Konrad (2007) - Strategy In Contests-An Introduction
Contents
Reference(s)
Konrad, Kai A. (2007), "Strategy in Contests-An Introduction", WZB-Markets and Politics Working Paper No. SP II 2007-01, link pdf
Abstract
Competition in which goods or rents are allocated as a function of the various efforts expended by players in trying to win these goods or rents is a very common phenomenon. A subset of examples comes from marketing, litigation, relative reward schemes or promotion tournaments in internal labor markets, beauty contests, influence activities, education filters, R&D contests, electoral competition in political markets, military conflict and sports. I survey here this type of competition which is sometimes called contest or tournament. I focus on the role of its various design aspects, such as prize structure, sequencing, nesting, repetition, elimination contests and many others. Some key insights about the nature and properties of this type of competition emerge from this analysis.
Keywords: Survey of contests, tournaments, conflict, strategic aspects
Summary
The paper provides a series of descriptions of setting in which contests might be applied, and then introduces three types of contests, followed by a detailed review of a number of other specific contest models with advanced assumptions.
The three key models (detailed below) are:
- First price all-pay auctions
- Additive noise contests
- The Tullock contest
The advanced assumption models (not detailed below - see the paper) include:
- Timing and Participation Models:
- Endogeneous Timing
- Voluntary Particpation
- Exclusion
- Cost and Prize Structure
- Choice of Cost
- Multiple Prizes
- Endogenous Prizes
- Delegation
- Externalities
- Joint Ownership
- Sabotage
- Information Externalities
- Public Goods and Free Riding
- Grand Contests
- Nested Contests
- Exogenous Sharing Rules
- The Choice of Sharing Rules
- Intra-group conflict
- Alliances
- Repeated Battles
- Nested Contests
Three Key Models
The following section provides the basic three models.
The general set up is as follows:
- There are [math]N\,[/math] players: [math]N=\{1,\ldots,n\}\,[/math]
- Players can exert effort [math]x_i\,[/math], [math]X=\{x_1,\ldots,x_n\}\,[/math]. Implicitly [math]x_i \in \mathbb{R}_+ \forall i\,[/math]
- There is, unless otherwise stated, a single prize [math]B\,[/math]
- Players have values [math]v_i(B)\,[/math] and costs [math]C_i(x_i)\,[/math]
The "contest success function" maps effort into probabilities:
- [math]p_i = p_i(x), p_i: \mathbb{R}_{+} \to [0,1]\,[/math]
Players have profit functions (i.e. utilities):
- [math]\pi_i(x)=p_i(x)\cdot v_i(B) - C_i(x_i)\,[/math]
All-pay auctions
To make things interesting but tractable we assume:
- 2 Players
- [math]C_i = C \forall i\,[/math]
- [math]v_1 \ge v_2 \ge 0\,[/math]
- Valuations and costs are common knowledge
The contest function for player 1 (for player 2 it is [math]1-p_1\,[/math]) is:
[math]
p_1(x) =
\begin{cases}
1 & x_i \gt x_2 \\
\frac{1}{2} & x_1 = x_2 \\
0 & x_1 \lt x_2
\end{cases}
[/math]
There is no equilibrium in pure strategies. This can be proved by contradiction, starting from the assumptions that [math](x_1,x_2)\,[/math] is a pure strategy NE. In mixed strategies there is an equilibrium described by the CDF of the distributions of effort as follows:
[math]
F_1(x_1) =
\begin{cases}
\frac{x_1}{v_2} & x_1 \in [0,v_2] \\
1 & x_1 \gt x_2
\end{cases}
[/math]
[math]
F_2(x_2) =
\begin{cases}
\left(1-\frac{v_2}{v_1}\right)+\frac{x_2}{v_1} & x_2 \in [0,v_2] \\
1 & x_1 \gt x_2
\end{cases}
[/math]
Contestant 2 will never expend effort greater than [math]v_2\,[/math], and contestant 1 can win for sure is he expends an epsilon more, so this constraints the effort of both parties to lie on the range [math][0,v_2]\,[/math]. If contestant 1 expends an epsilon more than [math]v_2\,[/math] he gaurantees a payoff of [math]v_1 - v_2\,[/math]; this is his equilibrium payoff. If he does this then player 2's payoff is zero; this is his equilibrium payoff! Under mixed strategies a player must be indifferent at all points, we use this indifference to solve the game:
- [math]\pi_1(x_1) = F_2(x_1)v_1 - x_1 = v_1-v_2\,[/math]
and
- [math]\pi_2(x_2) = F_1(x_2)v_2 - x_2 = 0\,[/math]
Together this can be used to solve for [math]F_1\,[/math] and [math]F_2\,[/math], yielding the results above. Using the standard expected value formula (and differentiating the CDFs to get PDFs first):
- [math]\mathbb{E}(X) = \int_{-\infty}^\infty x f(x)dx\,[/math]
We get that:
- [math]\mathbb{E}x_1 = \frac{v_2}{2}\,[/math] and [math]\mathbb{E}x_2=\frac{v_2^2}{2v_1}\,[/math]
Furthermore, we note that:
- [math](\mathbb{E}x_1 + \mathbb{E}x_2) = \frac{v_1 v_2 + v_2^2}{2v_1}\,[/math]
which under the assumption that [math]v_1 \ge v_2\,[/math], we have [math](\mathbb{E}x_1 + \mathbb{E}x_2) \le v_2\,[/math].
So not only is the auction in efficient in that it doesn't extract the value of the second highest player, but it is also inefficient in the sense that it does not necessarily allocate the good to the individual with the highest valuation.
Tullock Contests
The Tullock contest success function is:
[math] p_i(x) = \begin{cases} \frac{x_i^r}{\sum_{j=1}^n x_j^r & \mbox{ if } max \{x\} \gt 0 \\ \frac{1}{N} & \mbox{otherwise} \end{cases} [/math]
For [math]r=1\,[/math] this is called the lottery contest. Generally it is assumed that the cost function [math]C_i(x_i)=x_i\,[/math]. With [math]r=1\,[/math] adding constants to the function or changing [math]C_i(x_i)\,[/math] for some [math]i\,[/math] introduces handicaps.
There exists a pure strategy Nash equilibrium iff [math]r \le \frac{n}{n-1}\,[/math], though this generally leads to rent dissapation. For [math]r \gt \frac{n}{n-1}\,[/math] there are mixed strategy NE. In this case it is possible that [math]\sum x_i \gt B\,[/math], that is that there is excess total effort, but [math]\mathbb{E}(\sum x_i) \le B\,[/math].
For two players the first order conditions give:
- [math]\frac{rx_i^{r-1}x_j^r}{(x_i^r +x_j^r)}\cdot v_1 = 1\,[/math]
Plugging in for [math]v_i\,[/math], [math]v_j\,[/math] and [math]r\,[/math] provides a set of reaction functions.
For any number of players one can solve for the symmetry pure strategy Nash equilibrium by assuming [math]x_i = x_j \forall i\,[/math] and rearranging to get:
- [math]x_i = \frac{r(n-1)}{n^2}\cdot B \,[/math]
Taking the summation gives:
- [math]\sum x_i = \frac{r(n-1)}{n}\cdot B\,[/math]
And so we can see that with [math]r=1\,[/math] the rents will be fullly dissapated as [math]n \to \infty\,[/math].