Resource allocations in the best-of-k ($k=2,3$) contests

Abstract

We study best-of-k ($k=2,3$) contests with two players who have heterogeneous resource budgets that decrease within the stages proportionally to the resource allocated in the previous stages such that for each resource unit that a player allocates, he loses $\alpha$ (the fatigue parameter) units of resource from his budget. We first analyze the players’ resource allocations and show that if the resource budget is the same, although the recycled resource is larger in the best-of-three contest than in the best-of-two contest, the total resource allocation in the best-of-two contest with the optimal asymmetric tie-breaking rule might be larger than in the best-of-three contest.

Appendix

1.1 Proof of Proposition 1

By (8) and (10), when $\frac{\delta }{(1-\delta )}\ge \alpha \ge \delta ,$

$$\begin{aligned} x_{1} & = \delta \frac{v_{1}}{\alpha }\ge (1-\delta )v_{1}=x_{2} \\ y_{1} & = \delta \frac{w_{1}}{\alpha }\ge (1-\delta )w_{1}=y_{2}. \end{aligned}$$

Otherwise, if $\alpha >\frac{\delta }{(1-\delta )}$, we obtain that $x_{1}<x_{2}$ and $y_{1}<y_{2}.$ Similarly, by (9) and (10), when $\alpha<$ $\frac{\delta }{(1-\delta )},$

$$\begin{aligned} x_{1} & = v_{1}>v_{1}(1-\alpha )=x_{2} \\ y_{1} & = w_{1}>w_{1}(1-\alpha )=y_{2}. \end{aligned}$$

Thus, if $\alpha >\frac{\delta }{(1-\delta )}$, then $x_{1}<x_{2}$ and $y_{1}<y_{2}$, and if $\alpha <\frac{\delta }{(1-\delta )}$, then $x_{1}>x_{2}$ and $y_{1}>y_{2}.$ In particular, if $\delta \ge 0.5$, we obtain that $x_{1}>x_{2}$ and $y_{1}>y_{2}.$ $\ Q.E.D.$

1.2 Proof of Proposition 3

Assume that $r=1$. When we divide the equations (15) by each other, we obtain that

$$\begin{aligned} \frac{\alpha _{1}x_{1}}{\alpha _{2}y_{1}}=\frac{(v_{1}-\alpha _{1}x_{1})}{ (w_{1}-\alpha _{2}y_{1})}=\frac{v_{1}}{w_{1}}, \end{aligned}$$

(39)

and when we insert (39) into (15) we obtain that

$$\begin{aligned} & \delta \frac{x_{1}}{(x_{1}+y_{1})^{2}}-(1-\delta )\frac{\alpha _{1}x_{1}}{ \alpha _{1}x_{1}+\alpha _{2}y_{1}}\frac{\alpha _{2}}{(v_{1}-\alpha _{1}x_{1}+w_{1}-\alpha _{2}y_{1})} \\ & \quad =\frac{x_{1}}{\left( x_{1}+y_{1}\right) ^{2}\left( \alpha _{1}x_{1}+\alpha _{2}y_{1}\right) \left( v_{1}+w_{1}-\alpha _{1}x_{1}-\alpha _{2}y_{1}\right) }\cdot \\ & \qquad \left( \begin{array}{c} \delta \alpha _{1}\alpha _{2}x_{1}^{2}-\alpha _{1}\alpha _{2}y_{1}^{2}-\delta \alpha _{1}^{2}x_{1}^{2}-\delta \alpha _{2}^{2}y_{1}^{2}-\alpha _{1}\alpha _{2}x_{1}^{2}+\delta \alpha _{1}\alpha _{2}y_{1}^{2}+\delta \alpha _{1}v_{1}x_{1} \\ +\delta \alpha _{1}w_{1}x_{1}+\delta \alpha _{2}v_{1}y_{1}+\delta \alpha _{2}w_{1}y_{1}-2\alpha _{1}\alpha _{2}x_{1}y_{1} \end{array} \right) \\ & \quad =0 \end{aligned}$$

Since $y_{1}=\frac{w_{1}\alpha _{1}x_{1}}{v_{1}\alpha _{2}}$, we obtain

$$\begin{aligned} & \delta \alpha _{1}\alpha _{2}x_{1}^{2}-\alpha _{1}\alpha _{2}\left( \frac{ w_{1}\alpha _{1}x_{1}}{v_{1}\alpha _{2}}\right) ^{2}-\delta \alpha _{1}^{2}x_{1}^{2}-\delta \alpha _{2}^{2}\left( \frac{w_{1}\alpha _{1}x_{1}}{ v_{1}\alpha _{2}}\right) ^{2}-\alpha _{1}\alpha _{2}x_{1}^{2}\\ & \qquad +\delta \alpha _{1}\alpha _{2}\left( \frac{w_{1}\alpha _{1}x_{1}}{v_{1}\alpha _{2}}\right) ^{2} \\ & \qquad +\delta \alpha _{1}v_{1}x_{1}+\delta \alpha _{1}w_{1}x_{1}+\delta \alpha _{2}v\left( \frac{w_{1}\alpha _{1}x_{1}}{v_{1}\alpha _{2}}\right) +\delta \alpha _{2}w_{1}\left( \frac{w_{1}\alpha _{1}x_{1}}{v_{1}\alpha _{2}}\right) \\ & \qquad -2\alpha _{1}\alpha _{2}x_{1}\left( \frac{w_{1}\alpha _{1}x_{1}}{v_{1}\alpha _{2}}\right) \\ & \quad =0. \end{aligned}$$

The solution of the last equation gives us player 1’s resource allocation,

$$\begin{aligned} x_{1}=\frac{\delta \alpha _{2}v_{1}\left( v_{1}+w_{1}\right) ^{2}}{\alpha _{1}\alpha _{2}(\delta v_{1}^{2}+\delta w_{1}^{2})+(1-\delta )((v_{1}\alpha _{2})^{2}+(w_{1}\alpha _{1})^{2})+2\alpha _{1}\alpha _{2}v_{1}w_{1}}, \end{aligned}$$

and then player 2’s resource allocation is

$$\begin{aligned} y_{1}=\frac{\delta \alpha _{1}w_{1}\left( v_{1}+w_{1}\right) ^{2}}{\alpha _{1}\alpha _{2}(\delta v_{1}^{2}+\delta w_{1}^{2})+(1-\delta )((v_{1}\alpha _{2})^{2}+(w_{1}\alpha _{1})^{2})+2\alpha _{1}\alpha _{2}v_{1}w_{1}}. \end{aligned}$$

Note that when $\alpha =\alpha _{1}=\alpha _{2}$, we obtain the symmetric solution given by (8) as follows:

$$\begin{aligned} x_{1}=\frac{\delta v_{1}}{\alpha _{1}},{ }y_{1}=\frac{\delta w_{1}}{ \alpha _{2}}. \end{aligned}$$

Now, assume that $a_{2}\ge a_{1}$. Then we have

$$\begin{aligned} x_{1}\ge \frac{\delta \alpha _{2}v_{1}\left( v_{1}+w_{1}\right) ^{2}}{ (\alpha _{2})^{2}(\delta v_{1}^{2}+\delta w_{1}^{2})+(1-\delta )(\alpha _{2})^{2}((v_{1})^{2}+(w_{1})^{2})+2(\alpha _{2})^{2}v_{1}w_{1}}=\frac{ \delta v_{1}}{\alpha _{2}}, \end{aligned}$$

and

$$\begin{aligned} x_{1}\le \frac{\delta \alpha _{2}v_{1}\left( v_{1}+w_{1}\right) ^{2}}{ (\alpha _{1})^{2}(\delta v_{1}^{2}+\delta w_{1}^{2})+(1-\delta )(\alpha _{1})^{2}((v_{1})^{2}+(w_{1})^{2})+2(\alpha _{1})^{2}v_{1}w_{1}}=\frac{ \alpha _{2}\delta v_{1}}{(\alpha _{1})^{2}}. \end{aligned}$$

Similarly,

$$\begin{aligned} y_{1}\ge \frac{\delta \alpha _{1}w_{1}\left( v_{1}+w_{1}\right) ^{2}}{ (\alpha _{2})^{2}(\delta v_{1}^{2}+\delta w_{1}^{2})+(1-\delta )(\alpha _{2})^{2}((v_{1})^{2}+(w_{1})^{2})+2(\alpha _{2})^{2}v_{1}w_{1}}=\frac{ \alpha _{1}\delta w_{1}}{(\alpha _{2})^{2}}, \end{aligned}$$

and

$$\begin{aligned} y_{1}\le \frac{\delta \alpha _{1}w_{1}\left( v_{1}+w_{1}\right) ^{2}}{ (\alpha _{1})^{2}(\delta v_{1}^{2}+\delta w_{1}^{2})+(1-\delta )(\alpha _{1})^{2}((v_{1})^{2}+(w_{1})^{2})+2(\alpha _{1})^{2}v_{1}w_{1}}=\frac{ \delta w_{1}}{\alpha _{1}}. \end{aligned}$$

Thus, if $\frac{\alpha _{2}\delta }{(\alpha _{1})^{2}}\le 1$, the resource allocation of both players in the first stage are given by (16) and if $\frac{\alpha _{1}\delta }{ (\alpha _{2})^{2}}>1$, both players allocate their resource budgets in the first stage. Q.E.D.

1.3 Proof of Proposition 4

According to the players’ resource allocations in the first stage given by (31), (32), and in the second stage given by (33), (34), (35), and (36), we have

• 1. If $\alpha \le \frac{1}{3}$, and $\beta \le \frac{1}{2}$

  $$\begin{aligned} x_{1} & = v_{1}>x_{2}=v_{1}(1-\alpha )>x_{3}=v_{1}(1-\alpha )(1-\beta ) \\ y_{1} & = w_{1}>y_{2}=w_{1}(1-\alpha )>y_{3}=w_{1}(1-\alpha )(1-\beta ). \end{aligned}$$

• 2. If $\alpha \le \frac{1}{3}$, and $\beta >\frac{1}{2}$

  $$\begin{aligned} x_{1} & = v_{1}>x_{2}=\frac{v_{1}(1-\alpha )}{2\beta }>x_{3}=\frac{ v_{1}(1-\alpha )}{2} \\ y_{1} & = w_{1}>y_{2}=\frac{w_{1}(1-\alpha )}{2\beta }>y_{3}=\frac{ w_{1}(1-\alpha )}{2}. \end{aligned}$$

• 3. If $\alpha >\frac{1}{3}$, and $\beta \le \frac{1}{2}$

  $$\begin{aligned} x_{1} & = \frac{v_{1}}{3\alpha }>x_{2}=\frac{2}{3}v_{1}>x_{3}=\frac{2}{3} v_{1}(1-\beta ) \\ y_{1} & = \frac{w_{1}}{3\alpha }>y_{2}=\frac{2}{3}w_{1}>x_{3}=\frac{2}{3} w_{1}(1-\beta ). \end{aligned}$$

• 4. If $\alpha >\frac{1}{3}$, and $\beta >\frac{1}{2}$

  $$\begin{aligned} x_{1} & = \frac{v_{1}}{3\alpha }>\frac{2}{3}v_{1}(1-\beta )\text { and }x_{2}= \frac{v_{1}}{3\beta }>x_{3}=\frac{v_{1}}{3} \\ y_{1} & = \frac{w_{1}}{3\alpha }>y_{3}=\frac{w_{1}}{3}\text { and }y_{2}=\frac{ w_{1}}{3\beta }>y_{3}=\frac{w_{1}}{3}. \end{aligned}$$

In all of the above four cases, the resource allocations in the third stage are smaller than in the previous ones. However, in cases 1–3, independent of the values of the fatigue parameters $\alpha$ and $\beta$, the resource allocations of the first stage are larger than that of the second one. However, in case 4, the resource allocation of the first stage is larger than that of the second one if the fatigue parameter of the first stage $\alpha$ is smaller than the fatigue parameter of the second stage $\beta$; namely, $\alpha >\frac{1}{3}$, and $\beta >\frac{1}{2}$ and $\alpha <\beta .$ $\ Q.E.D.$

1.4 Proof of Proposition 5

By (12), player 1’s expected payoff in the best-of-two contest is

$$\begin{aligned} u_{best2}^{1}=\frac{v_{1}^{r}}{v_{1}^{r}+w_{1}^{r}}, \end{aligned}$$

and by (37), player 1’s expected payoff in the best-of-three contest is

$$\begin{aligned} u_{best3}^{1}=\frac{v_{1}^{2r}}{\left( v_{1}^{r}+w_{1}^{r}\right) ^{3}} \left( v_{1}^{r}+3w_{1}^{r}\right) . \end{aligned}$$

Thus, we have

$$\begin{aligned} u_{best2}^{1}-u_{best3}^{1}=\frac{v_{1}^{r}}{v_{1}^{r}+w_{1}^{r}} -\frac{v_{1}^{2r}}{\left( v_{1}^{r}+w_{1}^{r}\right) ^{3}}\left( v_{1}^{r}+3w_{1}^{r}\right) =-v_{1}^{r}\frac{w_{1}^{r}}{\left( v_{1}^{r}+w_{1}^{r}\right) ^{3}}\left( v_{1}^{r}-w_{1}^{r}\right) . \end{aligned}$$

We can see that $u_{best3}^{1}\ge u_{best2}^{1}$ iff $v_{1}\ge w_{1}$. The result is similar for player 2. Q.E.D.

1.5 Proof of Proposition 6

We assume that the fatigue parameter is the same for both stages of the best-of-three contest and that it is equal to the fatigue parameter of the best-of-two contest. We denote this parameter by $\alpha$. According to the analysis of the resource allocations in the previous sections, the players’ total effort in the best-of-two contest is maximized when $\delta \ge \alpha$ and is equal to

$$\begin{aligned} R_{best2}=(v_{1}+w_{1})(2-\alpha ). \end{aligned}$$

Then, we have the following three cases:

• 1. When $\alpha \ge \frac{1}{2}$, the resource constraints in all the stages of the best-of-three contest are not binding. Then, the players’ total resource allocations in the best-of-three contest is

  $$\begin{aligned} R_{best3}=\frac{v_{1}+w_{1}}{3\alpha }+\frac{v_{1}+w_{1}}{3\alpha }+\frac{ v_{1}+w_{1}}{3}. \end{aligned}$$

  The difference between the total resource in both contests is

  $$\begin{aligned} R_{best3}-R_{best2}=\frac{1}{3\alpha }\left( v_{1}+w_{1}\right) \left( 3\alpha ^{2}-5\alpha +2\right) \end{aligned}$$

  Thus, $R_{best3}-R_{best2}>0$ if $0\le \alpha <\frac{2}{3}$; otherwise, if $\alpha \ge \frac{2}{3}$ we obtain that $R_{best3}-R_{best2}\le 0.$

• 2. When $\frac{1}{3}\le \alpha <\frac{1}{2}$, the resource constraint in the first stage of the best-of-three contest is binding but in the second stage is not. Then, the players’ total resource allocation is

  $$\begin{aligned} R_{best3}=\frac{v_{1}+w_{1}}{3\alpha }+\frac{2(v_{1}+w_{1})}{3}+\frac{ 2(v_{1}+w_{1})(1-\alpha )}{3}. \end{aligned}$$

  The difference between the total resource in both contests is

  $$\begin{aligned} R_{best3}-R_{best2}=\frac{1}{3\alpha }\left( \alpha -1\right) ^{2}\left( v_{1}+w_{1}\right) >0. \end{aligned}$$

• 3. When $\alpha <\frac{1}{3}$, the resource constraint in the best-of-three contest is binding and the players’ total resource allocation is

  $$\begin{aligned} R_{best3}=(v_{1}+w_{1})+(v_{1}+w_{1})(1-\alpha )+(v_{1}+w_{1})(1-\alpha )^{2}. \end{aligned}$$

  Then

  $$\begin{aligned} R_{best3}-R_{best2}=\left( \alpha -1\right) ^{2}\left( v_{1}+w_{1}\right) >0. \end{aligned}$$

  Q.E.D.