Optimality of winner-take-all contests: the role of attitudes toward risk

Abstract

This paper studies the role of risk attitudes in determining the optimality of winner-take-all contests. We compare the typical single-winner lottery contest with two alternatives, both spreading the rewards to more players: through holding multiple prize-giving lottery competitions or through guaranteeing a bottom prize for the losers. In the first comparison, we find that the multiple-competition contest is as effective as the winner-take-all contest when the contestants are risk neutral, but the former induces more effort than the latter when the contestants are both risk averse and prudent. In the second comparison, we find that the contest with a bottom prize is always dominated by the winner-take-all contest when the contestants are risk neutral, but the former could have an advantage over the latter when the contestants are both risk averse and prudent, and it is more likely so as the contestants become more prudent.

Appendices

Appendix 1: Condition 1 under the ratio-form contest success function and the CARA or CRRA utility function

Let $p_{i} (x_{1} , \ldots ,x_{n} ) = \frac{{x_{i}^{r} }}{{\sum\nolimits_{k = 1}^{n} {x_{k}^{r} } }}$, we have

$$\frac{{\partial p_{i} (x_{1} , \ldots ,x_{n} )}}{{\partial x_{i} }} = \frac{{\left( {rx_{i}^{r - 1} } \right)\left( {\sum\nolimits_{k = 1}^{n} {x_{k}^{r} } - x_{i}^{r} } \right)}}{{\left( {\sum\nolimits_{k = 1}^{n} {x_{k}^{r} } } \right)^{2} }}.$$

Therefore,

$$p_{x} (x) \equiv \frac{{\partial p_{i} (x, \ldots ,x)}}{{\partial x_{i} }} = \frac{(n - 1)r}{{n^{2} }} \cdot \frac{1}{x}.$$

(13)

Note that the derivative of $- \left[ {\tfrac{1}{n}u^{\prime}(w + b - x) + (1 - \tfrac{1}{n})u^{\prime}(w - x)} \right]$ in F(x) is non-positive when $u^{\prime\prime}( \cdot ) \le 0$. As a result, a sufficient condition for Condition 1 to hold is

$$\frac{{dp_{x} (x)}}{dx}\left[ {u(w + b - x) - u(w - x)} \right] - p_{x} (x)\left[ {u^{\prime}(w + b - x) - u^{\prime}(w - x)} \right] < 0,$$

or equivalently (according to (13))

$$- \frac{1}{x}\left[ {u(w + b - x) - u(w - x)} \right] - \left[ {u^{\prime}(w + b - x) - u^{\prime}(w - x)} \right] < 0.$$

(14)

• (i) The Case of CARA: $u(y) = - e^{ - \lambda y}$, $\lambda \ge 0$, $y > 0$

  In this case, $\lambda$ is the (constant) absolute risk aversion measure, and (14) is equivalent to

  $$\lambda x - 1 < 0.$$

  (15)

  That is, Condition 1 is satisfied as long as x is sufficiently small (given the value of $\lambda$).

• (ii) The Case of CRRA: $u(y) = \frac{{y^{1 - \gamma } }}{1 - \gamma }$, $\gamma \ge 0{\text{ and}}\;\gamma \ne 1$, $y > 0$

  In this case, $\gamma$ is the (constant) relative risk aversion measure, and (14) is equivalent to

  $$- \frac{1}{x}\frac{1}{1 - \gamma }\left[ {(w + b - x)^{1 - \gamma } - (w - x)^{1 - \gamma } } \right] - \left[ {(w + b - x)^{ - \gamma } - (w - x)^{ - \gamma } } \right] < 0.$$

  (16)

Note that the LHS of (16) is zero when b = 0. So for (16) to hold when b > 0, it is sufficient that the derivative of the LHS of (16) with respect to b is negative, or

$$- \frac{w + b - x}{x} + \gamma < 0.$$

(17)

That is, Condition 1 is satisfied as long as x is sufficiently small (given the value of $\gamma$).

Appendix 2: Derivation of (4′)

The first order condition for player i’s problem to maximize expected utility $\sum\nolimits_{k = 0}^{m} {\left[ {\left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)p_{i}^{m - k} (1 - p_{i} )^{k} u\left( {w + (m - k)\frac{b}{m} - x_{i} } \right)} \right]}$, where $\left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right) = \frac{m(m - 1) \cdots (m - k + 1)}{{k!}}$, by choosing $x_{i}$ is

$$\begin{aligned} \frac{{\partial p_{i} }}{{\partial x_{i} }}&\biggl\{ \sum\limits_{k = 1}^{m - 1} \left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)\left[ \left( {m - k} \right)p_{i}^{m - 1 - k} (1 - p_{i} )^{k} - kp_{i}^{m - k} (1 - p_{i} )^{k - 1} \right]u\left( {w + (m - k)\frac{b}{m} - x_{i} } \right)\\&+ mp_{i}^{m - 1} u\left( {w + b - x_{i} } \right) - m(1 - p_{i} )^{m - 1} u\left( {w - x_{i} } \right) \biggr\} \hfill \\ & - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)p_{i}^{m - k} (1 - p_{i} )^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x_{i} } \right)} = 0. \hfill \end{aligned}$$

Imposing symmetry—$x_{i} = x{\text{ for all }}i$—on the first order condition, the symmetric interior Nash equilibrium satisfies

$$\begin{aligned} G_{m} (x) &\equiv p_{x} \left\{ \begin{aligned}& \sum\limits_{k = 1}^{m - 1} {\left( {\begin{array}{*{20}l} m \\ k \\ \end{array} } \right)\left[ {\left( {m - k} \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} - k\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k - 1} } \right]u\left( {w + (m - k)\frac{b}{m} - x} \right)} \hfill \\& + m\left( \frac{1}{n} \right)^{m - 1} u\left( {w + b - x} \right) - m\left( {1 - \frac{1}{n}} \right)^{m - 1} u\left( {w - x} \right) \hfill \\ \end{aligned} \right\} \hfill \\&\quad - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{*{20}l} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} = 0, \hfill \end{aligned}$$

where $p_{x} (x) \equiv \tfrac{{\partial p_{i} }}{{\partial x_{i} }}(x, \ldots ,x) \ge 0$. Note that

$$\begin{aligned} &\sum\limits_{k = 1}^{m - 1} {\left( {\begin{array}{*{20}l} m \\ k \\ \end{array} } \right)\left[ {\left( {m - k} \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} - k\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k - 1} } \right]u\left( {w + (m - k)\frac{b}{m} - x} \right)} \hfill \\ &\quad = \sum\limits_{k = 1}^{m - 1} {\left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)\left( {m - k} \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - k)\frac{b}{m} - x} \right)} - \\&\quad\quad\sum\limits_{k = 1}^{m - 1} {\left( {\begin{array}{*{20}l} m \\ k \\ \end{array} } \right)} k\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k - 1} u\left( {w + (m - k)\frac{b}{m} - x} \right) \hfill \\ &\quad = \sum\limits_{k = 1}^{m - 1} {m\left( {\begin{array}{*{20}l} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - k)\frac{b}{m} - x} \right)} -\\&\quad\quad \sum\limits_{{k^{\prime} = 0}}^{m - 2} {\left( {\begin{array}{*{20}l} m \\ {k^{\prime} + 1} \\ \end{array} } \right)} (k^{\prime} + 1)\left( \frac{1}{n} \right)^{{m - 1 - k^{\prime}}} \left( {1 - \frac{1}{n}} \right)^{{k^{\prime}}} u\left( {w + (m - 1 - k^{\prime})\frac{b}{m} - x} \right) \hfill \\ &\quad = \sum\limits_{k = 1}^{m - 1} {m\left( {\begin{array}{*{20}l} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - k)\frac{b}{m} - x} \right)} -\\&\quad\quad \sum\limits_{{k^{\prime} = 0}}^{m - 2} {m\left( {\begin{array}{*{20}c} {m - 1} \\ {k^{\prime}} \\ \end{array} } \right)} \left( \frac{1}{n} \right)^{{m - 1 - k^{\prime}}} \left( {1 - \frac{1}{n}} \right)^{{k^{\prime}}} u\left( {w + (m - 1 - k^{\prime})\frac{b}{m} - x} \right) \hfill \\ &\quad = \sum\limits_{k = 1}^{m - 1} {m\left( {\begin{array}{*{20}l} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - k)\frac{b}{m} - x} \right)} -\\&\quad\quad \sum\limits_{k = 0}^{m - 2} {m\left( {\begin{array}{*{20}l} {m - 1} \\ k \\ \end{array} } \right)} \left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - 1 - k)\frac{b}{m} - x} \right) \hfill \end{aligned}$$

Therefore, we have (4′)

$$\begin{aligned} G_{m} (x)& \equiv p_{x} \left\{ {\sum\limits_{k = 0}^{m - 1} {m\left( {\begin{array}{*{20}c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - k)\frac{b}{m} - x} \right)}}\right.\\&\quad\left.{ - \sum\limits_{k = 0}^{m - 1} {m\left( {\begin{array}{*{20}c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - 1 - k)\frac{b}{m} - x} \right)} } \right\} \hfill \\ &\quad - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} = 0. \hfill \end{aligned}$$

(4′)

Appendix 3: Proof of Proposition 3′

The symmetric equilibrium effort is determined by (2) when m = 1 and by (4′) when $m \ge 2$.

• (i) Players are risk neutral: $u^{\prime\prime}( \cdot ) = 0$.

  In this case, $F(x) = p_{x} b - 1$, and

  $$\begin{aligned} G_{m} (x) &= p_{x} \left\{ {\sum\limits_{k = 0}^{m - 1} {m\left( {\begin{array}{*{20}c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} \left( {w + (m - k)\frac{b}{m} - x} \right)}}\right.\\&\quad\left.{ - \sum\limits_{k = 0}^{m - 1} {m\left( {\begin{array}{*{20}c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} \left( {w + (m - 1 - k)\frac{b}{m} - x} \right)} } \right\} \hfill \\&\quad- \sum\limits_{k = 0}^{m} {\left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} } \hfill = p_{x} \left\{ {m\left( {w - x + \frac{b}{m} + \frac{m - 1}{n}\frac{b}{m}} \right)}\right.\\&\quad\left.{- m\left( {w - x + \frac{m - 1}{n}\frac{b}{m}} \right)} \right\} - \left( {\frac{1}{n} + 1 - \frac{1}{n}} \right)^{m} = p_{x} b - 1. \hfill \end{aligned}$$

  Therefore, m has no effect on the equilibrium effort.

• (ii) Players are both risk averse and prudent: $u^{\prime\prime}( \cdot ) < 0$ and $u^{\prime\prime\prime}( \cdot ) > 0$.

  In this case,

  $$\begin{aligned} & G_{m} (x) - F(x) = p_{x} (m + 1)\\& \left\{ \begin{gathered} \left[ {\frac{m}{m + 1}\sum\limits_{k = 0}^{m - 1} {\left( {\begin{array}{*{20}c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - k)\frac{b}{m} - x} \right)} + \frac{1}{m + 1}u(w - x)} \right] \hfill \\ - \left[ {\frac{m}{m + 1}\sum\limits_{k = 0}^{m - 1} {\left( {\begin{array}{*{20}c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - 1 - k)\frac{b}{m} - x} \right) + \frac{1}{m + 1}u(w + b - x)} } \right] \hfill \\ \end{gathered} \right\} \hfill \\&+ \left[ {\tfrac{1}{n}u^{\prime}(w + b - x) + (1 - \tfrac{1}{n})u^{\prime}(w - x)} \right] - \left[ {\sum\limits_{k = 0}^{m} {\left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} } \right]. \hfill \end{aligned}$$

  The two bracketed items in the curly parentheses are the expected utility of two wealth distributions respectively, and the second distribution is a Rothschild-Stiglitz risk increase over the first, both having a mean of $w - x + \frac{b}{m + 1}\left( {1 + \frac{m - 1}{n}} \right)$. So their difference is positive since $u^{\prime\prime}( \cdot ) < 0$. In addition, the last two bracketed items above are the expected utility—with $u^{\prime}( \cdot )$ being the utility function—of two wealth distributions respectively, and the first distribution is a Rothschild-Stiglitz risk increase over the second, both having a mean of $w - x + \frac{b}{n}.$ So their difference is also positive since $u^{\prime\prime\prime}( \cdot ) > 0$. Therefore, $G_{m} (x) - F(x) > 0$ for all x. This, together with Condition 1, suggests that the equilibrium effort level increases from m = 1 to $m \ge 2$. □

Appendix 4: Proof of Proposition 5

• (i) This is true because for risk-neutral players, the symmetric equilibrium effort level is determined by $p_{x} b - 1 = 0$ under both the shared-prize contest and the m-round multiple-competition contest.

• (ii) Suppose $u^{\prime\prime}( \cdot ) < 0$ and $u^{\prime\prime\prime}( \cdot ) > 0$. The symmetric equilibrium effort level is determined by $p_{x} b - 1 = 0$ under the shared-prize contest, and it is determined by (4′) under the m-round multiple-competition contest. To prove that the solution to $p_{x} b - 1 = 0$ is larger than the solution to (4′), it is sufficient to show that at the effort level x where $p_{x} b - 1 = 0$, $G_{m} (x) < 0$. Indeed, substituting $p_{x} = 1/b$ into the expression of $G_{m} (x)$ in (4′), we have

  $$\begin{aligned} G_{m} (x) = &\frac{m}{b}\biggl\{ \sum\limits_{k = 0}^{m - 1} {\left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - k)\frac{b}{m} - x} \right)} \\&- \sum\limits_{k = 0}^{m - 1} {\left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} u\left( {w + (m - 1 - k)\frac{b}{m} - x} \right)} \biggr\}\\& - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} \\ &< \sum\limits_{k = 0}^{m - 1} \left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} \biggl[ \frac{1}{2}u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)\\ &+ \frac{1}{2}u^{\prime}\left( {w + (m - 1 - k)\frac{b}{m} - x} \right) \biggr] - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} \\ & \le \sum\limits_{k = 0}^{m - 1} \left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k} \biggl[ \frac{1}{n}u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right) \\ &+ \left( {1 - \frac{1}{n}} \right)u^{\prime}\left( {w + (m - 1 - k)\frac{b}{m} - x} \right) \biggr] - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} \\ =& \sum\limits_{k = 0}^{m - 1} \left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right) \\ &+ \sum\limits_{k = 0}^{m - 1} {\left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - 1 - k} \left( {1 - \frac{1}{n}} \right)^{k + 1} u^{\prime}\left( {w + (m - 1 - k)\frac{b}{m} - x} \right)} \\ & - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} \\ =& \sum\limits_{k = 0}^{m - 1} \left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right) \\ &+ \sum\limits_{k = 1}^{m} {\left( {\begin{array}{c} {m - 1} \\ {k - 1} \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} \\ & - \sum\limits_{k = 0}^{m} {\left( {\begin{array}{c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)_{{}}^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} \\=& \left( \frac{1}{n} \right)_{{}}^{m} u^{\prime}\left( {w + b - x} \right) + \sum\limits_{k = 1}^{m - 1} {\left( {\begin{array}{c} {m - 1} \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} \\ &\times\sum\limits_{k = 1}^{m - 1} {\left( {\begin{array}{c} {m - 1} \\ {k - 1} \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} + \left( {1 - \frac{1}{n}} \right)^{m} u^{\prime}\left( {w - x} \right) \\ &- \sum\limits_{k = 0}^{m} {\left( {\begin{array}{c} m \\ k \\ \end{array} } \right)\left( \frac{1}{n} \right)^{m - k} \left( {1 - \frac{1}{n}} \right)^{k} u^{\prime}\left( {w + (m - k)\frac{b}{m} - x} \right)} = 0, \end{aligned}$$

  where the first inequality is due to Lemma 2, the second inequality due to $n \ge 2$ and $u^{\prime\prime}( \cdot ) < 0$, and the last equality due to Pascal’s rule, namely $\left( {\begin{array}{*{20}c} {m - 1} \\ k \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} {m - 1} \\ {k - 1} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} m \\ k \\ \end{array} } \right)$.

  □