Accuracy in contests: players’ perspective

Abstract

We propose a political theory for the slow adoption of technology in sports and other contests. We investigate players’ preferences for new technology that improves contest accuracy. Modeling accuracy as the elasticity of “production” in a standard Tullock contest, we show that players may dislike accuracy if heterogeneity among them is: (1) sufficiently low; (2) moderate when the initial accuracy is low; or (3) high when the initial accuracy is high. We apply our results to the recent adoption of goal-line technology by major European soccer leagues.

Appendix

Proof of Proposition 3

The derivative of $\theta ^{*}$ with respect to $r$ is given by

$$\begin{aligned} \frac{\partial \theta ^{*}}{\partial r}=-\frac{\partial g(.)/\partial r}{\partial g(.)/\partial \theta ^{*}}, \end{aligned}$$

(7)

where $g(.)$ is defined as in (6). Routine algebra yields

$$\begin{aligned} \frac{\partial \theta ^{*}}{\partial r}=\frac{(-\theta ^{*}\ln \theta ^{*})\left( 2(m-1)(\theta ^{*})^{r}-mc\theta ^{*}+n\right) }{r(2(m-1)(\theta ^{*})^{r}-mc\theta ^{*}+n)-mc\theta ^{*}-(n-1)c(\theta ^{*})^{1-r}}. \end{aligned}$$

(8)

Note that (6) implies

$$\begin{aligned} (\theta ^{*})^{r}=\frac{mc\theta ^{*}-n+\sqrt{(mc\theta ^{*}-n)^{2}+4(m-1)(n-1)c\theta ^{*}}}{2(m-1)}. \end{aligned}$$

(9)

Inserting (9) in the numerator and observing that (6) also implies $mc\theta ^{*}+(n-1)c(\theta ^{*})^{1-r}=(m-1)(\theta ^{*})^{r}+n$, (8) can be rewritten as

$$\begin{aligned} \dfrac{\partial \theta ^{*}}{\partial r}=\dfrac{(-\theta ^{*}\ln \theta ^{*})\sqrt{(mc\theta ^{*}-n)^{2}+4(m-1)(n-1)c\theta ^{*}} }{r(2(m-1)(\theta ^{*})^{r}-mc\theta ^{*}+n)-(m-1)(\theta ^{*})^{r}-n}. \end{aligned}$$

Clearly, the numerator has a positive sign because $\theta ^{*}<1$ by Proposition 1. In contrast, the denominator has a negative sign because $0<r<1$ implies

$$\begin{aligned}&r(2(m-1)(\theta ^{*})^{r}-mc\theta ^{*}+n)-(m-1)(\theta ^{*})^{r}-n\\&\quad <(m-1)(\theta ^{*})^{r}-mc\theta ^{*}, \end{aligned}$$

or, employing (9),

$$\begin{aligned} <\dfrac{-(mc\theta ^{*}+n)+\sqrt{(mc\theta ^{*}-n)^{2}+4(m-1)(n-1)c\theta ^{*}}}{2}<0. \end{aligned}$$

Therefore, $\frac{\partial \theta ^{*}}{\partial r}<0$ by (7), proving part (a). To prove part (b), note that the derivative of $\pi _{U}^{*}$ with respect to $r$ is given by

$$\begin{aligned} \frac{d\pi _{U}^{*}}{dr}=\frac{\partial \pi _{U}^{*}}{\partial r}+ \frac{\partial \pi _{U}^{*}}{\partial \theta }\dfrac{\partial \theta ^{*}}{\partial r}, \end{aligned}$$

(10)

where simple calculations yield

$$\begin{aligned} \dfrac{\partial \pi _{U}^{*}}{\partial r}&= -\frac{(m-1)(\theta ^{*})^{2r}+n(\theta ^{*})^{r}}{(m(\theta ^{*})^{r}+n)^{2}}V,\\ \dfrac{\partial \pi _{U}^{*}}{\partial \theta }&= \dfrac{r(\theta ^{*})^{r-1}[(1-r)(m(\theta ^{*})^{r}+n)+2r(\theta ^{*})^{r}]n}{\left( m(\theta ^{*})^{r}+n\right) ^{3}}V. \end{aligned}$$

Because, $0<\theta ^{*}$ if $0<r<1$ by Proposition 1 and $1\le m$ by assumption, $\frac{\partial \pi _{U}^{*}}{\partial r}<0$ and $\frac{\partial \pi _{U}^{*}}{\partial \theta }>0$. Moreover, $\frac{\partial \theta ^{*}}{\partial r}<0$ by part (a). Thus, both the first and second terms in (10) have negative signs, proving part (b). $\square$

Lemma 2

$\lim _{r\rightarrow 0}\theta ^{*}=\frac{1}{c}$ and

$$\begin{aligned} \lim \nolimits _{r\rightarrow 1}\theta ^{*}=\left\{ \begin{array}{lll} \dfrac{n-(n-1)c}{m(c-1)+1} &{} \quad \textit{if} &{} \; n>(n-1)c \\ &{} &{} \\ 0 &{} \quad \textit{if} &{} \; n\le (n-1)c. \end{array} \right. \end{aligned}$$

Proof of Lemma 2

Let $\theta _{0}^{*}\equiv \lim _{r\rightarrow 0}\theta ^{*}$ and $\theta _{1}^{*}\equiv \lim _{r\rightarrow 1}\theta ^{*}$. As $r\rightarrow 0$, (6) reduces to

$$\begin{aligned} (m-1)-mc\theta _{0}^{*}+n-(n-1)c\theta _{0}^{*}=0, \end{aligned}$$

which gives $\theta _{0}^{*}=\frac{1}{c}$, as claimed. Likewise, as $r\rightarrow 1$, (6) simplifies to

$$\begin{aligned} \theta _{1}^{*}\times \left[ \frac{n-(n-1)c}{mc-(m-1)}-\theta _{1}^{*}\right] =0. \end{aligned}$$

Clearly, for $c\ge \frac{n}{n-1}$, we have $\theta _{1}^{*}=0$. Suppose that $c<\frac{n}{n-1}$ and by way of contradiction $\theta _{1}^{*}=0$. Then, as $r\rightarrow 1$, $x_{U}^{*}\rightarrow 0$ and the first-order conditions in (4) reduce to

$$\begin{aligned} \frac{\partial \pi _{F}^{*}}{\partial x_{F}}=\dfrac{1}{nx_{F}^{*}}\left[ 1- \dfrac{1}{n}\right] V-c_{F}=0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial \pi _{U}}{\partial x_{U}}=\dfrac{1}{nx_{F}^{*}}V-c_{U}=0, \end{aligned}$$

which, together, imply $c=\frac{n}{n-1}\nless \frac{n}{n-1}$, which, in turn, implies $\theta ^{*}_{1}\ne 0$. Thus, $\theta ^{*}_{1}=\frac{n-(n-1)c}{mc-(m-1)}$ for $c<\frac{ n}{n-1}$, as desired. $\square$

Proof of Proposition 4

Using Proposition 1(c), the derivative of $\pi _{F}^{*}$ with respect to $r$ can be written as

$$\begin{aligned} \dfrac{d\pi _{F}^{*}}{dr}=\dfrac{dp_{F}^{*}}{dr}(1-r+2rp_{F}^{*})V-p_{F}^{*}(1-p_{F}^{*})V, \end{aligned}$$

(11)

where $p_{F}^{*}=\frac{1}{n+m\times (\theta ^{*})^{r}}$ by Proposition 1(a). Noting that $\frac{dp_{F}^{*}}{dr}=\frac{\partial p_{F}^{*}}{\partial r}+\frac{\partial p_{F}^{*}}{\partial \theta ^{*}}\frac{\partial \theta ^{*}}{\partial r}$ and employing (8), routine algebra yields

$$\begin{aligned} \frac{dp_{F}^{*}}{dr}=\frac{\dfrac{(m(\theta ^{*})^{r}+n-1)}{\left( m(\theta ^{*})^{r}+n\right) ^{2}}mc(\theta ^{*})^{1+r}(\ln \theta ^{*})}{2r(m-1)(\theta ^{*})^{2r}-(1+r)mc(\theta ^{*})^{1+r}+nr(\theta ^{*})^{r}-(n-1)c\theta ^{*}}. \end{aligned}$$

(12)

Observe that $\lim _{r\rightarrow 0}p_{F}^{*}=\frac{1}{m+n}$ and $\lim _{r\rightarrow 0}\frac{dp_{F}^{*}}{dr}=\frac{m\ln c}{(m+n)^{2}}$ because $\lim _{r\rightarrow 0}\theta ^{*}=\frac{1}{c}$ by Lemma 2. Because, by (11), we have

$$\begin{aligned} \lim \nolimits _{r\rightarrow 0}\dfrac{d\pi _{F}^{*}}{dr} =\lim \nolimits _{r\rightarrow 0}\left[ \dfrac{dp_{F}^{*}}{dr} V\right] -\lim \nolimits _{r\rightarrow 0}[p_{F}^{*}(1-p_{F}^{*})V], \end{aligned}$$

it follows

$$\begin{aligned} \lim \nolimits _{r\rightarrow 0}\frac{d\pi _{F}^{*}}{dr}=\frac{m\ln c-(m+n-1)}{\left( m+n\right) ^{2}}V. \end{aligned}$$

This proves part (a) because it readily implies that $\lim \nolimits _{r \rightarrow 0}\frac{d\pi _{F}^{*}}{dr}<0$ if $c<e^{1+(n-1)/m}$ and $\lim \nolimits _{r\rightarrow 0}\frac{d\pi _{F}^{*}}{dr}>0$ if $c>e^{1+(n-1)/m}$. The same steps in part (a) along with Lemma 2 yield

$$\begin{aligned} \lim \nolimits _{r\rightarrow 1}\dfrac{d\pi _{F}^{*}}{dr}=-\dfrac{2mc\ln \left( \dfrac{n-(n-1)c}{mc-(m-1)}\right) +mc+n}{\dfrac{(cm+n)^{3}}{\left( mc-m+1\right) \left( m+n-1\right) }}\text { if }c<\frac{n}{n-1}, \end{aligned}$$

(13)

and

$$\begin{aligned} \lim \nolimits _{r\rightarrow 1}\frac{d\pi _{F}^{*}}{dr}=-\frac{n-1}{n^{2}} V\text { if }c>\frac{n}{n-1}. \end{aligned}$$

Obviously, $\lim \nolimits _{r\rightarrow 1}\frac{d\pi _{F}^{*}}{dr}<0$ if$\ c>\frac{n}{n-1}$. To examine the sign of $\frac{d\pi _{F}^{*}}{dr}$ for $c<\frac{n}{n-1}$, let $f(c)\equiv 2mc\ln (\frac{n-(n-1)c}{mc-(m-1)} )+mc+n$. Observe that $f(1)=m+n>0$ and $\lim _{c\rightarrow \frac{n}{n-1} }f(c)=-\infty <0$, which, together with the continuity, implies that there exists $\widetilde{c}(n,m)\in (1,\frac{n}{n-1})$ such that $f(\widetilde{c} (n,m))=0$. To show uniqueness, it is enough to show that $f(c)$ is monotonic over $(1,\frac{n}{n-1})$. To this end, calculate the second derivative to obtain

$$\begin{aligned} \dfrac{\partial ^{2}f(c)}{\partial c^{2}}=\dfrac{2m\left( m+n-1\right) }{ \dfrac{\left( n-(n-1)c\right) ^{2}\left( mc-(m-1)\right) ^{2}}{ 1+m(n-1)+n(m-1)}}\left( \frac{2n\left( m-1\right) }{1+m(n-1)+n(m-1)}-c\right) . \end{aligned}$$

Observe that $\tfrac{\partial ^{2}f(c)}{\partial c^{2}}>0$ for $\frac{ 2n\left( m -1\right) }{1 + m(n - 1) + n(m - 1)}>c$ and $\tfrac{\partial ^{2}f(c)}{ \partial c^{2}}<0$ for $\frac{2n\left( m-1\right) }{1+m(n-1)+n(m-1)}<c$. Also, observe that

$$\begin{aligned} \left\{ \begin{array}{lll} 1<\frac{2n\left( m-1\right) }{1+m(n-1)+n(m-1)}<\frac{n}{n-1} &{}\quad \textit{if} &{} \quad m-1>n \\ &{} &{} \\ \frac{2n\left( m-1\right) }{1+m(n-1)+n(m-1)}<1 &{} \quad \textit{if} &{} \quad m-1<n. \end{array} \right. \end{aligned}$$

Therefore, if $m-1<n$, $\tfrac{\partial f(c)}{\partial c}$ is monotonically decreasing over $(1,\frac{n}{n-1})$, it thus takes its supremum at $c=1$. Because

$$\begin{aligned} \frac{\partial f(c)}{\partial c}|_{c=1}={-}\frac{2m\left( n{-}(m{-}1)\right) (m+n{-}{1})\left( m+n{-}2mn{-}1\right) ^{2}}{\left( n^{2}+mn{-}m+1\right) ^{2}}<0\text { for }m{-}1<n, \end{aligned}$$

it implies that, if $m-1<n$, $\frac{\partial f(c)}{\partial c}<0$ for all $c\in$ $(1,\frac{n}{n-1})$. If $m-1>n$, however, $\frac{\partial f(c)}{ \partial c}$ takes its maximum at $c=\frac{2n\left( m-1\right) }{ 1+m(n-1)+n(m-1)}$ over $(1,\frac{n}{n-1})$. Because

$$\begin{aligned} \frac{\partial f(c)}{\partial c}|_{c=\frac{2n\left( m-1\right) }{ 1+m(n-1)+n(m-1)}}=-\frac{m\left( 5\left( n-1\right) \left( m-1\right) +2mn\right) }{m+n-1}+2m\ln \frac{n}{m-1}<0 \end{aligned}$$

for $m-1>n,$ it implies that, if $m-1>n$, $\frac{\partial f(c)}{\partial c}<0$ for all $c\in (1,\frac{n}{n-1})$. As a result, $f(c)$ is monotonically decreasing over $(1, \frac{n}{n-1})$, which shows that $\widetilde{c}(n,m)$ is unique. Moreover, as $f(c)$ is decreasing, referring to (13), this shows

$$\begin{aligned} 0>\frac{d\pi _{F}^{*}}{dr}\text { if }c\in (1,\widetilde{c}(n,m)) \text { and }0<\frac{d\pi _{F}^{*}}{dr}\text { if }c\in \left( \widetilde{c }(n,m),\frac{n}{n-1}\right) , \end{aligned}$$

which completes the proof. $\square$

Proof of Proposition 5

By (11),

$$\begin{aligned} \frac{d\pi _{F}^{*}}{dr}=\frac{dp_{F}^{*}}{dr}(1-r+2rp_{F}^{*})V-p_{F}^{*}(1-p_{F}^{*})V. \end{aligned}$$

And by definition of total payoff, $\frac{d(\alpha N\times \pi _{F}^{*})}{dr}=\alpha N\frac{d\pi _{F}^{*}}{dr}$. As $N\rightarrow \infty$, we therefore have

$$\begin{aligned}&\lim _{N\rightarrow \infty }\frac{d\left( \alpha N\times \pi _{F}^{*}\right) }{dr}\nonumber \\&\quad =\lim _{N\rightarrow \infty }\left( \alpha N\times \frac{dp_{F}^{*}}{dr} (1-r+2rp_{F}^{*})V\right) -\lim _{N\rightarrow \infty }(\alpha N\times p_{F}^{*}(1-p_{F}^{*})V). \end{aligned}$$

As $p_{F}^{*}=\frac{1}{N(\alpha +(\theta ^{*})^{r})}$ from Proposition 1(a), the second limit is

$$\begin{aligned}&\lim _{N\rightarrow \infty }\left( \alpha N\times p_{F}^{*}(1-p_{F}^{*})V\right) \\&\quad =\lim _{N\rightarrow \infty }\left( \alpha N\times \frac{1}{N\left( \alpha + (\theta ^{*})^{r}\right) }\left( 1-\frac{1}{N\left( \alpha +(\theta ^{*})^{r}\right) }\right) V\right) =\frac{\alpha }{\alpha +c\theta ^{*}}V. \end{aligned}$$

By (12), we have

$$\begin{aligned} \frac{dp_{F}^{*}}{dr}=\dfrac{\dfrac{(N(\theta ^{*})^{r}+\alpha N-1)}{ \left( N(\theta ^{*})^{r}+\alpha N\right) ^{2}}Nc(\theta ^{*})^{1+r}(\ln \theta ^{*})}{2r(N-1)(\theta ^{*})^{2r}-(1+r)Nc(\theta ^{*})^{1+r}+\alpha Nr(\theta ^{*})^{r}-(\alpha N-1)c\theta ^{*}}. \end{aligned}$$

From here, it follows

$$\begin{aligned} \lim _{N\rightarrow \infty }\left( \alpha N\times \frac{dp_{F}^{*}}{dr} (1-r+2rp_{F}^{*})V\right) =-\lim _{N\rightarrow \infty }\frac{\alpha c\theta ^{*}\ln \theta ^{*}}{(\alpha +c\theta ^{*})^{2}}V. \end{aligned}$$

Thus,

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{d(\alpha N\times \pi _{F}^{*})}{dr} =-\lim _{N\rightarrow \infty }\alpha \dfrac{\alpha +c\theta ^{*}(1+\ln \theta ^{*})}{\left( \alpha +c\theta ^{*}\right) ^{2}}V. \end{aligned}$$

(14)

Next, for $n=\alpha N$ and $m=N$, (6) reduces to

$$\begin{aligned} (\theta ^{*})^{2r}-\frac{N}{N-1}c(\theta ^{*})^{r+1}+\frac{\alpha N}{ N-1}(\theta ^{*})^{r}-\frac{\alpha N-1}{N-1}c\theta ^{*}=0. \end{aligned}$$

From here, as $\theta ^{*}>0$ for $r\in (0,1)$, it readily follows that $\lim _{N\rightarrow \infty }\theta ^{*}=c^{-1/(1-r)}$. Using this, (14) becomes

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{d(\alpha N\times \pi _{F}^{*})}{dr} =-\alpha \dfrac{\alpha +c^{-\frac{r}{1-r}}\left( 1-\frac{\ln c}{1-r}\right) }{\left( \alpha +c^{-\frac{r}{1-r}}\right) ^{2}}V. \end{aligned}$$

Using Proposition 1(c) and replicating the same lines of algebra, we also find

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{d\left( N\times \pi _{U}^{*}\right) }{dr}=-\dfrac{ \alpha c^{-\frac{r}{1-r}}\left( c^{-\frac{r}{1-r}}+\alpha \left( 1+\frac{\ln c}{1-r}\right) \right) }{ \left( \alpha +c^{-\frac{r}{1-r}}\right) ^{2}}V. \end{aligned}$$

It is easy to see that $\lim _{N\rightarrow \infty }\frac{d(N\times \pi _{U}^{*})}{dr}<0$. To examine the sign of $\lim _{N\rightarrow \infty } \frac{d(\alpha N\times \pi _{F}^{*})}{dr}$, let $f(c)\equiv \alpha +c^{- \frac{r}{1-r}}(1-\frac{\ln c}{1-r})$. Obviously, $\lim _{N\rightarrow \infty } \frac{d(\alpha N\times \pi _{F}^{*})}{dr}$ and $f(c)$ have the opposite signs. Now, observe that $f(c)$ takes its unique minimum at $c=e^{(1-r^{2})/r}$ because

$$\begin{aligned} \frac{\partial f(c)}{\partial c}=\frac{c^{\frac{1}{r-1}}}{\left( r-1\right) ^{2}}\left( r^{2}+\left( \ln c\right) r-1\right) \left\{ \begin{array}{lll} >0 &{} \quad \textit{if} &{} \, c>e^{\frac{1-r^{2}}{r}} \\ <0 &{} \quad \textit{if} &{} \, c<e^{\frac{1-r^{2}}{r}}, \end{array} \right. \end{aligned}$$

where $f(e^{\frac{1-r^{2}}{r}})=\alpha -\frac{1}{re^{1+r}}<0$ if $\alpha < \frac{1}{re^{1+r}}$ and $f(e^{\frac{1-r^{2}}{r}})>0$ if $\alpha >\frac{1}{ re^{1+r}}$. Because $f(1)=\alpha +1>0$, $\lim _{c\rightarrow \infty }f(c)=\alpha >0$, and $\lim _{N\rightarrow \infty }\frac{d(\alpha N\times \pi _{F}^{*})}{dr}$ and $f(c)$ have the opposite signs, it follows

$$\begin{aligned} \tfrac{d(\alpha N\pi _{F}^{*})}{dr}\left\{ \begin{array}{ll} \ge 0 &{} \quad \hbox {if } \alpha \le \frac{1}{re^{1+r}} \hbox { and } \underline{c}(\alpha ,r)<c<\overline{c}(\alpha ,r)\\ <0 &{} \quad \hbox {otherwise}, \end{array} \right. \end{aligned}$$

where $\underline{c}(\alpha ,r)$ and $\overline{c}(\alpha ,r)$ solve $f(c)=0$, which completes the proof. $\square$

Proof of Proposition 6

As there are $n$ favorites and $m$ underdogs, total effort ($TE$) is given by $TE=nx_{F}^{*}+mx_{U}^{*}$. Employing $\theta ^{*}=\frac{ x_{U}^{*}}{x_{F}^{*}}$ and Proposition 1(a) and 1(b), it can be written explicitly as

$$\begin{aligned} TE=\frac{V}{c_{U}}\frac{r(\theta ^{*})^{r-1}}{\left( m(\theta ^{*})^{r}+n\right) ^{2}}\left( m(\theta ^{*})+n\right) ((m-1)(\theta ^{*})^{r}+n), \end{aligned}$$

or, more conveniently, as

$$\begin{aligned} TE=\frac{V}{c_{F}}\underset{=h(r,\theta ^{*})}{\underbrace{\frac{r\left( m\theta ^{*}+n\right) (m(\theta ^{*})^{r}+n-1)}{\left( m(\theta ^{*})^{r}+n\right) ^{2}}},} \end{aligned}$$

where we have employed that $\theta ^{*}=\frac{\left( (m-1)(\theta ^{*})^{r}+n\right) (\theta ^{*})^{r}}{(m(\theta ^{*})^{r}+n-1)c}$ by (6) and $c=\frac{c_{U}}{c_{F}}$ by definition.

As the derivative of $TE$ with respect to $r$ is given by

$$\begin{aligned} \frac{d(TE)}{dr}=\frac{V}{c_{F}}\left( \frac{\partial h(r,\theta ^{*})}{ \partial r}+\frac{\partial h(r,\theta ^{*})}{\partial \theta }\frac{ \partial \theta ^{*}}{\partial r}\right) , \end{aligned}$$

routine algebra, along with Lemma 2, yields

$$\begin{aligned} \lim \nolimits _{r\rightarrow 0}\frac{d(TE)}{dr}&= \frac{V}{c_{F}}\frac{ (m+cn)\left( m+n-1\right) \left( m+n\right) }{\left( m+n\right) ^{3}c},\end{aligned}$$

(15)

$$\begin{aligned} \lim \nolimits _{r\rightarrow 1}\dfrac{d(TE)}{dr}&= \left\{ \begin{array}{ll} \dfrac{V}{c_{F}}\dfrac{cm+n+mn(c-1)\ln \left( \frac{c+n-cn}{cm-m+1}\right) }{\frac{ \left( cm+n\right) ^{2}}{(m+n-1)}} &{} \;\textit{if} \; c<\frac{n}{n-1}\ \\ \dfrac{V}{c_{F}}\dfrac{n-1}{n} &{} \; \textit{if} \; c\ge \frac{n}{n-1}. \end{array} \right. \end{aligned}$$

(16)

From (15), it is straightforward to see that $\lim _{r\rightarrow 0}\frac{d(TE)}{dr}>0$, proving part (a). Likewise, if $c\ge \frac{n}{n-1}$, it is easy to see from (16) that $\lim _{r\rightarrow 1}\frac{d(TE)}{dr}>0$. Suppose now $c<\frac{n}{n-1}$ and let $f(c)\equiv mc+n+mn\left( c-1\right) \ln (\frac{n-(n-1)c}{mc-(m-1)})$ . As all other terms are always positive, the sign of $\lim \limits _{r \rightarrow 1}\frac{d(TE)}{dr}$ is the same as that of $f(c)$. Note that $f(1)=m$ and $\lim \limits _{c\rightarrow \frac{n}{n-1}}f(c)=-\infty$, which together with the continuity assures that there exists $\widehat{c}(n,m)\in (1,\frac{n}{n-1})$ such that $f(\widehat{c}(n,m))=0$. To show uniqueness, note that the first and second derivatives of $f(c)$ are given as

$$\begin{aligned} \frac{\partial f(c)}{\partial c}&= {-}\dfrac{m\left( m\left( n{-}1\right) c^{2}{-}\left( m{-}n{-}1\right) \left( n{-}1\right) c{-}n^{2}\right) }{\left( n{-}(n{-}1)c\right) \left( mc{-}(m{-}1)\right) }+mn\ln \left( \dfrac{n{-}(n{-}1)c}{mc{-}(m{-}1)}\right) ,\\ \frac{\partial ^{2}f(c)}{\partial c^{2}}&= -\dfrac{mn\left( m+n-1\right) \left( \left( m-n+1\right) c+\left( n-m+1\right) \right) }{\left( n-(n-1)c\right) ^{2}\left( cm-(m-1)\right) ^{2}}<0. \end{aligned}$$

Obviously, $f(c)$ is concave as $\frac{\partial ^{2}f(c)}{\partial c^{2}}<0$, which together with

$$\begin{aligned} \frac{\partial f(c)}{\partial c}|_{c=1}=m, f(c=1)=m\text { and } \frac{\partial f(c)}{\partial c}|_{c=\frac{n}{n-1}}=-\infty , \lim _{c\rightarrow \frac{n}{n-1}}f(c)=-\infty , \end{aligned}$$

assures that there exists a unique $\widehat{c}(n,m)\in (1,\frac{n}{n-1})$ such that $f(\widehat{c}(n,m))=0$. Moreover, because $f(1)=m>0$ and the sign of $\frac{d(TE)}{dr}$ is the same as that of $f(c)$, we have

$$\begin{aligned} \frac{d(TE)}{dr}\left\{ \begin{array}{lll} >0 \quad \textit{if} \quad c\in (1,\widehat{c}(n,m)) \\ <0 \quad \textit{if} \quad c\in (\widehat{c}(n,m),\frac{n}{n-1}), \end{array} \right. \end{aligned}$$

which completes the proof. $\square$

Proof of Proposition 8

Let $\pi _{ij}^{*}$ denote the payoff of type $i$ in a pairwise contest when his opponent is of type $j$. Also, let $E[\pi _{i}^{*}]$ denote the expected payoff of type $i$. Because the opponent is chosen randomly, the underdog’s expected payoff is given by

$$\begin{aligned} E[\pi _{U}^{*}]=\frac{m-1}{m+n-1}\pi _{UU}^{*}+\frac{n}{m+n-1}\pi _{UF}^{*}. \end{aligned}$$

Differentiating it with respect to $r$ gives

$$\begin{aligned} \frac{dE[\pi _{U}^{*}]}{dr}=\frac{m-1}{m+n-1}\frac{d\pi _{UU}^{*}}{dr }+\frac{n}{m+n-1}\frac{d\pi _{UF}^{*}}{dr}. \end{aligned}$$

Note that $\frac{d\pi _{UF}^{*}}{dr}<0$ by Proposition 2(b) and employing Proposition 1(a) and (c), it immediately follows that $\frac{d\pi _{UU}^{*}}{dr}=-\frac{V}{4}<0$. Because $\frac{dE[\pi _{U}^{*}]}{dr}$ is a convex combination of $\frac{d\pi _{UU}^{*}}{dr}<0$ and $\frac{d\pi _{UF}^{*}}{dr}<0$, it must be that $\frac{dE[\pi _{U}^{*}]}{dr}<0$, which proves part (a). To prove part (b), note that the favorite’s expected payoff is given by

$$\begin{aligned} E[\pi _{F}^{*}]=\frac{n-1}{m+n-1}\pi _{FF}^{*}+\frac{m}{m+n-1}\pi _{FU}^{*}. \end{aligned}$$

Differentiating it with respect to $r$ yields

$$\begin{aligned} \frac{dE[\pi _{F}^{*}]}{dr}=\frac{n-1}{m+n-1}\frac{d\pi _{FF}^{*}}{dr }+\frac{m}{m+n-1}\frac{d\pi _{FU}^{*}}{dr}, \end{aligned}$$

where $\pi _{FF}^{*}=\dfrac{2-r}{4}V$ and $\pi _{FU}^{*}=\dfrac{ c^{r}(c^{r}-r+1)}{\left( c^{r}+1\right) ^{2}}V$ are obtained by using Proposition 1(a) and (c). Routine algebra yields

$$\begin{aligned} \frac{dE[\pi _{F}^{*}]}{dr}=\frac{n{-}1}{m+n{-}1}\left( -\frac{V}{4}\right) +\frac{m}{ m+n{-}1}\frac{c^{r}((1+c^{r}{-}r+rc^{r})\ln c-(1+c^{r}))}{\left( 1+c^{r}\right) ^{3}}V. \end{aligned}$$

Taking now its limit as $r\rightarrow 0$ gives

$$\begin{aligned} \lim _{r\rightarrow 0}\frac{dE[\pi _{F}^{*}]}{\partial r}=\frac{m\ln c-(m+n-1)}{4(m+n-1)}V. \end{aligned}$$

Clearly, $\frac{dE[\pi _{F}^{*}]}{\partial r}<0$ if $c<e^{1+(n-1)/m}$ and $\frac{dE[\pi _{F}^{*}]}{\partial r}>0$ if $c>e^{1+(n-1)/m}$. Finally, taking the limit as $r\rightarrow 1$ yields

$$\begin{aligned} \lim _{r\rightarrow 1}\frac{dE[\pi _{F}^{*}]}{\partial r}=m(c+1)^{3}\left( \dfrac{4c(2c\ln c-c-1)}{(c+1)^{3}}-\dfrac{n-1}{m}\right) V. \end{aligned}$$

Similarly, $\lim \limits _{r\rightarrow 1}\frac{\partial (E[\pi _{F}^{*}]) }{\partial r}>0$ if $\frac{4c(2c\ln c-c-1)}{(c+1)^{3}}>\frac{n-1}{m}$ and $\lim \limits _{r\rightarrow 1}\frac{\partial (E[\pi _{F}^{*}])}{\partial r} <0$ if $\frac{4c(2c\ln c-c-1)}{(c+1)^{3}}<\frac{n-1}{m}$. Because $\frac{ 4c(2c\ln c-c-1)}{(c+1)^{3}}$ has only one critical point over $(1,\infty )$ that turns out to be its maximizer, $\frac{4c(2c\ln c-c-1)}{(c+1)^{3}}>\frac{ n-1}{m}$ if and only if $s>\dfrac{n-1}{m}$ and $\underline{c}(m,n)<c< \overline{c}(m,n)$, where $s\approx 1.06$ uniquely solves $\max \limits _{c\in (1,\infty )}\{\frac{4c(2c\ln c-c-1)}{(c+1)^{3}}\}$ and $\underline{c}(m,n)< \overline{c}(m,n)$ solve $\frac{4c(2c\ln c-c-1)}{(c+1)^{3}}-\frac{n-1}{m}=0$ . This completes the proof because these findings clearly imply that $\lim \limits _{r\rightarrow 1}\frac{\partial (E[\pi _{F}^{*}])}{\partial r} >0$ if $\frac{n-1}{m}<s$ and $\underline{c}(m,n)$ $<c<\overline{c}(m,n)$ and $\lim \limits _{r\rightarrow 1}\frac{\partial (E[\pi _{F}^{*}])}{\partial r} <0$ if otherwise. $\square$