Sabotage in team contests

Serhat Doğan; Kerim Keskin; Çağrı Sağlam

doi:10.1007/s11127-019-00643-1

Public Choice

September 2019, Volume 180, Issue 3–4, pp 383–405 | Cite as

Sabotage in team contests

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Serhat Doğan
Kerim Keskin
Çağrı Sağlam

Article

First Online: 01 February 2019

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Abstract

In the contest literature, sabotage is defined as a deliberate and costly activity that damages the opponent’s likelihood of winning the contest. Most of the existing results suggest that, anticipating a possible sabotage, contestants would be discouraged from exerting high efforts. In this paper we investigate the act of sabotage in a team contest wherein team members exert costly efforts as a contribution to their team’s aggregate effort, which in turn determines the contest’s outcome. For the baseline model with no sabotage, there exists a corner equilibrium implying a free-rider problem in each team. As for the model with sabotage, our characterization of Nash equilibrium reveals two important results: (i) a unique interior equilibrium exists so that the free-rider problem no longer is a concern and (ii) the discouragement effect of sabotage vanishes for some players. On top of those conclusions, we investigate the team owner’s problems of prize allocation and team formation with the objective being to maximize his team’s winning probability.

Keywords

Team contests Sabotage Tullock contests Free riding Discouragement effect Encouragement effect

JEL Classification

C72 D74

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Notes

Acknowledgements

We would like to thank the editor and an anonymous reviewer, as well as seminar participants at ADA University, Bilgi University, Bosphorus Workshop on Economic Design, and Koç University Winter Workshop in Economics. The usual disclaimer applies.

Appendix 1

Proof of Proposition 1

Given an aggregate effort ${\mathcal {E}}_2$ for team 2, the first-order condition with respect to $e_1^a$ for player a in team 1 is

$$\begin{aligned} \frac{\gamma _1^a {\mathcal {E}}_2}{({\mathcal {E}}_1 + {\mathcal {E}}_2)^2} V_1^a - 1 = 0 . \end{aligned}$$

For player d in team 1, a symmetric first-order condition can be written as

$$\begin{aligned} \frac{\gamma _1^d {\mathcal {E}}_2}{({\mathcal {E}}_1 + {\mathcal {E}}_2)^2} V_1^d - 1 = 0 . \end{aligned}$$

Accordingly, it must be that

$$\begin{aligned} \gamma _1^a V_1^a = \gamma _1^d V_1^d \end{aligned}$$

(1)

in the equilibrium, which is not necessarily true. That leads to a corner solution such that if the right-hand side exceeds the left-hand side, then only the attackers exert positive productive effort in the equilibrium, and vice versa. Considering a symmetric result for the other team, and under the assumption that $\gamma _1^j V_1^j > \gamma _1^{j'} V_1^{j'}$ and $\gamma _2^k V_2^k > \gamma _2^{k'} V_2^{k'}$, the respective equilibrium efforts are

$$\begin{aligned} e_1^j = \frac{\gamma _1^j \gamma _2^k \left( V_1^j\right)^2 V_2^k}{\left( \gamma _1^j V_1^j + \gamma _2^k V_2^k\right) ^2}, \quad e_1^{j'} = 0 , \\ e_2^k = \frac{\gamma _1^j \gamma _2^k V_1^j \left( V_2^k\right)^2}{\left( \gamma _1^j V_1^j + \gamma _2^k V_2^k\right) ^2}, \quad e_2^{k'} = 0. \end{aligned}$$

so that

$$\begin{aligned} {\mathcal {E}}^*_1 = \frac{\left( \gamma _1^j\right)^2 \gamma _2^k \left( V_1^j\right)^2 V_2^k}{\left( \gamma _1^j V_1^j + \gamma _2^k V_2^k\right) ^2}, \quad \hbox {and }\quad {\mathcal {E}}^*_2 = \frac{\gamma _1^j \left( \gamma _2^k\right)^2 V_1^j \left( V_2^k\right)^2}{\left( \gamma _1^j V_1^j + \gamma _2^k V_2^k\right) ^2} . \end{aligned}$$

(2)

Finally, for the sake of completeness, we must note that if Eq. (1) indeed holds for team $i \in \{1,2\}$, then multiple equilibria exist such that both members of team i exert non-negative productive efforts reaching an aggregate effort of ${\mathcal {E}}_i^*$. $\square$

Proof of Proposition 2

Consider the maximization problems for players d in team 1 and a in team 2. The corresponding first-order conditions can be written as

$$\begin{aligned}&\frac{\partial {U_1^d}}{\partial {e_1^d}}&= \dfrac{{\gamma _1^d}}{1+{s_2^a}} \dfrac{E_2}{(E_1+E_2)^2}{V_1^d} - 1 =0 \end{aligned}$$

(3)

$$\begin{aligned}&\frac{\partial {U_1^d}}{\partial {s_1^d}}&= \dfrac{{\gamma _2^a}}{\left( 1+{s_1^d}\right) ^2} \dfrac{ {e_2^a}E_1}{(E_1+E_2)^2}{V_1^d} - {\mu _1^d} =0 \end{aligned}$$

(4)

$$\begin{aligned}&\frac{\partial {U_2^a}}{\partial {e_2^a}}&= \dfrac{{\gamma _2^a}}{1+{s_1^d}} \dfrac{E_1}{(E_1+E_2)^2}{V_2^a} - 1 =0 \end{aligned}$$

(5)

$$\begin{aligned}&\frac{\partial {U_2^a}}{\partial {s_2^a}}&= \dfrac{{\gamma _1^d}}{\left( 1+{s_2^a}\right) ^2} \dfrac{ {e_1^d}E_2}{(E_1+E_2)^2}{V_2^a} - {\mu _2^a} =0 \end{aligned}$$

(6)

From (3) we get

$$\begin{aligned} \frac{{\gamma _1^d} E_2 {V_1^d}}{(E_1+E_2)^2} = 1+{s_2^a} ; \end{aligned}$$

and from (6) we get

$$\begin{aligned} \frac{{\gamma _1^d} E_2 {V_2^a}}{(E_1+E_2)^2} = {\mu _2^a}\frac{\left( 1+{s_2^a}\right) ^2}{{e_1^d}} . \end{aligned}$$

Thus

$$\begin{aligned} {e_1^d} = \frac{{\mu _2^a}{\gamma _1^d} E_2 \left( {V_1^d}\right) ^2}{{V_2^a}(E_1+E_2)^2} . \end{aligned}$$

In a similar manner, we can write all productive and sabotage efforts in terms of each $\gamma _i^j$, $V_i^j$, $\mu _i^j$, and $E_i$. The values are

$$\begin{aligned} 1+{s_1^d}= & {} \frac{{\gamma _2^a} E_1 {V_2^a}}{(E_1+E_2)^2}, \quad 1+{s_1^a} = \frac{{\gamma _2^d} E_1 {V_2^d}}{(E_1+E_2)^2} , \end{aligned}$$

(7)

$$\begin{aligned} 1+{s_2^d}= & {} \frac{{\gamma _1^a} E_2 {V_1^a}}{(E_1+E_2)^2}, \quad 1+{s_2^a} = \frac{{\gamma _1^d} E_2 {V_1^d}}{(E_1+E_2)^2} , \end{aligned}$$

(8)

$$\begin{aligned} {e_1^d}= & {} \frac{{\mu _2^a}{\gamma _1^d} E_2 \left( {V_1^d}\right) ^2}{{V_2^a}(E_1+E_2)^2}, \quad {e_1^a} = \frac{{\mu _2^d}{\gamma _1^a} E_2 \left( {V_1^a}\right) ^2}{{V_2^d}(E_1+E_2)^2} , \end{aligned}$$

(9)

$$\begin{aligned} {e_2^d}= & {} \frac{{\mu _1^a}{\gamma _2^d} E_1 ({V_2^d})^2}{{V_1^a}(E_1+E_2)^2}, \quad {e_2^a} = \frac{{\mu _1^d}{\gamma _2^a} E_1 \left( {V_2^a}\right) ^2}{{V_1^d}(E_1+E_2)^2} . \end{aligned}$$

(10)

Notice that we also have

$$\begin{aligned} \frac{{e_2^a}}{(1+{s_1^d})} ={\mu _1^d}\frac{{V_2^a}}{{V_1^d}}; \end{aligned}$$

and by symmetry,

$$\begin{aligned} \frac{{e_1^a}}{\left( 1+{s_2^d}\right) } ={\mu _2^d}\frac{{V_1^a}}{{V_2^d}} \hbox { ; }\frac{{e_2^d}}{\left( 1+{s_1^a}\right) } ={\mu _1^a}\frac{{V_2^d}}{{V_1^a}} \quad \hbox {and }\frac{{e_1^d}}{\left( 1+{s_2^a}\right) } ={\mu _2^a}\frac{{V_1^d}}{{V_2^a}} . \end{aligned}$$

Thus,

$$\begin{aligned} E^*_1 = {\gamma _1^d}\frac{{e_1^d}}{1+ {s_2^a}} + {\gamma _1^a}\frac{{e_1^a}}{1+ {s_2^d}} = {\gamma _1^d} {\mu _2^a}\frac{{V_1^d}}{{V_2^a}} + {\gamma _1^a}{\mu _2^d}\frac{{V_1^a}}{{V_2^d}} \end{aligned}$$

and

$$\begin{aligned} E^*_2 = {\gamma _2^d}\frac{{e_2^d}}{1+ {s_1^a}} + {\gamma _2^a}\frac{{e_2^a}}{1+ {s_1^d}} = {\gamma _2^d} {\mu _1^a}\frac{{V_2^d}}{{V_1^a}} + {\gamma _2^a}{\mu _1^d}\frac{{V_2^a}}{{V_1^d}} . \end{aligned}$$

By replacing those $E^*_1$ and $E^*_2$ values into Eqs. (7)–(10), we can write the equilibrium values of all productive and sabotage efforts.

Finally, from Eqs. (7)–(10), we can derive the necessary and sufficient conditions for the existence of an interior equilibrium: Given positive aggregate efforts for both teams, an interior equilibrium exists if and only if for every player $j \in \{a,d\}$ in team $i \in \{1,2\}$, we have $s_i^j > 0$, i.e.,

$$\begin{aligned} \frac{{\gamma _i^j} E_{-i} {V_i^j}}{(E_1+E_2)^2} > 1. \end{aligned}$$

And those inequalities are satisfied easily if all of the winning prizes are sufficiently large.¹⁴ This completes the proof. $\square$

Proof of Proposition 3

For given values of $\gamma _i^j$ and $\mu _i^j$ for every player $j \in \{a,d\}$ in team $i \in \{1,2\}$, the owner of team 1 aims to maximize

$$\begin{aligned} P_1(E_1, E_2) = \frac{E_1}{E_1+E_2} = \frac{{\gamma _1^d} {\mu _2^a}\frac{{V_1^d}}{{V_2^a}} + {\gamma _1^a}{\mu _2^d}\frac{{V_1^a}}{{V_2^d}}}{\left( {\gamma _1^d} {\mu _2^a}\frac{{V_1^d}}{{V_2^a}} + {\gamma _1^a}{\mu _2^d}\frac{{V_1^a}}{{V_2^d}}\right) + \left( {\gamma _2^d} {\mu _1^a}\frac{{V_2^d}}{{V_1^a}} + {\gamma _2^a}{\mu _1^d}\frac{{V_2^a}}{{V_1^d}}\right) } . \end{aligned}$$

The first-order condition for team 1 yields¹⁵

$$\begin{aligned} \frac{\partial E_1}{\partial {V_1^d}}E_2 = \frac{\partial E_2}{\partial {V_1^d}} E_1 . \end{aligned}$$

Now, considering that ${V_1^a}$ and ${V_1^d}$ are dependent variables, we have

$$\begin{aligned} \left( {\gamma _1^d}{\mu _2^a}\frac{1}{{V_2^a}} - {\gamma _1^a} {\mu _2^d} \frac{1}{{V_2^d}}\right) E_2 = \left( {\gamma _2^d}{\mu _1^a}\frac{{V_2^d}}{\left( {V_1^a}\right) ^2} - {\gamma _2^a} {\mu _1^d} \frac{{V_2^a}}{\left( {V_1^d}\right) ^2}\right) E_1 . \end{aligned}$$

Similarly, for team 2 we get

$$\begin{aligned} \left( {\gamma _2^d}{\mu _1^a}\frac{1}{{V_1^a}} - {\gamma _2^a} {\mu _1^d} \frac{1}{{V_1^d}}\right) E_1 = \left( {\gamma _1^d}{\mu _2^a}\frac{{V_1^d}}{\left( {V_2^a}\right) ^2} - {\gamma _1^a} {\mu _2^d} \frac{{V_1^a}}{\left( {V_2^d}\right) ^2}\right) E_2 . \end{aligned}$$

After arranging terms, we are left with

$$\begin{aligned} \left[ {\gamma _1^d}{\gamma _2^d}{\mu _1^a}{\mu _2^a}\left( {V_1^d}{V_2^d}\right) ^2 + {\gamma _1^a}{\gamma _2^a}{\mu _1^d}{\mu _2^d}\left( {V_1^a}{V_2^a}\right) ^2\right] \left( {V_1^d}{V_2^d} - {V_1^a}{V_2^a}\right) = 0 . \end{aligned}$$

Since the first term is positive, it must be that ${V_1^d}{V_2^d} = {V_1^a}{V_2^a}$. From this equation we find that

$$\begin{aligned} \frac{{V_1^d}}{{V_2^a}} = \frac{{V_1^a}}{{V_2^d}} = \frac{{V_1^d} + {V_1^a}}{{V_2^d} + {V_2^a}} = \frac{V_1}{V_2} . \end{aligned}$$

Then, for the sake of expositional simplicity, we set

$$\begin{aligned} \rho _{1}^d ={\gamma _1^d} {\mu _2^a}, \quad \rho _{2}^d = {\gamma _2^d} {\mu _1^a}, \quad \rho _{1}^a = {\gamma _1^a} {\mu _2^d}, \quad \hbox { and } \rho _{2}^a = {\gamma _2^a} {\mu _1^d} . \end{aligned}$$

Now we can write

$$\begin{aligned} E_1 = {\gamma _1^d} {\mu _2^a}\frac{{V_1^d}}{{V_2^a}} + {\gamma _1^a}{\mu _2^d}\frac{{V_1^a}}{{V_2^d}} = \frac{V_1}{V_2} \left( \rho _{1}^d + \rho _{1}^a\right) \end{aligned}$$

and

$$\begin{aligned} E_2 = {\gamma _2^d} {\mu _1^a}\frac{{V_2^d}}{{V_1^a}} + {\gamma _2^a}{\mu _1^d}\frac{{V_2^a}}{{V_1^d}} = \frac{V_2}{V_1} \left( \rho _{2}^d + \rho _{2}^a\right) . \end{aligned}$$

Following some algebraic operations, the first-order condition for team 1 can be rewritten as

$$\begin{aligned} \left( \rho _{1}^d \frac{V_1}{V_2 {V_1^d}} - \rho _{1}^a \frac{V_1}{V_2 {V_1^a}}\right) \frac{V_2}{V_1}\left( \rho _{2}^d + \rho _{2}^a\right) = \left( \rho _{2}^d \frac{V_2}{V_1 {V_1^a}} - \rho _{2}^a \frac{V_2}{V_1 {V_1^d}}\right) \frac{V_1}{V_2}\left( \rho _{1}^d + \rho _{1}^a\right) . \end{aligned}$$

Canceling out all $V_1$ and $V_2$, we have

$$\begin{aligned} \frac{\rho _{1}^d\left( \rho _{2}^d + \rho _{2}^a\right) + \rho _{2}^a \left( \rho _{1}^d + \rho _{1}^a\right) }{{V_1^d}}= \frac{\rho _{2}^d \left( \rho _{1}^d + \rho _{1}^a\right) + \rho _{1}^a \left( \rho _{2}^d + \rho _{2}^a\right) }{{V_1^a}} . \end{aligned}$$

From this equality we find

$$\begin{aligned} {V_1^d} = \frac{V_1 \left[ \rho _{1}^d \left( \rho _{2}^d + \rho _{2}^a\right) + \rho _{2}^a \left( \rho _{1}^d + \rho _{1}^a\right) \right] }{2\left( \rho _{2}^d + \rho _{2}^a\right) \left( \rho _{1}^d + \rho _{1}^a\right) } = \frac{V_1}{2}\left( \frac{\rho _{1}^d}{\rho _{1}^d + \rho _{1}^a} + \frac{ \rho _{2}^a }{\rho _{2}^d + \rho _{2}^a}\right) . \end{aligned}$$

Finally, returning back to the standard notation, we have

$$\begin{aligned} {V_1^{d*}} = \frac{V_1}{2}\left( \frac{{\gamma _1^d} {\mu _2^a}}{{\gamma _1^d} {\mu _2^a} + {\gamma _1^a} {\mu _2^d}} + \frac{ {\gamma _2^a} {\mu _1^d} }{{\gamma _2^d} {\mu _1^a} + {\gamma _2^a} {\mu _1^d} }\right) . \end{aligned}$$

The optimal share for the other player is $V_1^{a*} = V_1 - V_1^{d*}$. $\square$

Proof of Proposition 4

Now that the strategic interaction is absent, the analysis turns out to be much simpler. The first-order condition with respect to $\gamma _1^d$ is

$$\begin{aligned} {\mu ^a}\frac{{V_1^d}}{{V_2^a}} + {\mu ^d}\frac{{V_1^a}}{{V_2^d}} \frac{\partial \gamma _1^a}{\partial \gamma _1^d} = 0 . \end{aligned}$$

Moreover, from the derivative of the budget constraint, it follows that

$$\begin{aligned} \alpha \left( \gamma _1^a\right) ^{\alpha -1}\frac{\partial \gamma _1^a}{\partial \gamma _1^d} + \alpha \left( \gamma _1^d\right) ^{\alpha -1} = 0 \end{aligned}$$

which implies

$$\begin{aligned} \frac{\partial \gamma _1^a}{\partial \gamma _1^d} = -\,\left( \frac{\gamma _1^d}{\gamma _1^a}\right) ^{\alpha -1} . \end{aligned}$$

Using this information in the first-order condition above, we have

$$\begin{aligned} {\mu ^a}\frac{{V_1^d}}{{V_2^a}} - {\mu ^d}\frac{{V_1^a}}{{V_2^d}} \left( \frac{\gamma _1^d}{\gamma _1^a}\right) ^{\alpha -1} = 0 \end{aligned}$$

so that

$$\begin{aligned} \frac{\gamma _1^d}{\gamma _1^a} = \left( \frac{{\mu ^a}{V_1^d}{V_2^d}}{\mu ^d{V_2^a}{V_1^a}} \right) ^{\frac{1}{\alpha -1}} . \end{aligned}$$

Then putting this finding into the budget constraint, we have

$$\begin{aligned} \gamma _1^{a*} = \left( \dfrac{\Gamma _1}{1 + \left( \dfrac{{\mu ^a}{V_1^d}{V_2^d}}{\mu ^d{V_2^a}{V_1^a}} \right) ^{\frac{\alpha }{\alpha -1}}}\right) ^\frac{1}{\alpha } \quad \hbox {and }\quad \gamma _1^{d*} = \left( \dfrac{\Gamma _1}{1 + \left( \dfrac{\mu ^d{V_2^a}{V_1^a}}{{\mu ^a}{V_1^d}{V_2^d}} \right) ^{\frac{\alpha }{\alpha -1}}}\right) ^\frac{1}{\alpha } . \end{aligned}$$

$\square$

Appendix 2

The alternative model with restricted sabotage

In this paper, we have considered directionally restricted sabotage allowing each member of a team to sabotage only a particular member of the opposing team. Here we relax that assumption and analyze the case of directed sabotage: each team member can sabotage any member of the opposing team. Similar to our original model, $e_i^j$ denotes the productive effort exerted by player $j\in \{a,d\}$ in team $i\in \{1,2\}$. As for the sabotage efforts, we need a new notation including the origin and the destination of sabotage. Let $s_i^{jk}$ denote the sabotage made by player $j \in \{a,d\}$ in team $i \in \{1,2\}$ against player $k \in \{a,d\}$ in the opposing team. Also let $\mu _i^{jk}$ denote the corresponding marginal cost of such sabotage activity.

Then, we can write the respective aggregate effort functions as follows:

$$\begin{aligned} E_1 = \frac{\gamma _1^a}{1+s_2^{aa}+s_2^{da}}e_1^a + \frac{\gamma _1^d}{1+s_2^{ad}+s_2^{dd}}e_1^d \end{aligned}$$

and

$$\begin{aligned} E_2 = \frac{\gamma _2^a}{1+s_1^{aa}+s_1^{da}}e_2^a + \frac{\gamma _2^d}{1+s_1^{ad}+s_1^{dd}}e_2^d . \end{aligned}$$

Assuming that

$$\begin{aligned} C_i^j\left( e_i^j,s_i^{ja},s_i^{jd}\right) = e_i^j + \mu _i^{ja}s_i^{ja} + \mu _i^{jd}s_i^{jd}, \end{aligned}$$

we say that player $j\in \{a,d\}$ in team $i\in \{1,2\}$ maximizes

$$\begin{aligned} U_i^j \left( \left( e_i^j,s_i^{ja},s_i^{jd}\right) , \cdot \right) = \frac{E_i}{E_1+E_2}V_i^j -e_i^j - \mu _i^{ja}s_i^{ja} - \mu _i^{jd}s_i^{jd} . \end{aligned}$$

Utilizing the first-order conditions with respect to $s_1^{aa}$ and $s_1^{da}$, we get

$$\begin{aligned} \frac{\partial {U_1^a}}{\partial {s_1^{aa}}}&= \dfrac{{\gamma _2^a}}{\left( 1+{s_1^{da} + s_1^{aa}}\right) ^2} \dfrac{e_2^a E_1}{\left( E_1+E_2\right) ^2}{V_1^a} - \mu _1^{aa} =0 \\ \frac{\partial {U_1^d}}{\partial {s_1^{da}}}&= \dfrac{{\gamma _2^a}}{\left( 1+{s_1^{da} + s_1^{aa}}\right) ^2} \dfrac{e_2^a E_1}{\left( E_1+E_2\right) ^2}{V_1^d} - \mu _1^{da} =0 . \end{aligned}$$

In order for these first-order conditions to be satisfied simultaneously, it must be that

$$\begin{aligned} \frac{V_1^a}{\mu _1^{aa}} = \frac{V_1^d}{\mu _1^{da}}. \end{aligned}$$

Otherwise, we must have a corner solution. To put it differently, unless the last equality is satisfied, directionally restricted sabotage would be observed in the equilibrium. Below we elaborate further on that issue.

Note that if any of the derivatives with respect to $s_1^{aa}$ and $s_1^{da}$ are positive, then a marginal increment in the corresponding variable would be a possible deviation. Therefore, none can be positive at an equilibrium. The result implies that if one of the first-order conditions is satisfied, then the derivative with respect to the other variable should be negative, which corresponds to a corner solution for that variable. For more concrete arguments, assume without loss of generality that

$$\begin{aligned} \frac{V_1^a}{\mu _1^{aa}} < \frac{V_1^d}{\mu _1^{da}}. \end{aligned}$$

If the former first-order condition is satisfied, then the derivative with respect to $s_1^{da}$ would be positive. That cannot happen in equilibrium. Then, it must be that the latter first-order condition is satisfied, meaning that a corner solution exists for $s_1^{aa}$, which is $s_1^{aa}=0$.

Finally, given our model’s interpretation, it is reasonable to assume that $\mu _1^{aa} > \mu _1^{da}$. That is because the defenders in team 1 are located closer to the attackers in team 2 than the attackers in team 1, so that if the attackers in team 2 are to be sabotaged, the defenders in team 1 should have a lower cost than the attackers in team 1.¹⁶ Accordingly, for a wide range of $V_1^a$, $V_1^d$, $V_2^a$, and $V_2^d$ values, the current model would reduce to our original model with directionally restricted sabotage in the equilibrium.

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Serhat Doğan
- 1
Kerim Keskin
- 2
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Çağrı Sağlam
- 1

1.Department of EconomicsBilkent UniversityAnkaraTurkey
2.Department of EconomicsKadir Has UniversityIstanbulTurkey

Sabotage in team contests

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