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The optimal design of round-robin tournaments with three players

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Abstract

We study the optimal design of round-robin tournaments with three symmetric players. We characterize the subgame-perfect equilibrium in these tournaments with either one or two prizes. Our results show that the players who wish to maximize their expected payoffs or their probabilities of winning have different preferences about the order of games under tournaments with one or two prizes. We analyze the optimal allocations of players for a designer who wishes to maximize the players’ expected total effort in the tournaments with one and two prizes, and by comparing between them, it is demonstrated that in order to maximize the players’ expected total effort the designer should allocate only one prize.

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Notes

  1. Applications of the all-pay contest have been made to rent-seeking and lobbying in organizations, R&D races, political contests, promotions in labor markets, trade wars, military and biological wars of attrition (see, for example, Hillman and Riley 1989; Baye et al. 1993; Amman and Leininger 1996; Krishna and Morgan 1997; Che and Gale 1998; Siegel 2009).

  2. In each match, the probability of a tie in our model is zero, since each player’s strategy set is infinite. Note that it has been suggested that for the FIFA football tournaments starting from 2026 there will be no draws, and group matches that end in draws will be decided by penalty shootouts (https://www.ibtimes.co.uk/how-48-team-world-cup-2026-will-work-what-left-be-decided-1600342).

  3. In real life, there are many kinds of three-player games. For example, Leagues A and B of the 2018 UEFA Nations League consist of three teams (https://en.wikipedia.org/wiki/2018%E2%80%9319_UEFA_Nations_League#League_A). Other examples can be found in tennis tournaments between nations, such as men’s Davis Cup and women’s Fed Cup. For instance, in 2015 three pools of Group 3 of the European Zone of Davis Cup consisted of three teams (https://www.daviscup.com/en/draws-results/group-iii/europe/2015.aspx). The same structure also appeared in two pools of the 2018 Fed Cup European Zone (https://www.fedcup.com/en/draws-results/group-i/europe-africa/2018.aspx).

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Appendices

Appendix A: A random allocation of players in the round-robin tournament with three players and two prizes

In the following, we provide in every possible vertex (game) of Figure 1, the players’ mixed strategies, their expected payoffs and their probabilities of winning. These results are summarized in Table 6. The complete analysis can be found in Online-Appendix A.

Table 6 Players’ expected payoffs and winning probabilities in Vertexes A–F of Figure 1

The players’ expected payoffs

$$ \begin{aligned} E(u_{1} ) & = \frac{89}{864} = 0.103 \\ E(u_{2} ) & = \frac{13}{96} = 0.135 \\ E(u_{3} ) & = \frac{2,645}{13,824} = 0.19 \\ \end{aligned} $$

The players’ probabilities of winning

Player 1’s probability to win a prize is

$$ p_{1} = p_{12}^{F} \cdot p_{13}^{E} \cdot p_{23}^{C} + p_{12}^{F} \cdot p_{13}^{E} \cdot p_{32}^{C} + \frac{2}{3} \cdot \left( {p_{12}^{F} \cdot p_{31}^{E} \cdot p_{23}^{B} } \right) + p_{12}^{F} \cdot p_{31}^{E} \cdot p_{32}^{B} + p_{21}^{F} \cdot p_{13}^{D} \cdot p_{23}^{A} + \frac{2}{3} \cdot \left( {p_{21}^{F} \cdot p_{13}^{D} \cdot p_{32}^{A} } \right) = 0.6417 $$

Player 2’s probability to win a prize is

$$ p_{2} = p_{12}^{F} \cdot p_{13}^{E} \cdot p_{23}^{C} + \frac{2}{3} \cdot \left( {p_{12}^{F} \cdot p_{31}^{E} \cdot p_{23}^{B} } \right) + p_{21}^{F} \cdot p_{13}^{D} \cdot p_{23}^{A} + \frac{2}{3} \cdot \left( {p_{21}^{F} \cdot p_{13}^{D} \cdot p_{32}^{A} } \right) + p_{21}^{F} \cdot p_{31}^{D} = 0.6473 $$

And, player 3’s probability to win a prize is

$$ p_{3} = p_{12}^{F} \cdot p_{13}^{E} \cdot p_{32}^{C} + \frac{2}{3} \cdot \left( {p_{12}^{F} \cdot p_{31}^{E} \cdot p_{23}^{B} } \right) + p_{12}^{F} \cdot p_{31}^{E} \cdot p_{32}^{B} + \frac{2}{3} \cdot \left( {p_{21}^{F} \cdot p_{13}^{D} \cdot p_{32}^{A} } \right) + p_{21}^{F} \cdot p_{31}^{D} = 0.7109 $$

The players’ expected total effort

The expected total effort in the tournament is

$$ {\text{TE}} = {\text{TE}}^{F} + p_{12}^{F} \cdot {\text{TE}}^{E} + p_{21}^{F} \cdot {\text{TE}}^{D} + p_{12}^{F} \cdot p_{13}^{E} \cdot {\text{TE}}^{C} + p_{12}^{F} \cdot p_{31}^{E} \cdot {\text{TE}}^{B} + p_{21}^{F} \cdot p_{13}^{D} \cdot {\text{TE}}^{A} = 0.57 $$

The length of the tournament

The probability that the winners of the tournament will be determined before the last round is

$$ p_{21}^{F} \cdot p_{31}^{D} = 0.181 $$

Appendix B: The winner of the first round competes in the second round of the round-robin tournament with three players and two prizes

In the following, we provide in every possible vertex (game) of Figure 2, the players’ mixed strategies, their expected payoffs and their probabilities of winning. These results are summarized in Table 7. The complete analysis can be found in Online-Appendix B.

Table 7 Players’ expected payoffs and winning probabilities in Vertexes A–G of Figure 2

The players’ expected payoffs

$$ \begin{aligned} E(u_{1} ) = \frac{13}{96} = 0.135 \hfill \\ E(u_{2} ) = \frac{13}{96} = 0.135 \hfill \\ E(u_{3} ) = \frac{5}{24} = 0.208 \hfill \\ \end{aligned} $$

The players’ probabilities of winning

Player 1’s probability to win a prize is

$$ p_{1} = p_{12}^{G} \cdot p_{13}^{F} \cdot p_{23}^{D} + p_{12}^{G} \cdot p_{13}^{F} \cdot p_{32}^{D} + \frac{2}{3} \cdot (p_{12}^{G} \cdot p_{31}^{F} \cdot p_{23}^{C} ) + p_{12}^{G} \cdot p_{31}^{F} \cdot p_{32}^{C} + p_{21}^{G} \cdot p_{23}^{E} \cdot p_{13}^{B} + \frac{2}{3} \cdot (p_{21}^{G} \cdot p_{32}^{E} \cdot p_{13}^{A} ) = 0.648 $$

Player 2’s probability to win a prize is

$$ p_{2} = p_{12}^{G} \cdot p_{13}^{F} \cdot p_{23}^{D} + \frac{2}{3} \cdot (p_{12}^{G} \cdot p_{31}^{F} \cdot p_{23}^{C} ) + p_{21}^{G} \cdot p_{23}^{E} \cdot p_{13}^{B} + p_{21}^{G} \cdot p_{23}^{E} \cdot p_{31}^{B} + \frac{2}{3} \cdot \left( {p_{21}^{G} \cdot p_{32}^{E} \cdot p_{13}^{A} } \right) + p_{21}^{G} \cdot p_{32}^{E} \cdot p_{31}^{A} = 0.648 $$

And, player 3’s probability to win a prize is

$$ p_{3} = p_{12}^{G} \cdot p_{13}^{F} \cdot p_{32}^{D} + \frac{2}{3} \cdot \left( {p_{12}^{G} \cdot p_{31}^{F} \cdot p_{23}^{C} } \right) + p_{12}^{G} \cdot p_{31}^{F} \cdot p_{32}^{C} + p_{21}^{G} \cdot p_{23}^{E} \cdot p_{31}^{B} + \frac{2}{3} \cdot \left( {p_{21}^{G} \cdot p_{32}^{E} \cdot p_{13}^{A} } \right) + p_{21}^{G} \cdot p_{32}^{E} \cdot p_{31}^{A} = 0.704 $$

The players’ expected total effort

The expected total effort in the tournament is

$$ \begin{aligned} {\text{TE}} & = {\text{TE}}^{G} + p_{12}^{G} \cdot {\text{TE}}^{F} + p_{21}^{G} \cdot {\text{TE}}^{E} + p_{12}^{G} \cdot p_{13}^{F} \cdot {\text{TE}}^{D} + p_{12}^{G} \cdot p_{31}^{F} \cdot {\text{TE}}^{C} \\ & \quad + p_{21}^{G} \cdot p_{23}^{E} \cdot {\text{TE}}^{B} + p_{21}^{G} \cdot p_{32}^{E} \cdot {\text{TE}}^{A} = 0.52 \\ \end{aligned} $$

The length of the tournament

The probability that the winners of the tournament will be determined before the last round is equal to zero since there is no possibility that one of the players lost twice in the first two rounds.

Appendix C: The winner of the first round competes in the third round of the round-robin tournament with three players and two prizes

In the following, we provide in every possible vertex (game) of Figure 3, the players’ mixed strategies, their expected payoffs and their probabilities of winning. These results are summarized in Table 8. The complete analysis can be found in Online-Appendix C.

Table 8 Players’ expected payoffs and winning probabilities in Vertexes A–E of Figure 3

The players’ expected payoffs

$$ \begin{aligned} E(u_{1} ) & = \frac{1}{24} = 0.042 \\ E(u_{2} ) & = \frac{1}{24} = 0.042 \\ E(u_{3} ) & = \frac{1}{6} = 0.166 \\ \end{aligned} $$

The players’ probabilities of winning

Player 1’s probability to win a prize is

$$ p_{1} = p_{12}^{E} \cdot p_{23}^{D} \cdot p_{13}^{B} + \frac{2}{3} \cdot \left( {p_{12}^{E} \cdot p_{23}^{D} \cdot p_{31}^{B} } \right) + p_{12}^{E} \cdot p_{32}^{D} + p_{21}^{E} \cdot p_{13}^{C} \cdot p_{23}^{A} + \frac{2}{3} \cdot \left( {p_{21}^{E} \cdot p_{13}^{C} \cdot p_{32}^{A} } \right) = 0.639 $$

Player 2’s probability to win a prize is

$$ p_{2} = p_{12}^{E} \cdot p_{23}^{D} \cdot p_{13}^{B} + \frac{2}{3} \cdot \left( {p_{12}^{E} \cdot p_{23}^{D} \cdot p_{31}^{B} } \right) + p_{21}^{E} \cdot p_{13}^{C} \cdot p_{23}^{A} + \frac{2}{3} \cdot \left( {p_{21}^{E} \cdot p_{13}^{C} \cdot p_{32}^{A} } \right) + p_{21}^{E} \cdot p_{31}^{C} = 0.639 $$

And, player 3’s probability to win a prize is

$$ p_{3} = \frac{2}{3} \cdot \left( {p_{12}^{E} \cdot p_{23}^{D} \cdot p_{31}^{B} } \right) + p_{12}^{E} \cdot p_{32}^{D} + \frac{2}{3} \cdot \left( {p_{21}^{E} \cdot p_{13}^{C} \cdot p_{32}^{A} } \right) + p_{21}^{E} \cdot p_{31}^{C} = 0.722 $$

The players’ expected total effort

The expected total effort in the tournament is

$$ {\text{TE}} = {\text{TE}}^{E} + p_{12}^{E} \cdot {\text{TE}}^{D} + p_{21}^{E} \cdot {\text{TE}}^{C} + p_{12}^{E} \cdot p_{23}^{D} \cdot {\text{TE}}^{B} + p_{21}^{E} \cdot p_{13}^{C} \cdot {\text{TE}}^{A} = 0.75 $$

The length of the tournament

The probability that the winners of the tournament will be determined before the last round is

$$ p_{12}^{E} \cdot p_{32}^{D} + p_{21}^{E} \cdot p_{31}^{C} = 0.44 $$

Appendix D: A random allocation of players in the round-robin tournament with three players and one prize

This case which is presented in Figure 4 was already analyzed by Krumer et al. (2017). In the following, we summarize the main results.

The players’ expected payoffs

$$ \begin{aligned} E(u_{1} ) & = \frac{1}{12} = 0.083 \\ E(u_{2} ) & = \frac{5}{12} = 0.416 \\ E(u_{3} ) & = 0 \\ \end{aligned} $$

The players’ probabilities of winning

$$ \begin{aligned} p_{1} & = p_{12}^{F} \cdot p_{13}^{E} + \frac{{p_{12}^{F} \cdot p_{31}^{E} \cdot p_{23}^{C} }}{3} + \frac{{p_{21}^{F} \cdot p_{13}^{D} \cdot p_{32}^{B} }}{3} = 0.193 \\ p_{2} & = \frac{{p_{12}^{F} \cdot p_{31}^{E} \cdot p_{23}^{C} }}{3} + p_{21}^{F} \cdot p_{13}^{D} \cdot p_{23}^{B} + \frac{{p_{21}^{F} \cdot p_{13}^{D} \cdot p_{32}^{B} }}{3} + p_{21}^{F} \cdot p_{31}^{D} \cdot p_{23}^{A} = 0.683 \\ p_{3} & = \frac{{p_{12}^{F} \cdot p_{31}^{E} \cdot p_{23}^{C} }}{3} + p_{12}^{F} \cdot p_{31}^{E} \cdot p_{32}^{C} + \frac{{p_{21}^{F} \cdot p_{13}^{D} \cdot p_{32}^{B} }}{3} + p_{21}^{F} \cdot p_{31}^{D} \cdot p_{32}^{A} = 0.124 \\ \end{aligned} $$

The players’ expected total effort

The expected total effort in the tournament is

$$ {\text{TE}} = {\text{TE}}^{F} + p_{12}^{F} \cdot {\text{TE}}^{E} + p_{21}^{F} \cdot {\text{TE}}^{D} + p_{12}^{F} \cdot p_{31}^{E} \cdot {\text{TE}}^{C} + p_{21}^{F} \cdot p_{13}^{D} \cdot {\text{TE}}^{B} + p_{21}^{F} \cdot p_{31}^{D} \cdot {\text{TE}}^{A} = 0.5 $$

The length of the tournament

The probability that the winner of the tournament will be determined before the last round is

$$ p_{12}^{F} \cdot p_{13}^{E} = 0.119 $$

Appendix E: The winner of the first round competes in the second round of the round-robin tournament with three players and one prize

In the following, we provide in every possible vertex (game) of Figure 5, the players’ mixed strategies, their expected payoffs and their probabilities of winning. These results are summarized in Table 9. The complete analysis can be found in Online-Appendix E.

Table 9 Players’ expected payoffs and winning probabilities in Vertexes A–E of Figure 5

The players’ expected payoffs

$$ \begin{aligned} E(u_{1} ) & = 0 \\ E(u_{2} ) & = 0 \\ E(u_{3} ) & = 0 \\ \end{aligned} $$

The players’ probabilities of winning

Player 1’s probability to win the prize is

$$ p_{1} = p_{12}^{E} \cdot p_{13}^{D} + \frac{{p_{12}^{E} \cdot p_{31}^{D} \cdot p_{23}^{B} }}{3} + \frac{{p_{21}^{E} \cdot p_{32}^{C} \cdot p_{13}^{A} }}{3} = 0.348 $$

Player 2’s probability to win the prize is

$$ p_{2} = \frac{{p_{12}^{E} \cdot p_{31}^{D} \cdot p_{23}^{B} }}{3} + p_{21}^{E} \cdot p_{23}^{C} + \frac{{p_{21}^{E} \cdot p_{32}^{C} \cdot p_{13}^{A} }}{3} = 0.348 $$

And, player 3’s probability to win the prize is

$$ p_{3} = \frac{{p_{12}^{E} \cdot p_{31}^{D} \cdot p_{23}^{B} }}{3} + p_{12}^{E} \cdot p_{31}^{D} \cdot p_{32}^{B} + \frac{{p_{21}^{E} \cdot p_{32}^{C} \cdot p_{13}^{A} }}{3} + p_{21}^{E} \cdot p_{32}^{C} \cdot p_{31}^{A} = 0.303 $$

The players’ expected total effort

The expected total effort in the tournament is

$$ {\text{TE}} = {\text{TE}}^{E} + p_{12}^{E} \cdot {\text{TE}}^{D} + p_{21}^{E} \cdot {\text{TE}}^{C} + p_{12}^{E} \cdot p_{31}^{D} \cdot {\text{TE}}^{B} + p_{21}^{E} \cdot p_{32}^{C} \cdot {\text{TE}}^{A} = 1 $$

The length of the tournament

The probability that the winner of the tournament will be determined before the last round is

$$ p_{12}^{E} \cdot p_{13}^{D} + p_{21}^{E} \cdot p_{23}^{C} = 0.636 $$

Appendix F: The winner of the first round competes in the third round of the round-robin tournament with three players and one prize

In the following, we provide in every possible vertex (game) of Figure 6, the players’ mixed strategies, their expected payoffs and their probabilities of winning. These results are summarized in Table 10. The complete analysis can be found in Online-Appendix F.

Table 10 Players’ expected payoffs and winning probabilities in Vertexes A–G of Figure 6

The players’ expected payoffs

$$ \begin{aligned} E(u_{1} ) & = \frac{1}{12} = 0.083 \\ E(u_{2} ) & = \frac{1}{12} = 0.083 \\ E(u_{3} ) & = 0 \\ \end{aligned} $$

The players’ probabilities of winning

Player 1’s probability to win the prize is

$$ p_{1} = p_{12}^{G} \cdot p_{23}^{F} \cdot p_{13}^{D} + \frac{{p_{12}^{G} \cdot p_{23}^{F} \cdot p_{31}^{D} }}{3} + p_{12}^{G} \cdot p_{32}^{F} \cdot p_{13}^{C} + \frac{{p_{21}^{G} \cdot p_{13}^{E} \cdot p_{32}^{B} }}{3} = 0.458 $$

Player 2’s probability to win the prize is

$$ p_{2} = \frac{{p_{12}^{G} \cdot p_{23}^{F} \cdot p_{31}^{D} }}{3} + p_{21}^{G} \cdot p_{13}^{E} \cdot p_{23}^{B} + \frac{{p_{21}^{G} \cdot p_{13}^{E} \cdot p_{32}^{B} }}{3} + p_{21}^{G} \cdot p_{31}^{E} \cdot p_{23}^{A} = 0.458 $$

And, player 3’s probability to win the prize is

$$ p_{3} = \frac{{p_{12}^{G} \cdot p_{23}^{F} \cdot p_{31}^{D} }}{3} + p_{12}^{G} \cdot p_{32}^{F} \cdot p_{31}^{C} + \frac{{p_{21}^{G} \cdot p_{13}^{E} \cdot p_{32}^{B} }}{3} + p_{21}^{G} \cdot p_{31}^{E} \cdot p_{32}^{A} = 0.083 $$

The players’ expected total effort

The expected total effort in the tournament is

$$ \begin{aligned} {\text{TE}} & = {\text{TE}}^{G} + p_{12}^{G} \cdot {\text{TE}}^{F} + p_{21}^{G} \cdot {\text{TE}}^{E} + p_{12}^{G} \cdot p_{23}^{F} \cdot {\text{TE}}^{D} + p_{12}^{G} \cdot p_{32}^{F} \cdot {\text{TE}}^{C} \\ & \quad + p_{21}^{G} \cdot p_{13}^{E} \cdot {\text{TE}}^{B} + p_{21}^{G} \cdot p_{31}^{E} \cdot {\text{TE}}^{A} = 0.83 \\ \end{aligned} $$

The length of the tournament

The probability that the winner of the tournament will be determined before the last round is equal to zero since there is no possibility that one of the players won twice in the first two rounds.

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Krumer, A., Megidish, R. & Sela, A. The optimal design of round-robin tournaments with three players. J Sched 23, 379–396 (2020). https://doi.org/10.1007/s10951-019-00624-8

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