculations we have

=

=

=

=

=

=

i=2

n

kX n s

(k 1) 2 (

(n s + 1)

s=2

n

(k 1) 2 (k 1) +

sX

i=2

kX 1

(

(n s + 1)

s=2

k+1

n X 1

sX

(k 1) 2 (

(n s + 2) +

s=3

k+1

1

X

(k 1) 2 k + n

sX

s=2 i=2

i=2

1

(n i + 2)

k+1

1

(k 1) 2 ( k + n X k i + 2

(n i + 2) ) + 1

i=2

k+1

1 X

(k 1) 2 (n k i + 2

n i + 2

i=2

1) + 1

!

1

(n i + 2) ) + n + k 2

(k 1)

sX

i=2

!

1

(n i + 2) ) + n + k 2

(k 1)

!

1

(n i + 2) ) + k 2

k 1

!

+ 1

The following result shows the e¤ect of the number of players on the optimal head start.

Proposition 6 In the two-stage elimination contest the value of the optimal heat start decreases in the

number of players n.

Proof. By (13), we obtain that

=

=

(n + 1)

(n)

k+1

1 X

(k 1) 2 (n + 1) k i + 2

n i + 3

i=2

n k i + 2

n i + 2

k+1

1 X k i + 2

(k 1) 2 (n i + 3)(n i + 2) ( i + 2) < 0

i=2

Now we …x the number of the players and change only the number of the …nalists. Accordingly, Figure 3

presents the optimal head start as a function of the number of …nalists k when the number of players is

15

Figure 3: The optimal as a function of the number of …nalists k

n = 100. [Figure 3 to be here]. We can see by the …gure that the optimal head start …rst decreases and

later increases in the number of …nalists k: The minimal value of the optimal head start is obtained when

the number of …nalists is k = 17, and the maximal value of the optimal head start is obtained when the

number of …nalists is either k = 2 or k = n = 100: The last result can be generalized as follows:

Proposition 7 In the two-stage elimination contest the optimal head start has the same value if the number

of …nalists is either two or is equal to the number of players.

Proof. By (6) we have

and by (13) we also have

(k = 2) = 2n 1

n 1

n+1

1 X

(k = n) =

(n 1) 2 (n n i + 2

n i + 2

=

i=2

1) + 1

n+1

1 X

(n 1) 2 (n 1) + 1 = n

n 1 + 1 = 2n 1

n 1

i=2

16

Figure 3 showed that the optimal head start (k) may either increase or decrease in the number of

…nalists. However, if the number of players is su¢ ciently high we obtain that

Proposition 8 In the two-stage elimination contest if the number of players approaches in…nity then the

optimal head start decreases in the number of …nalists.

Proof. By (13) we have

lim

n!1 = lim

n!1

k+1

1 X

(k 1) 2 (n k i + 2

n i + 2

i=2

1) + 1

By L’hopital’s rule we obtain

lim

n!1 =

1

(k 1) 2 (k2 1

= 3k 2

2k 2

kX

i) + 1

i=2

Thus, when the number of players n converges to in…nity, the optimal head start decreases in the number

of …nalists k.

So far we have analyzed the optimal head start for any number of …nalists k; but we can also calculate

the optimal number of …nalists k given the optimal head start. Suppose …rst that the number of players is

n = 3: Then by Proposition 7 and (6), the optimal head start is the same for either two or three …nalists

and is equal to = 2:5: Thus, by (5) the total e¤ort with two …nalists is T E(k = 2) = 0:7483, while by

(11) the total e¤ort with three …nalists is T E(k = 3) = 0:7546. Thus we have

Proposition 9 In the two-stage elimination contest with three players it is optimal to advance all the players

to the …nal stage when the winner in the …rst stage has the optimal head start = 2:5:

We can now numerically analyze the optimal number of …nalists for any number of players n > 3.

Consider …rst a two-stage elimination contest without a head start, i.e., = 1. Then, the optimal number

of …nalists for di¤erent values of the number of players n are

n : 3 4 5 6 7 8 9 10 15 20 30 40 50 60 100 150 300 500

k : 2 2 3 3 3 4 4 4 5 6 8 10 11 12 16 20 29 37

17

Figure 4: The optimal total e¤ort as a function of the number of …nalists k:

We can see that the optimal number of …nalists increases in the number of players n.

The optimal

numbers of …nalists k for di¤erent values of the head start when the number of players is n = 100 are

: 1 1:1 1:3 1:5 1:7 2 3 4 5 10 15 50 100 200 300

k : 16 19 23 26 28 29 32 34 35 36 37 37 37 37 37

We can see that the optimal number of …nalists k () increases in the value of the head start , but from

a relatively large …nite value of , the optimal number of …nalists almost does not change. By our analysis,

we can …nd the optimal two-stage elimination contest for any number of players, as shown in Figure 4 where

we …nd the optimal head start and then calculate the players’total e¤ort for any number of …nalists. [Figure

4 to be here]. We can see that if the number of players is n = 100, the highest total e¤ort is obtained when

the number of …nalists is 27. Then the optimal head start is = 1:57:

18

5 Concluding remarks

In contrast to the one-stage contest in which the designer can a¤ect the players’performance (e¤ort) mostly

by determining the prize structure, in multi-stage contests he has other tools to a¤ect these performances.

In our two-stage elimination contest, for instance, the designer determines the number of …nalists and the

head start given to the winner in the …rst stage. We showed that, independent of the number of …nalists,

it is optimal for the designer to maximize the players’total e¤ort by giving a head start to the winner in

the …rst stage such that he will have some advantage in the competition in the …nal stage. Moreover, we

showed that the contest designer can increase the players’ total e¤ort by the optimal combination of the

number of …nalists and the head start given to the winner in the …rst stage. We found that the optimal

head start is monotonically decreasing in the number of players but is not a monotonic function of the

number of the …nalists, while the number of …nalists is monotonically increasing in the number of players.

We demonstrated that by controlling some structural parameters of the two-stage elimination contest the

designer can signi…cantly increase the contestants’performance compared to the one-stage contest.

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