On the optimal allocation of prizes in best-of-three all-pay auctions

Abstract

We study best-of-three all-pay auctions with two players who compete in three stages with a single match per stage. The first player to win two matches wins the contest. We assume that a prize sum is given, and show that if players are symmetric, the allocation of prizes does not have any effect on the players’ expected total effort. On the other hand, if players are asymmetric, in order to maximize the players’ expected total effort, independent of the players’ types, it is not optimal to allocate a single final prize to the winner. Instead, it is optimal to allocate intermediate prizes in the first stage or/and in the second stage in addition to the final prize. When the asymmetry of the players’ types is sufficiently high, it is optimal to allocate intermediate prizes in both two first stages and a final prize to the winner.

Appendix

1.1 Proof of Proposition 2

1. If condition 5 holds, the expected total effort given by (11) is

$$\begin{aligned} TE^{(B1)}= & {}\; TE^{(1B1)}+p_{1}^{(1B1)}TE^{(2)}+p_{2}^{(1B1)}TE^{(2B1)}+p_{1}^{(1B1)}p_{2}^{(2)}TE^{(3)}\nonumber \\&+p_{2}^{(1B1)}p_{1}^{(2B1)}TE^{(3)} \nonumber \\= & {}\; \frac{\alpha v_{2}}{2}\left( 1+\frac{\alpha v_{2}}{(v_{1}-\beta v_{2})-(v_{1}(1-\alpha )-v_{2}(2-2\alpha -\beta ))}\right) \nonumber \\&\quad +\left( 1-\frac{\alpha v_{2}}{2((v_{1}-\beta v_{2})-(v_{1}(1-\alpha )-v_{2}(2-2\alpha -\beta )))}\right) \frac{\beta v_{2}}{2}\nonumber \\&\times \left( 1+\frac{\beta v_{2}}{ v_{1}(1-\alpha )-(v_{1}-v_{2})(1-\alpha -\beta )}\right) \nonumber \\&\quad +\frac{\alpha v_{2}}{2((v_{1}-\beta v_{2})-(v_{1}(1-\alpha )-v_{2}(2-2\alpha -\beta )))}\frac{v_{2}(1-\alpha )}{2}\nonumber \\&\times \left( 1+\frac{ v_{2}(1-\alpha )}{v_{1}(1-\alpha )-v_{2}(1-\alpha -\beta )}\right) \nonumber \\&\quad +\left( 1-\frac{\alpha v_{2}}{2((v_{1}-\beta v_{2})-(v_{1}(1-\alpha )-v_{2}(2-2\alpha -\beta )))}\right) \nonumber \\&\times \frac{\beta v_{2}}{2(\beta v_{1}+v_{2}(1-\alpha -\beta ))}\frac{v_{2}(1-\alpha -\beta )(v_{1}+v_{2})}{ 2v_{1}} \nonumber \\&\quad +\frac{\alpha v_{2}}{2((v_{1}-\beta v_{2})-(v_{1}(1-\alpha )-v_{2}(2-2\alpha -\beta )))}\nonumber \\&\times \left( 1-\frac{v_{2}(1-\alpha )}{2(v_{1}(1-\alpha )-v_{2}(1-\alpha -\beta ))}\right) \frac{v_{2}(1-\alpha -\beta )(v_{1}+v_{2})}{ 2v_{1}} \end{aligned}$$

(13)

Let $v_{1}=v,v_{2}=1$ and $\beta =0.$ Then we have

$$\begin{aligned} TE_{\alpha }^{(B1)}= & {} \;\frac{\alpha }{2}+\frac{\alpha ^{2}}{2\alpha v+4-4\alpha }+\frac{\alpha -\alpha ^{2}}{2(2\alpha v+4-4\alpha )}\\&+\left( \frac{ \alpha }{2\alpha v+4-4\alpha }\right) \left( \frac{(1-\alpha )^{2}}{2v-2\alpha v-2+2\alpha }\right) \\&+\left( \frac{v+1-\alpha v-\alpha }{2v}\right) \left( \frac{\alpha }{2(\alpha v+2-2\alpha } \right) \left( 1-\frac{1-\alpha }{2v-2\alpha v-2+2\alpha }\right) \end{aligned}$$

The marginal effect of $\alpha$ on the expected total effort is

$$\begin{aligned} \frac{dTE_{\alpha }^{(B1)}}{d\alpha }=\frac{4\alpha ^{2}v^{3}+16(\alpha -\alpha ^{2})v^{2}+(13\alpha ^{2}-32\alpha +24)v+6\alpha ^{2}-12\alpha +6}{ 8\alpha ^{2}v^{3}+32(\alpha -\alpha ^{2})v^{2}+32(\alpha ^{2}-2\alpha +1)v} \end{aligned}$$

(14)

when $\alpha$ approaches zero we obtain that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{dTE_{\alpha }^{(B1)}}{d\alpha }=\frac{24v+6 }{32v} \end{aligned}$$

Thus, if condition 5 holds, allocating a sufficiently small prize of $\alpha$ in the first stage, increases the players’ expected total effort.

2. If condition 7 holds, the expected total effort given by (12) is

$$\begin{aligned} TE_{\alpha }^{(B2)}= & {} \frac{v_{2}(2-\alpha -\beta )-v_{1}(1-\alpha )}{2}\left( 1+ \frac{v_{2}(2-\alpha -\beta )-v_{1}(1-\alpha )}{v_{1}-\beta v_{2}}\right) \nonumber \\&\quad +\left( 1-\frac{v_{2}(2-\alpha -\beta )-v_{1}(1-\alpha )}{2(v_{1}-\beta v_{2})}\right) \frac{\beta v_{2}}{2}\nonumber \\&\times \left( 1+\frac{\beta v_{2}}{v_{1}(1-\alpha )-(v_{1}-v_{2})(1-\alpha -\beta )}\right) \nonumber \\& \quad +\frac{v_{2}(2-\alpha -\beta )-v_{1}(1-\alpha )}{2(v_{1}-\beta v_{2})} \frac{v_{1}(1-\alpha )-v_{2}(1-\alpha -\beta )}{2}\nonumber \\&\times \left( 1+\frac{v_{1}(1-\alpha )-v_{2}(1-\alpha -\beta }{v_{2}(1-\alpha )}\right) \nonumber \\&\quad +\left( 1-\frac{v_{2}(2-\alpha -\beta )-v_{1}(1-\alpha )}{2(v_{1}-\beta v_{2})}\right) \frac{\beta v_{2}}{2(\beta v_{1}+v_{2}(1-\alpha -\beta ))}\nonumber \\&\times \frac{ v_{2}(1-\alpha -\beta )(v_{1}+v_{2})}{2v_{1}} \nonumber \\&\quad +\frac{v_{2}(2-\alpha -\beta )-v_{1}(1-\alpha )}{2(v_{1}-\beta v_{2})} \frac{v_{1}(1-\alpha )-v_{2}(1-\alpha -\beta )}{2v_{2}(1-\alpha )}\nonumber \\&\times \frac{ v_{2}(1-\alpha -\beta )(v_{1}+v_{2})}{2v_{1}} \end{aligned}$$

(15)

Let $v_{1}=v,v_{2}=1$ and $\beta =0.$ Then we have

$$\begin{aligned} TE_{\alpha }^{(B2)}= & {}\; \frac{2-\alpha -v(1-\alpha )}{2}+\frac{(2-\alpha -v(1-\alpha ))^{2}}{2v} \\&+\frac{(2-\alpha -v(1-\alpha ))}{2v}\left( \frac{v(1-\alpha )-1+\alpha }{2}+ \frac{(v(1-\alpha )-1+\alpha )^{2}}{2(1-\alpha )}\right) \\&+\frac{(2-\alpha -v(1-\alpha ))}{2v}\left( \frac{v(1-\alpha )-1+\alpha }{ 2(1-\alpha )}\right) \left( \frac{v-\alpha v+1-\alpha }{2v}\right) \end{aligned}$$

The marginal effect of $\alpha$ on the expected total effort is

$$\begin{aligned} \frac{dTE_{\alpha }^{(B2)}}{d\alpha }=\frac{4(\alpha -1)v^{4}+(12-14\alpha )v^{3}+(18-23)v^{2}+(18-10\alpha )v+2\alpha -3}{8v^{2}} \end{aligned}$$

(16)

When $\alpha$ approaches zero we obtain that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{dTE_{\alpha }^{(B2)}}{d\alpha }=-\frac{ -4v^{4}+12v^{3}-23v^{2}+18v-3}{8v^{2}} \end{aligned}$$

It can be verified that the last term is positive for all $v>1.$ Thus, if condition 7 holds, compared to the best-of-three all-pay auction with a single final prize, allocating a sufficiently small prize of $\alpha$ in the first stage increases the players’ expected total effort.

1.2 Proof of Proposition 3

1. Let $v_{1}=v,v_{2}=1,$ and $\alpha =0$. Then, if condition 5 holds, the expected total effort given by (13) is

$$\begin{aligned} TE_{\beta }^{(B1)}=\frac{\beta }{2}+\frac{\beta ^{2}}{2\beta v+2-2\beta }+ \frac{(1-\beta )(v+1)}{2v}\frac{\beta }{2\beta v+2-2\beta } \end{aligned}$$

The marginal effect of $\beta$ on the expected total effort is

$$\begin{aligned} \frac{dTE_{\beta }^{(B1)}}{d\beta }=\frac{2\beta ^{2}v^{3}+(4\beta -3\beta ^{2})v^{2}+(3-2\beta )v+\beta ^{2}-2\beta +1}{4\beta ^{2}v^{3}+8(\beta -\beta ^{2})v^{2}+4(\beta ^{2}-2\beta +1)v} \end{aligned}$$

(17)

when $\beta$ approaches zero we obtain that

$$\begin{aligned} \lim _{\beta \rightarrow 0}\frac{dTE_{\beta }^{(B1)}}{d\beta }=\frac{3v+1}{4} \end{aligned}$$

Thus, if condition 5 holds, allocating a sufficiently small prize of $\beta$ in the second stage increases the players’ expected total effort.

2. Let $v_{1}=v,v_{2}=1,$ and $\alpha =0$. Then if condition 7 holds, the expected total effort given by (15) is

$$\begin{aligned} TE_{\beta }^{(B2)}= & {} \;\frac{2-\beta -v}{2}+\frac{(2-\beta -v)^{2}}{2(v-\beta )} \\&+\frac{3v-\beta -2}{2(v-\beta )}\left( \frac{\beta }{2}+\frac{\beta ^{2}}{ 2(\beta v+1-\beta )}\right) \\&+\frac{2-\beta -v}{2(v-\beta )}\left( \frac{v-1+\beta }{2}+\frac{(v-1+\beta )^{2} }{2}\right) \\&+\frac{3v-\beta -2}{2(v-\beta )}\frac{\beta }{2(\beta v+1-\beta )}\frac{ v+1-\beta v-\beta }{2v} \\&+\frac{2-\beta -v}{2(v-\beta )}\frac{v-1+\beta }{2}\frac{v+1-\beta v-\beta }{2v} \end{aligned}$$

The marginal effect of $\beta$ on the expected total effort is

$$\begin{aligned} \frac{dTE_{\beta }^{(B2)}}{d\beta }=\frac{A}{B} \end{aligned}$$

(18)

where

$$\begin{aligned} A= & {}\; -7\beta ^{2}v^{6}-(8\beta ^{3}-41\beta ^{2}+14\beta )v^{5}-(-\beta ^{4}-36\beta ^{3}+116\beta ^{2}-68\beta +7)v^{4} \nonumber \\&-(-2\beta ^{5}+9\beta ^{4}+56\beta ^{3}-18\beta ^{2}+140\beta -30)v^{3}-(6\beta ^{5}-24\beta ^{4}-24\beta ^{3}\nonumber \\&+144\beta ^{2}-136\beta +41)v^{2} \nonumber \\&-(-6\beta ^{5}+25\beta ^{4}-16\beta ^{3}-40\beta ^{2}+54\beta -20)v-(2\beta ^{5}-9\beta ^{4}\nonumber \\&+12\beta ^{3}-3\beta ^{2}-4\beta +2) \end{aligned}$$

(19)

and

$$\begin{aligned} B= & {}\; \beta ^{2}v^{5}+16(-\beta ^{3}-\beta ^{2}+\beta )v^{4}+8(\beta ^{4}+4\beta ^{3}-3\beta ^{2}-2\beta +1)v^{3} \nonumber \\&+16(-\beta ^{4}+2\beta ^{2}-\beta )v^{2}+8(\beta ^{4}-2\beta ^{3}+\beta ^{2})v \end{aligned}$$

(20)

When $\beta$ approaches zero we obtain that

$$\begin{aligned} \lim _{\beta \rightarrow 0}\frac{dTE_{\beta }^{(B2)}}{d\beta }=\frac{ -7v^{4}+30v^{3}-41v^{2}+20v-2}{8v^{3}} \end{aligned}$$

It can be verified that the last term is positive for all $1<v\le 2.$ Since condition 7 holds if $1\le v<2-\beta$, we obtain that compared to the best-of-three all-pay auction with a single final prize, allocating a sufficiently small prize of $\beta$ in the second stage increases the players’ expected total effort.

1.3 Proof of Proposition 4

1. Let $v_{1}=v,v_{2}=1,$ and $\alpha =\beta$. Then, if condition 5 holds, the expected total effort given by (13) is

$$\begin{aligned} TE_{\alpha \beta }^{(B1)}= & {}\; \frac{\alpha }{2}+\frac{\alpha ^{2}}{2(\alpha v+2-4\alpha )}+ (1-\frac{\alpha }{2(\alpha v+2-4\alpha )}(\frac{\alpha }{2}+ \frac{\alpha ^{2}}{2(\alpha v+1-2\alpha )}\\&+\frac{\alpha }{2(\alpha v+2-4\alpha )}\left( \frac{1-\alpha }{2}+\frac{ (1-\alpha )^{2}}{2(v-\alpha v-1+2\alpha )}\right) \\&+\frac{v+1-2\alpha v-2\alpha }{2v}\left( 1-\frac{\alpha }{2(\alpha v+2-4\alpha )} \right) \frac{\alpha }{2(\alpha v+1-2\alpha )} \\&+\frac{v+1-2\alpha v-2\alpha }{2v}\frac{\alpha }{2(\alpha v+2-4\alpha )}\left( 1- \frac{1-\alpha }{2(v-\alpha v-1+2\alpha )}\right) \end{aligned}$$

The marginal effect of $\alpha$ on the expected total effort is

$$\begin{aligned} \frac{dTE_{\alpha \beta }^{(B1)}}{d\alpha }=\frac{C}{D} \end{aligned}$$

(21)

where

$$\begin{aligned} C= & {}\; (8\alpha ^{6}-16\alpha ^{5}+8\alpha ^{4})v^{7}+(-132\alpha ^{6}+29\alpha ^{5}-212\alpha ^{4}+48\alpha ^{3})v^{6} \nonumber \\&+(872\alpha ^{6}-2144\alpha ^{5}+1930\alpha ^{4}-764\alpha ^{3}+114\alpha ^{2})v^{5} \nonumber \\&+(-2932\alpha ^{6}+7752\alpha ^{5}-8128\alpha ^{4}+4258\alpha ^{3}-1123\alpha ^{2}+120\alpha )v^{4} \nonumber \\&+(5192\alpha ^{6}-14560\alpha +16974\alpha ^{4}-10506\alpha ^{3}+3625\alpha ^{2}-656\alpha +48)v^{3} \nonumber \\&+(-4240\alpha ^{6}+12544\alpha ^{5}-15820\alpha ^{4}+10868\alpha ^{3}-4278\alpha ^{2}+912\alpha -82)v^{2} \nonumber \\&+(480\alpha ^{6}-1664\alpha ^{5}+2480\alpha ^{4}-1984\alpha ^{3}+886\alpha ^{2}-208\alpha +20)v \nonumber \\&+896\alpha ^{6}-2688\alpha ^{5}+3360\alpha ^{4}-2240\alpha ^{3}+840\alpha ^{2}-168\alpha +14 \end{aligned}$$

(22)

and

$$\begin{aligned} D= & {}\; (8\alpha ^{6}-16\alpha ^{5}+8\alpha ^{4})v^{7}+(-128\alpha ^{6}+288\alpha ^{5}-208\alpha ^{4}+48\alpha ^{3})v^{6} \nonumber \\&+(832\alpha ^{6}-2048\alpha ^{5}+1840\alpha ^{4}-720\alpha ^{3}+104\alpha ^{2})v^{5} \nonumber \\&+(-2816\alpha ^{6}+7424\alpha ^{5}-7680\alpha ^{4}+3904\alpha ^{3}-976\alpha ^{2}+96\alpha )v^{4} \nonumber \\&+(5248\alpha ^{6}-14592\alpha ^{5}+16576\alpha ^{4}-9792\alpha ^{3}+3144\alpha ^{2}-512\alpha +32)v^{3} \nonumber \\&+(-5120\alpha ^{6}+14848\alpha ^{5}-17920\alpha ^{4}+11520\alpha ^{3}-4160\alpha ^{2}+800\alpha -64)v^{2} \nonumber \\&+(2048\alpha ^{6}-6144\alpha ^{5}+7680\alpha ^{4}-5120\alpha ^{3}+1920\alpha ^{2}-384\alpha +32)v \end{aligned}$$

(23)

When $\alpha$ approaches zero we obtain that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{dTE_{\alpha \beta }^{(B1)}}{d\alpha }=\frac{ 48v^{3}-82v^{2}+20v+14}{32v^{3}-64v^{2}+32v} \end{aligned}$$

It can be verified that the last term is positive for all $v\ge 1$. Thus, if condition 5 holds, compared to the best-of-three all-pay auction with a single final prize, allocating a sufficiently small prize of $\alpha$ in both first stages increases the players’ expected total effort.

2. If condition 7 holds, $v_{1}=v,v_{2}=1,$ and $\alpha =0,$ then the expected total effort given by (15) is

$$\begin{aligned} TE_{\alpha \beta }^{(B2)}= & {} \;\frac{2-\beta -v}{2}+\frac{(2-\beta -v)^{2}}{ 2(v-\beta )} \\&+\frac{3v-\beta -2}{2(v-\beta )}\left( \frac{\beta }{2}+\frac{\beta ^{2}}{ 2(\beta v+1-\beta )}\right) \\&+\frac{2-\beta -v}{2(v-\beta )}\left( \frac{v-1+\beta }{2}+\frac{(v-1+\beta )^{2} }{2}\right) \\&+\frac{3v-\beta -2}{2(v-\beta )}\frac{\beta }{2(\beta v+1-\beta )}\frac{ v+1-\beta v-\beta }{2v} \\&+\frac{2-\beta -v}{2(v-\beta )}\frac{v-1+\beta }{2}\frac{v+1-\beta v-\beta }{2v} \end{aligned}$$

The marginal effect of $\beta$ on the expected total effort is

$$\begin{aligned} \frac{dTE_{\alpha \beta }^{(B2)}}{d\beta }=\frac{E}{F} \end{aligned}$$

(24)

where

$$\begin{aligned} E= & {}\; (-4\alpha ^{3}+4\alpha ^{2})v^{7}+(2\alpha ^{4}+40\alpha ^{3}-43\alpha ^{2}+8\alpha )v^{6} \nonumber \\&+(-20\alpha ^{4}-17\alpha ^{3}+206\alpha ^{2}-58\alpha +4)v^{5}+(88\alpha ^{4}+384\alpha ^{3}\nonumber \\&-559\alpha ^{2}+200\alpha -19)v^{4} \nonumber \\&+(-208\alpha ^{4}-312\alpha ^{3}+835\alpha ^{2}-386\alpha +53)v^{3}+(262\alpha ^{4}+120\alpha ^{3}\nonumber \\&-578\alpha ^{2}+340\alpha -59)v^{2} \nonumber \\&+(-156\alpha ^{4}+76\alpha ^{3}+135\alpha ^{2}-112\alpha +23)v+(32\alpha ^{4}-32\alpha ^{3}+8\alpha -2) \end{aligned}$$

(25)

and

$$\begin{aligned} F= & {}\; 8\alpha ^{2}v^{5}+(-16\alpha ^{3}-32\alpha ^{2}+16\alpha )v^{4}+(8\alpha ^{4}+64\alpha ^{3}-32\alpha +8)v^{3} \nonumber \\&+(-32\alpha ^{4}-48\alpha ^{3}+64\alpha ^{2}-16\alpha )v^{2}+(32\alpha ^{4}-32\alpha ^{3}+8\alpha ^{2})v \end{aligned}$$

(26)

When $\beta$ approaches zero we obtain that

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{dTE_{\alpha \beta }^{(B2)}}{d\alpha }=\frac{ 4v^{5}-19v^{4}+53v^{3}-59v^{2}+23v-2}{8v^{3}} \end{aligned}$$

It can be verified that the last term is positive for all $v>1.$ Since condition 7 holds if $1\le v<2-\beta$, we obtain that compared to the best0of-three all-pay auction with a single final prize, allocating a sufficiently small prize of $\beta$ in the second stage increases the players’ expected total effort.

1.4 Proof of Proposition 5

When players are sufficiently asymmetric, namely, $v_{1}=v>>v_{2}=1$, condition 5 holds. In that case, if we allocate a prize of $\alpha$ in the first stage, by (14), we obtain that the marginal effect of the prize on the expected total effort when v approaches infinity is

$$\begin{aligned} \lim _{v\rightarrow \infty }\frac{dTE_{\alpha }^{(B1)}}{d\alpha }=\frac{ 4\alpha ^{2}}{8\alpha ^{2}}=\frac{1}{2} \end{aligned}$$

If we allocate a prize of $\beta$ in the second stage, by (17), we obtain that the marginal effect of the prize on the expected total effort when v approaches infinity is

$$\begin{aligned} \lim _{v\rightarrow \infty }\frac{dTE_{\beta }^{(B1)}}{d\beta }=\frac{2\beta ^{2}}{4\beta ^{2}}=\frac{1}{2} \end{aligned}$$

and if we allocate a prize of $\alpha$ in both the first stages, by (21), (22) and (23), we obtain that the marginal effect of the prizes on the expected total effort when v approaches infinity is

$$\begin{aligned} \lim _{v\rightarrow \infty }\frac{dTE_{\alpha \beta }^{(B1)}}{d\alpha }=\frac{ 8\alpha ^{6}-16\alpha ^{5}+8\alpha ^{4}}{8\alpha ^{6}-16\alpha ^{5}+8\alpha ^{4}}=1 \end{aligned}$$

Thus, we obtain that

$$\begin{aligned} \lim _{v\rightarrow \infty }\frac{dTE_{\alpha \beta }^{(B1)}}{d\alpha } >\lim _{v\rightarrow \infty }\frac{dTE_{\alpha }^{(B1)}}{d\alpha } =\lim _{v\rightarrow \infty }\frac{dTE_{\beta }^{(B1)}}{d\beta } \end{aligned}$$

In other words, in order to maximize the players’ expected total effort. it is better to award intermediate prizes in both first stages than in only one of them .