Optimal seedings in interdependent contests

Abstract

We study a model of two interdependent contests and heterogeneous players with commonly known types. The winners of both contests have winning values that depend on the types (abilities) of both winners. Therefore, endogenous win probabilities in each match depend on the outcomes of the other contests through the identity of the winner. The designer seeds players according to their types in order to maximize (minimize) the total effort. For such interdependent contests, each of which includes two heterogeneous players, we consider two different types of a winning value function and demonstrate that for multiplicative value functions it is optimal to place the two highest type players in different contests. On the other hand, for additive value functions it is optimal to place the two highest type players in the same contest since otherwise they practically do not affect each other.

Appendix

1.1 Proof of Proposition 1

By (5), the second-order conditions (SOC) of the maximization problems (1)–(4) are

$$\begin{aligned} \textrm{soc}_{m_{h}}= & {} \frac{-2x_{l}}{\left( x_{h}+x_{l}\right) ^{3}}\left( f(m_{h},w_{h})\frac{y_{h}}{y_{h}+y_{l}}+f(m_{h},w_{l})\frac{y_{l}}{ y_{h}+y_{l}}\right) \\ \textrm{soc}_{m_{l}}= & {} \frac{-2x_{h}}{\left( x_{h}+x_{l}\right) ^{3}}\left( f(m_{l},w_{h})\frac{y_{h}}{y_{h}+y_{l}}+f(m_{l},w_{l})\frac{y_{l}}{ y_{h}+y_{l}}\right) \\ \textrm{soc}_{w_{h}}= & {} \frac{-2y_{l}}{\left( y_{h}+y_{l}\right) ^{3}}\left( g(m_{h},w_{h})\frac{x_{h}}{x_{h}+x_{l}}+g(m_{l},w_{h})\frac{x_{l}}{ x_{h}+x_{l}}\right) \\ \textrm{soc}_{w_{l}}= & {} \frac{-2y_{h}}{\left( y_{h}+y_{l}\right) ^{3}}\left( g(m_{h},w_{l})\frac{x_{h}}{x_{h}+x_{l}}+g(m_{l},w_{l})\frac{x_{l}}{ x_{h}+x_{l}}\right) . \end{aligned}$$

We can see that the SOC are negative as we need$\square$.

1.2 Proof of Proposition 2

The F.O.C. given by (5) can be rewritten as follows:

$$\begin{aligned} \frac{x_{l}m_{h}^{\alpha }}{\left( x_{h}+x_{l}\right) ^{2}}\left( w_{h}^{\beta }\frac{y_{h}}{y_{h}+y_{l}}+w_{l}^{\beta }\frac{y_{l}}{ y_{h}+y_{l}}\right)= & {} 1 \nonumber \\ \frac{x_{h}m_{l}^{\alpha }}{\left( x_{h}+x_{l}\right) ^{2}}\left( w_{h}^{\beta }\frac{y_{h}}{y_{h}+y_{l}}+w_{l}^{\beta }\frac{y_{l}}{ y_{h}+y_{l}}\right)= & {} 1 \nonumber \\ \frac{y_{l}w_{h}^{\beta }}{\left( y_{h}+y_{l}\right) ^{2}}\left( m_{h}^{\alpha }\frac{x_{h}}{x_{h}+x_{l}}+m_{l}^{\alpha }\frac{x_{l}}{ x_{h}+x_{l}}\right)= & {} 1 \nonumber \\ \frac{y_{h}w_{l}^{\beta }}{\left( y_{h}+y_{l}\right) ^{2}}\left( m_{h}^{\alpha }\frac{x_{h}}{x_{h}+x_{l}}+m_{l}^{\alpha }\frac{x_{l}}{ x_{h}+x_{l}}\right)= & {} 1. \end{aligned}$$

(21)

By dividing the first two F.O.C we obtain (7)

$$\begin{aligned} \frac{x_{h}}{x_{l}}=\frac{m_{h}^{\alpha }}{m_{l}^{\alpha }}. \end{aligned}$$

And, by dividing the last two F.O.C. we obtain (8)

$$\begin{aligned} \frac{y_{h}}{y_{l}}=\frac{w_{h}^{\beta }}{w_{l}^{\beta }}. \end{aligned}$$

We can now calculate the winning probability of each player using (8):

$$\begin{aligned} \frac{y_{h}}{y_{h}+y_{l}}=\frac{\frac{y_{h}}{y_{l}}}{\frac{y_{h}}{y_{l}}+1}= \frac{\frac{w_{h}^{\beta }}{w_{l}^{\beta }}}{\frac{w_{h}^{\beta }}{ w_{l}^{\beta }}+1}=\frac{w_{h}^{\beta }}{w_{h}^{\beta }+w_{l}^{\beta }}, \end{aligned}$$

(22)

and

$$\begin{aligned} \frac{y_{l}}{y_{h}+y_{l}}=\frac{1}{\frac{y_{h}}{y_{l}}+1}=\frac{1}{\frac{ w_{h}^{\beta }}{w_{l}^{\beta }}+1}=\frac{w_{l}^{\beta }}{w_{h}^{\beta }+w_{l}^{\beta }}. \end{aligned}$$

(23)

Similarly, by (7), we have

$$\begin{aligned} \frac{x_{h}}{x_{h}+x_{l}}=\frac{\frac{x_{h}}{x_{l}}}{\frac{x_{h}}{x_{l}}+1}= \frac{\frac{m_{h}^{\alpha }}{m_{l}^{\alpha }}}{\frac{m_{h}^{\alpha }}{ m_{l}^{\alpha }}+1}=\frac{m_{h}^{\alpha }}{m_{h}^{\alpha }+m_{l}^{\alpha }}, \end{aligned}$$

(24)

and

$$\begin{aligned} \frac{x_{l}}{x_{h}+x_{l}}=\frac{1}{\frac{x_{h}}{x_{l}}+1}=\frac{1}{\frac{ m_{h}^{\alpha }}{m_{l}^{\alpha }}+1}=\frac{m_{l}^{\alpha }}{m_{h}^{\alpha }+m_{l}^{\alpha }}. \end{aligned}$$

(25)

Now, inserting (22) and (23) into (21) yields

$$\begin{aligned} \frac{x_{l}m_{h}^{\alpha }}{\left( x_{h}+x_{l}\right) ^{2}}\left( w_{h}^{\beta }\frac{w_{h}^{\beta }}{w_{h}^{\beta }+w_{l}^{\beta }} +w_{l}^{\beta }\frac{w_{l}^{\beta }}{w_{h}^{\beta }+w_{l}^{\beta }}\right) =1, \end{aligned}$$

or, alternatively,

$$\begin{aligned}{} & {} \frac{x_{l}m_{h}^{\alpha }}{\left( x_{h}+x_{l}\right) ^{2}}\left( \frac{ w_{h}^{2\beta }+w_{l}^{2\beta }}{w_{h}^{\beta }+w_{l}^{\beta }}\right) \\{} & {} \quad =\frac{m_{l}^{\alpha }}{m_{h}^{\alpha }+m_{l}^{\alpha }}\frac{ m_{h}^{\alpha }}{x_{h}+x_{l}}\left( \frac{w_{h}^{2\beta }+w_{l}^{2\beta }}{ w_{h}^{\beta }+w_{l}^{\beta }}\right) =1. \end{aligned}$$

Then, by (7), we have

$$\begin{aligned}{} & {} \frac{m_{l}^{\alpha }}{m_{h}^{\alpha }+m_{l}^{\alpha }}\frac{m_{h}^{\alpha }}{x_{l}\frac{m_{h}^{\alpha }}{m_{l}^{\alpha }}+x_{l}}\left( \frac{ w_{h}^{2\beta }+w_{l}^{2\beta }}{w_{h}^{\beta }+w_{l}^{\beta }}\right) \\{} & {} \quad =\frac{m_{l}^{\alpha }}{m_{h}^{\alpha }+m_{l}^{\alpha }}\frac{ m_{h}^{\alpha }}{x_{l}(\frac{m_{h}^{\alpha }}{m_{l}^{\alpha }}+1)}\left( \frac{w_{h}^{2\beta }+w_{l}^{2\beta }}{w_{h}^{\beta }+w_{l}^{\beta }}\right) \\{} & {} \quad =\frac{m_{l}^{\alpha }}{m_{h}^{\alpha }+m_{l}^{\alpha }}\frac{ m_{h}^{\alpha }}{x_{l}(\frac{m_{h}^{\alpha }+m_{l}^{\alpha }}{m_{l}^{\alpha } })}\left( \frac{w_{h}^{2\beta }+w_{l}^{2\beta }}{w_{h}^{\beta }+w_{l}^{\beta }}\right) =1. \end{aligned}$$

Thus, the equilibrium effort of the low-type player from set M is

$$\begin{aligned} x_{l}=\frac{m_{h}^{\alpha }m_{l}^{2\alpha }\left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) }{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) ^{2}\left( w_{h}^{\beta }+w_{l}^{\beta }\right) }, \end{aligned}$$

and similarly, the other players’ equilibrium efforts are

$$\begin{aligned} x_{h}= & {} \frac{m_{h}^{2\alpha }m_{l}^{\alpha }\left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) }{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) ^{2}\left( w_{h}^{\beta }+w_{l}^{\beta }\right) } \\ y_{h}= & {} \frac{w_{h}^{2\beta }w_{l}^{\beta }\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) }{\left( w_{h}^{\beta }+w_{l}^{\beta }\right) ^{2}\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) } \\ y_{l}= & {} \frac{w_{h}^{\beta }w_{l}^{2\beta }\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) }{\left( w_{h}^{\beta }+w_{l}^{\beta }\right) ^{2}\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) }. \end{aligned}$$

$\square$

1.3 Proof of Proposition 3

By (9), the total effort in set M is

$$\begin{aligned} TE_{M}= & {} x_{h}+x_{l}=\frac{m_{h}^{2\alpha }m_{l}^{\alpha }\left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) }{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) ^{2}\left( w_{h}^{\beta }+w_{l}^{\beta }\right) }+ \frac{m_{h}^{\alpha }m_{l}^{2\alpha }\left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) }{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) ^{2}\left( w_{h}^{\beta }+w_{l}^{\beta }\right) } \\= & {} \frac{\left( m_{h}^{2\alpha }m_{l}^{\alpha }+m_{h}^{\alpha }m_{l}^{2\alpha }\right) \left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) }{ \left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) ^{2}\left( w_{h}^{\beta }+w_{l}^{\beta }\right) }=\frac{m_{l}^{\alpha }m_{h}^{\alpha }\left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) }{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) }, \end{aligned}$$

and in set W it is

$$\begin{aligned} TE_{W}= & {} y_{h}+y_{l}=\frac{\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) w_{h}^{2\beta }w_{l}^{\beta }}{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) ^{2}}+\frac{\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) w_{h}^{\beta }w_{l}^{2\beta }}{ \left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) ^{2}} \\= & {} \frac{\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) \left( w_{h}^{2\beta }w_{l}^{\beta }+w_{h}^{\beta }w_{l}^{2\beta }\right) }{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) ^{2}}=\frac{\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) w_{l}^{\beta }w_{h}^{\beta }}{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) }. \end{aligned}$$

Thus, the total effort in both sets is

$$\begin{aligned} TE= & {} TE_{M}+TE_{W}=x_{h}+x_{l}+y_{h}+y_{l} \\= & {} \frac{m_{h}^{\alpha }m_{l}^{\alpha }\left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) }{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) }+\frac{\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) w_{h}^{\beta }w_{l}^{\beta }}{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) } \\= & {} \frac{m_{l}^{\alpha }m_{h}^{\alpha }\left( w_{h}^{2\beta }+w_{l}^{2\beta }\right) +\left( m_{h}^{2\alpha }+m_{l}^{2\alpha }\right) w_{h}^{\beta }w_{l}^{\beta }}{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) } \\= & {} \frac{m_{l}^{\alpha }m_{h}^{\alpha }w_{h}^{2\beta }+m_{l}^{\alpha }m_{h}^{\alpha }w_{l}^{2\beta }+w_{h}^{\beta }w_{l}^{\beta }m_{h}^{2\alpha }+w_{h}^{\beta }w_{l}^{\beta }m_{l}^{2\alpha }}{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) } \\= & {} \frac{(m_{h}^{\alpha }w_{h}^{\beta }+m_{l}^{\alpha }w_{l}^{\beta })(m_{h}^{\alpha }w_{l}^{\beta }+m_{l}^{\alpha }w_{h}^{\beta })}{\left( m_{h}^{\alpha }+m_{l}^{\alpha }\right) \left( w_{h}^{\beta }+w_{l}^{\beta }\right) }. \end{aligned}$$

$\square$

1.4 Proof of Proposition 4

Below, we calculate the terms ${\tilde{w}}$ and ${\tilde{m}}$ that appear in the equilibrium efforts given by (15). By definition,

$$\begin{aligned} {\tilde{m}}= & {} \frac{x_{h}}{x_{h}+x_{l}}m_{h}+\frac{x_{l}}{x_{h}+x_{l}}m_{l} \\ {\tilde{w}}= & {} \frac{y_{h}}{y_{h}+y_{l}}w_{h}+\frac{y_{l}}{y_{h}+y_{l}}w_{l}. \end{aligned}$$

Inserting (13) and (14) yields

$$\begin{aligned} {\tilde{w}}= & {} \frac{y_{h}}{y_{h}+y_{l}}w_{h}+\frac{y_{l}}{y_{h}+y_{l}}w_{l}= \frac{\alpha {\tilde{m}}+\beta w_{h}}{\beta w_{h}+\beta w_{l}+2\alpha {\tilde{m}} }w_{h}+\left( 1-\frac{\alpha {\tilde{m}}+\beta w_{h}}{\beta w_{h}+\beta w_{l}+2\alpha {\tilde{m}}}\right) w_{l} \nonumber \\= & {} \frac{\beta \left( w_{h}^{2}+w_{l}^{2}\right) +\alpha {\tilde{m}}\left( w_{h}+w_{l}\right) }{\beta w_{h}+\beta w_{l}+2\alpha {\tilde{m}}}=\frac{\beta \left( w_{h}^{2}+w_{l}^{2}\right) +\alpha \left( \frac{x_{h}}{x_{h}+x_{l}} m_{h}+\frac{x_{l}}{x_{h}+x_{l}}m_{l}\right) \left( w_{h}+w_{l}\right) }{ \beta w_{h}+\beta w_{l}+2\alpha \left( \frac{x_{h}}{x_{h}+x_{l}}m_{h}+\frac{ x_{l}}{x_{h}+x_{l}}m_{l}\right) } \nonumber \\= & {} \frac{\alpha \left( w_{h}+w_{l}\right) \left( x_{h}m_{h}+x_{l}m_{l}\right) +\beta \left( w_{h}^{2}+w_{l}^{2}\right) \left( x_{h}+x_{l}\right) }{2\alpha \left( x_{h}m_{h}+x_{l}m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \left( x_{h}+x_{l}\right) }. \end{aligned}$$

(26)

By (15), we have

$$\begin{aligned} x_{h}+x_{l}= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) ^{2}\left( \alpha m_{l}+\beta {\tilde{w}}\right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}}+\frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) ^{2} }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}} \nonumber \\= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) ^{2}\left( \alpha m_{l}+\beta {\tilde{w}}\right) +\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) ^{2}}{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}} \nonumber \\= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) }{\alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}}, \end{aligned}$$

(27)

and

$$\begin{aligned} x_{h}m_{h}+x_{l}m_{l}= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) ^{2}\left( \alpha m_{l}+\beta {\tilde{w}}\right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}}m_{h}+\frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) ^{2} }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}}m_{l}\nonumber \\= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) \left( \alpha \left( m_{h}^{2}+m_{l}^{2}\right) +\beta {\tilde{w}}\left( m_{h}+m_{l}\right) \right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}}. \end{aligned}$$

(28)

Inserting (27) and (28) into (26) yields

$$\begin{aligned} {\tilde{w}}= & {} \frac{\alpha \left( w_{h}+w_{l}\right) \left( x_{h}m_{h}+x_{l}m_{l}\right) +\beta \left( w_{h}^{2}+w_{l}^{2}\right) \left( x_{h}+x_{l}\right) }{2\alpha \left( x_{h}m_{h}+x_{l}m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \left( x_{h}+x_{l}\right) } \\= & {} \frac{\alpha \left( w_{h}+w_{l}\right) \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) \left( \alpha \left( m_{h}^{2}+m_{l}^{2}\right) +\beta {\tilde{w}}\left( m_{h}+m_{l}\right) \right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}}+\beta \left( w_{h}^{2}+w_{l}^{2}\right) \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) }{ \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}}}{2\alpha \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}} \right) \left( \alpha \left( m_{h}^{2}+m_{l}^{2}\right) +\beta {\tilde{w}} \left( m_{h}+m_{l}\right) \right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}}+\beta \left( w_{h}+w_{l}\right) \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}} \right) }{\alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}}} \\= & {} \frac{\beta \left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}} \right) \left( w_{h}^{2}+w_{l}^{2}\right) +\alpha \left( \alpha \left( m_{h}^{2}+m_{l}^{2}\right) +\beta {\tilde{w}}\left( m_{h}+m_{l}\right) \right) \left( w_{h}+w_{l}\right) }{\beta \left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) \left( w_{h}+w_{l}\right) +2\alpha \left( \alpha \left( m_{h}^{2}+m_{l}^{2}\right) +\beta {\tilde{w}}\left( m_{h}+m_{l}\right) \right) } \\= & {} \frac{\alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}^{2}+w_{l}^{2}\right) +2\beta ^{2}{\tilde{w}}\left( w_{h}^{2}+w_{l}^{2}\right) +\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) \left( w_{h}+w_{l}\right) +\alpha \beta {\tilde{w}}\left( m_{h}+m_{l}\right) \left( w_{h}+w_{l}\right) }{\alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}+w_{l}\right) +2\beta ^{2}{\tilde{w}}\left( w_{h}+w_{l}\right) +2\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) +2\alpha \beta {\tilde{w}}\left( m_{h}+m_{l}\right) }. \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} \left( \left( \alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}+w_{l}\right) +2\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) \right) +\left( 2\alpha \beta \left( m_{h}+m_{l}\right) +2\beta ^{2}\left( w_{h}+w_{l}\right) \right) {\tilde{w}}\right) {\tilde{w}} \\{} & {} \quad =\left( \alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}^{2}+w_{l}^{2}\right) +\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) \left( w_{h}+w_{l}\right) \right) \\{} & {} \qquad +\,\left( 2\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) +\alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}+w_{l}\right) \right) {\tilde{w}}, \end{aligned}$$

or, alternatively,

$$\begin{aligned}{} & {} \left( \alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}+w_{l}\right) +2\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) -2\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) -\alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}+w_{l}\right) \right) {\tilde{w}} \\{} & {} \qquad +\,\left( 2\alpha \beta \left( m_{h}+m_{l}\right) +2\beta ^{2}\left( w_{h}+w_{l}\right) \right) {\tilde{w}}^{2}-\left( \alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}^{2}+w_{l}^{2}\right) \right. \\{} & {} \qquad \left. +\,\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) \left( w_{h}+w_{l}\right) \right) \\{} & {} \quad =0. \end{aligned}$$

Rearranging the last equation yields the following quardratic equation of the parameter ${\tilde{w}}$,

$$\begin{aligned}{} & {} \left( 2\alpha \beta \left( m_{h}+m_{l}\right) +2\beta ^{2}\left( w_{h}+w_{l}\right) \right) {\tilde{w}}^{2}+\left( 2\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) -2\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) \right) {\tilde{w}} \\{} & {} \qquad -\,\left( \alpha \beta \left( m_{h}+m_{l}\right) \left( w_{h}^{2}+w_{l}^{2}\right) +\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) \left( w_{h}+w_{l}\right) \right) \\{} & {} \quad =0. \end{aligned}$$

The solution of this equation is

$$\begin{aligned} {\tilde{w}}= & {} \frac{\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) -\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) }{2\beta \left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) } \\{} & {} \pm \frac{\sqrt{\left( \left( \alpha m_{h}+\beta w_{h}\right) ^{2}+\left( \alpha m_{l}+\beta w_{l}\right) ^{2}\right) \left( \left( \alpha m_{h}+\beta w_{l}\right) ^{2}+\left( \alpha m_{l}+\beta w_{h}\right) ^{2}\right) }}{ 2\beta \left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) }. \end{aligned}$$

Since ${\tilde{w}}$ is positive, we have only one feasible solution which is

$$\begin{aligned} {\tilde{w}}= & {} \frac{\sqrt{\left( \left( \alpha m_{h}+\beta w_{h}\right) ^{2}+\left( \alpha m_{l}+\beta w_{l}\right) ^{2}\right) \left( \left( \alpha m_{h}+\beta w_{l}\right) ^{2}+\left( \alpha m_{l}+\beta w_{h}\right) ^{2}\right) }}{2\beta \left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) } \\{} & {} +\,\frac{\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) -\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) }{2\beta \left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) }. \end{aligned}$$

Similarly, we obtain that

$$\begin{aligned} {\tilde{m}}= & {} \frac{\sqrt{\left( \left( \alpha m_{h}+\beta w_{h}\right) ^{2}+\left( \alpha m_{l}+\beta w_{l}\right) ^{2}\right) \left( \left( \alpha m_{h}+\beta w_{l}\right) ^{2}+\left( \alpha m_{l}+\beta w_{h}\right) ^{2}\right) }}{2\alpha \left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) } \\{} & {} +\,\frac{\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) -\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) }{2\alpha \left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) }. \end{aligned}$$

$\square$

1.5 Proof of Proposition 5

By (15), we obtain the total effort in set M is

$$\begin{aligned} x_{h}+x_{l}= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) ^{2}\left( \alpha m_{l}+\beta {\tilde{w}}\right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}}+\frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) ^{2} }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}} \\= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) ^{2}\left( \alpha m_{l}+\beta {\tilde{w}}\right) +\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) ^{2}}{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}} \\= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) \left( \left( \alpha m_{h}+\beta {\tilde{w}} \right) +\left( \alpha m_{l}+\beta {\tilde{w}}\right) \right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}} \\= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) \left( \alpha m_{h}+\alpha m_{l}+2\beta \tilde{w }\right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) ^{2}} \\= & {} \frac{\left( \alpha m_{h}+\beta {\tilde{w}}\right) \left( \alpha m_{l}+\beta {\tilde{w}}\right) }{\left( \alpha \left( m_{h}+m_{l}\right) +2\beta {\tilde{w}}\right) }. \end{aligned}$$

Inserting (16) into the above equation gives us

$$\begin{aligned} x_{h}+x_{l}=\frac{\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) +\alpha \beta (m_{h}+m_{l})w_{h}+\alpha \beta (m_{h}+m_{l})w_{l}+2\alpha ^{2}m_{h}m_{l}}{ 2\left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) }. \end{aligned}$$

Similarly, the total effort in set W is

$$\begin{aligned} y_{h}+y_{l}= & {} \frac{\left( \alpha {\tilde{m}}+\beta w_{h}\right) \left( \alpha {\tilde{m}}+\beta w_{l}\right) }{\left( 2\alpha {\tilde{m}}+\beta \left( w_{h}+w_{l}\right) \right) } \\= & {} \frac{\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) +\alpha \beta (w_{h}+w_{l})m_{h}+\alpha \beta (w_{h}+w_{l})m_{l}+2\beta ^{2}w_{h}w_{l}}{ 2\left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) } \end{aligned}$$

Thus, the total effort in both sets is

$$\begin{aligned} TE= & {} x_{h}+x_{l}+y_{h}+y_{l} \\= & {} \frac{\beta ^{2}\left( w_{h}^{2}+w_{l}^{2}\right) +\alpha \beta (m_{h}+m_{l})w_{h}+\alpha \beta (m_{h}+m_{l})w_{l}+2\alpha ^{2}m_{h}m_{l}}{ 2\left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) } \\{} & {} +\,\frac{\alpha ^{2}\left( m_{h}^{2}+m_{l}^{2}\right) +\alpha \beta (w_{h}+w_{l})m_{h}+\alpha \beta (w_{h}+w_{l})m_{l}+2\beta ^{2}w_{h}w_{l}}{ 2\left( \alpha \left( m_{h}+m_{l}\right) +\beta \left( w_{h}+w_{l}\right) \right) } \\= & {} \frac{\alpha m_{h}+\alpha m_{l}+\beta w_{h}+\beta w_{l}}{2}. \end{aligned}$$

$\square$

1.6 Proof of Proposition 7

Assume that $\beta$ is any constant and that $\alpha$ approaches infinity. Then, by (20), given that $v_{1}>v_{2}>v_{3}>v_{4}$, we obtain that

• (1)
  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{3}}{TE_{1}}=\lim _{\alpha \rightarrow \infty }\frac{\frac{\left( v_{1}^{\alpha }v_{3}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{1}^{\alpha }v_{2}^{\beta }+v_{4}^{\alpha }v_{3}^{\beta }\right) }{\left( v_{1}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{2}^{\beta }+v_{3}^{\beta }\right) }}{ \frac{\left( v_{1}^{\alpha }v_{4}^{\beta }+v_{2}^{\alpha }v_{3}^{\beta }\right) \left( v_{1}^{\alpha }v_{3}^{\beta }+v_{2}^{\alpha }v_{4}^{\beta }\right) }{\left( v_{1}^{\alpha }+v_{2}^{\alpha }\right) \left( v_{3}^{\beta }+v_{4}^{\beta }\right) }}=\frac{\left( v_{3}^{\beta }+v_{4}^{\beta }\right) v_{2}^{\beta }}{\left( v_{2}^{\beta }+v_{3}^{\beta }\right) v_{4}^{\beta }} >1. \end{aligned}$$
• (2)
  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{3}}{TE_{2}}=\lim _{\alpha \rightarrow \infty }\frac{\frac{\left( v_{1}^{\alpha }v_{3}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{1}^{\alpha }v_{2}^{\beta }+v_{4}^{\alpha }v_{3}^{\beta }\right) }{\left( v_{1}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{2}^{\beta }+v_{3}^{\beta }\right) }}{ \frac{\left( v_{1}^{\alpha }v_{4}^{\beta }+v_{3}^{\alpha }v_{2}^{\beta }\right) \left( v_{1}^{\alpha }v_{2}^{\beta }+v_{3}^{\alpha }v_{4}^{\beta }\right) }{\left( v_{1}^{\alpha }+v_{3}^{\alpha }\right) \left( v_{2}^{\beta }+v_{4}^{\beta }\right) }}=\frac{\left( v_{2}^{\beta }+v_{4}^{\beta }\right) v_{3}^{\beta }}{\left( v_{2}^{\beta }+v_{3}^{\beta }\right) v_{4}^{\beta }} >1. \end{aligned}$$
• (3)
  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{3}}{TE_{4}}=\lim _{\alpha \rightarrow \infty }\frac{\frac{\left( v_{1}^{\alpha }v_{3}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{1}^{\alpha }v_{2}^{\beta }+v_{4}^{\alpha }v_{3}^{\beta }\right) }{\left( v_{1}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{2}^{\beta }+v_{3}^{\beta }\right) }}{ \frac{\left( v_{3}^{\alpha }v_{1}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{4}^{\alpha }v_{1}^{\beta }+v_{3}^{\alpha }v_{2}^{\beta }\right) }{\left( v_{3}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{1}^{\beta }+v_{2}^{\beta }\right) }}=\lim _{\alpha \rightarrow \infty }\frac{ v_{1}^{\alpha }}{v_{3}^{\alpha }}\frac{\left( v_{1}^{\beta }+v_{2}^{\beta }\right) v_{3}^{\beta }v_{2}^{\beta }}{\left( v_{2}^{\beta }+v_{3}^{\beta }\right) v_{1}^{\beta }v_{2}^{\beta }}=\infty . \end{aligned}$$
• (4)
  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{3}}{TE_{5}}=\lim _{\alpha \rightarrow \infty }\frac{\frac{\left( v_{1}^{\alpha }v_{3}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{1}^{\alpha }v_{2}^{\beta }+v_{4}^{\alpha }v_{3}^{\beta }\right) }{\left( v_{1}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{2}^{\beta }+w_{3}^{\beta }\right) }}{ \frac{\left( v_{2}^{\alpha }v_{1}^{\beta }+v_{4}^{\alpha }v_{3}^{\beta }\right) \left( v_{4}^{\alpha }v_{1}^{\beta }+v_{2}^{\alpha }v_{3}^{\beta }\right) }{\left( v_{2}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{1}^{\beta }+v_{3}^{\beta }\right) }}=\lim _{\alpha \rightarrow \infty }\frac{ v_{1}^{\alpha }}{v_{2}^{\alpha }}\frac{\left( v_{1}^{\beta }+v_{3}^{\beta }\right) v_{3}^{\beta }v_{2}^{\beta }}{\left( v_{2}^{\beta }+w_{3}^{\beta }\right) v_{1}^{\beta }v_{3}^{\beta }}=\infty . \end{aligned}$$
• (5)
  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{3}}{TE_{6}}=\lim _{\alpha \rightarrow \infty }\frac{\frac{\left( v_{1}^{\alpha }v_{3}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{1}^{\alpha }v_{2}^{\beta }+v_{4}^{\alpha }v_{3}^{\beta }\right) }{\left( v_{1}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{2}^{\beta }+w_{3}^{\beta }\right) }}{ \frac{\left( v_{2}^{\alpha }v_{1}^{\beta }+v_{3}^{\alpha }v_{4}^{\beta }\right) \left( v_{3}^{\alpha }v_{1}^{\beta }+v_{2}^{\alpha }v_{4}^{\beta }\right) }{\left( v_{2}^{\alpha }+w_{3}^{\alpha }\right) \left( v_{1}^{\beta }+v_{4}^{\beta }\right) }}=\lim _{\alpha \rightarrow \infty }\frac{ v_{1}^{\alpha }}{v_{2}^{\alpha }}\frac{\left( v_{1}^{\beta }+v_{4}^{\beta }\right) v_{3}^{\beta }v_{2}^{\beta }}{\left( v_{2}^{\beta }+w_{3}^{\beta }\right) v_{1}^{\beta }v_{4}^{\beta }}=\infty . \end{aligned}$$

Thus, when $\alpha$ converges to infinity, the optimal seeding for a designer who wishes to maximize the players’ total effort is $M:1{-}4$, $W:2{-}3.$

Likewise, by (20), given that $v_{1}>v_{2}>v_{3}>v_{4}$, we obtain that

• (6)
  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{4}}{TE_{5}}=\lim _{\alpha \rightarrow \infty }\frac{\frac{\left( v_{3}^{\alpha }v_{1}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{4}^{\alpha }v_{1}^{\beta }+v_{3}^{\alpha }v_{2}^{\beta }\right) }{\left( v_{3}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{1}^{\beta }+v_{2}^{\beta }\right) }}{ \frac{\left( v_{2}^{\alpha }v_{1}^{\beta }+v_{4}^{\alpha }v_{3}^{\beta }\right) \left( v_{4}^{\alpha }v_{1}^{\beta }+v_{2}^{\alpha }v_{3}^{\beta }\right) }{\left( v_{2}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{1}^{\beta }+v_{3}^{\beta }\right) }}=\lim _{\alpha \rightarrow \infty }\frac{ v_{3}^{\alpha }}{v_{2}^{\alpha }}\frac{\left( v_{1}^{\beta }+v_{3}^{\beta }\right) v_{1}^{\beta }v_{2}^{\beta }}{\left( v_{1}^{\beta }+v_{2}^{\beta }\right) v_{1}^{\beta }v_{3}^{\beta }}=0. \end{aligned}$$
• (7)
  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{4}}{TE_{6}}=\lim _{\alpha \rightarrow \infty }\frac{\frac{\left( v_{3}^{\alpha }v_{1}^{\beta }+v_{4}^{\alpha }v_{2}^{\beta }\right) \left( v_{4}^{\alpha }v_{1}^{\beta }+v_{3}^{\alpha }v_{2}^{\beta }\right) }{\left( v_{3}^{\alpha }+v_{4}^{\alpha }\right) \left( v_{1}^{\beta }+v_{2}^{\beta }\right) }}{ \frac{\left( v_{2}^{\alpha }v_{1}^{\beta }+v_{3}^{\alpha }v_{4}^{\beta }\right) \left( v_{3}^{\alpha }v_{1}^{\beta }+v_{2}^{\alpha }v_{4}^{\beta }\right) }{\left( v_{2}^{\alpha }+w_{3}^{\alpha }\right) \left( v_{1}^{\beta }+v_{4}^{\beta }\right) }}=\lim _{\alpha \rightarrow \infty }\frac{ v_{3}^{\alpha }}{v_{2}^{\alpha }}\frac{\left( v_{1}^{\beta }+v_{4}^{\beta }\right) v_{1}^{\beta }v_{2}^{\beta }}{\left( v_{1}^{\beta }+v_{2}^{\beta }\right) v_{1}^{\beta }v_{4}^{\beta }}=0. \end{aligned}$$

  Since we already found that $\lim _{\alpha \rightarrow \infty }\frac{TE_{4}}{ TE_{3}}=0$, and similarly it can be verified that

  $$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{TE_{4}}{TE_{2}}=\lim _{\alpha \rightarrow \infty }\frac{TE_{4}}{TE_{1}}=0, \end{aligned}$$

  we obtain that when $\alpha$ converges to infinity, the optimal seeding for a designer who wishes to minimize the players’ total effort is $M:3{-}4$, $W:1{-}2.$

$\square$