All-pay 2 $\times$ 2 Hex: a multibattle contest with complementarities

Abstract

We examine a modified 2 $\times$ 2 game of Hex in which the winner of each cell is determined by a Tullock contest. The player establishing a winning path of cells in the game wins a fixed prize. Examining the polar cases of all cells being contested simultaneously versus all four cells being contested sequentially, we show that there is an increase in the total expected payoff for the players in the sequential case. We identify conditions under which players have identical and non-identical expected payoffs when the contest order is pre-specified. We also examine dissipation for random order contests. We thus provide a canonical model of a multibattle contest in which complementarities between battlefields are heterogeneous across both battlefields and players.

Appendices

Appendix 1: Polar cases: simultaneous versus sequential cases

For Proposition 1 we will start by computing the expected payoffs from the simultaneous and sequential cases. For the simultaneous case, we first prove a lemma which shows that in any pure strategy Nash equilibrium, the players will play identical strategies.

Case 1: All four cells simultaneously

Lemma 1

In any pure strategy Nash equilibrium of the simultaneous contest game, both players will play identical strategies.

Proof

Following Klumpp and Polborn (2006; p. 1103), given a set of strategies for players X and Y across all cells other than $j$, there is some probability $p_j \in (0,1)$ that cell $j$ is pivotal for both players. A pivotal cell is a cell such that winning that cell wins the overall contest, while losing the cell loses the overall contest. Recall that $X_j > 0$ and $Y_j > 0$ for each cell $j \in A$ (see footnote 4). From this it follows that the probability of winning any given cell $j$ by either player is $\frac{T_j}{Z_j} > 0$. Hence the probability of any particular division of winning cells between players is non-zero implying the probability that any given cell is pivotal is non-zero. This is true in spite of the complementarities involved in the game of Hex. The fact that the North and South cells can be pivotal for each player is easy to see. East is a pivotal cell for both players when X wins North and Y wins South. West is a pivotal cell if X wins South and Y wins North. As only one player can win the contest, any pivotal cell must be completing a winning path for both players. Clearly, the player who wins the pivotal cell will win the contest.

Given a set of strategies for the other cells generating a $p_j > 0$, player X’s conditional payoff function can be written as $U_X ( \cdot ) = \left( \frac{X_{j}}{Z_{j}}\right) p_j - X_j$, while player Y’s conditional payoff function can be written as $U_Y ( \cdot ) = \left( \frac{Y_{j}}{Z_{j}}\right) p_j - Y_j$. This gives the following first order conditions

$$\begin{aligned} \frac{\partial U_X}{\partial X_j}&= \left( \frac{Y_{j}}{Z_{j}^{2}}\right) p_j - 1 = 0 \\ \frac{\partial U_Y}{\partial Y_j}&= \left( \frac{X_{j}}{Z_{j}^{2}}\right) p_j - 1 = 0 \end{aligned}$$

From this we obtain $\left( \frac{X_{j}}{Z_{j}^{2}}\right) = \left( \frac{Y_{j}}{Z_{j}^{2}}\right)$, and hence $X_j = Y_j$. Since the cell $j$ had been chosen arbitrarily, it follows that in equilibrium players X and Y will play the same strategies, i.e. $X_N = Y_N;\; X_S = Y_S;\, X_E = Y_E$ and $X_W = Y_W$. This completes the proof. $\square$

Computing payoffs for the simultaneous case:

Recall that we can write player X’s utility function as

$$\begin{aligned} U_X ( \cdot )&= \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}}\right) + \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) \left( \frac{X_{E}}{Z_{E}}\right) + \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}}\right) \left( \frac{X_{W}}{Z_{W}}\right) \\&- X_N - X_S - X_E - X_W \end{aligned}$$

Taking the first derivatives gives us

$$\begin{aligned} \frac{\partial U_X}{\partial X_N}&= \left( \frac{Y_{N}}{Z_{N}^{2}}\right) \left( \frac{X_{S}}{Z_{S}}\right) + \left( \frac{Y_{N}}{Z_{N}^{2}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) \left( \frac{X_{E}}{Z_{E}}\right) - \left( \frac{Y_{N}}{Z_{N}^{2}}\right) \left( \frac{X_{S}}{Z_{S}}\right) \left( \frac{X_{W}}{Z_{W}}\right) - 1 \\ \frac{\partial U_X}{\partial X_E}&= \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) \left( \frac{Y_{E}}{Z_{E}^{2}}\right) - 1 \\ \frac{\partial U_X}{\partial X_W}&= \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}}\right) \left( \frac{Y_{W}}{Z_{W}^{2}}\right) - 1 \\ \frac{\partial U_X}{\partial X_S}&= \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}^{2}}\right) - \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}^{2}}\right) \left( \frac{X_{E}}{Z_{E}}\right) + \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}^{2}}\right) \left( \frac{X_{W}}{Z_{W}}\right) - 1 \end{aligned}$$

Recall that we can write player Y’s expected value function as

$$\begin{aligned} U_Y ( \cdot )&= \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) + \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}}\right) \left( \frac{Y_{W}}{Z_{W}}\right) + \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) \left( \frac{Y_{E}}{Z_{E}}\right) \\&- Y_N - Y_S - Y_E -Y_W \end{aligned}$$

whose first derivatives are

$$\begin{aligned} \frac{\partial U_Y}{\partial Y_N}&= \left( \frac{X_{N}}{Z_{N}^{2}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) + \left( \frac{X_{N}}{Z_{N}^{2}}\right) \left( \frac{X_{S}}{Z_{S}}\right) \left( \frac{Y_{W}}{Z_{W}}\right) - \left( \frac{X_{N}}{Z_{N}^{2}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) \left( \frac{Y_{E}}{Z_{E}}\right) - 1 \\ \frac{\partial U_Y}{\partial Y_E}&= \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{Y_{S}}{Z_{S}}\right) \left( \frac{X_{E}}{Z_{E}^{2}}\right) - 1 \\ \frac{\partial U_Y}{\partial Y_W}&= \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}}\right) \left( \frac{X_{W}}{Z_{W}^{2}}\right) - 1 \\ \frac{\partial U_Y}{\partial Y_S}&= \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}^{2}}\right) - \left( \frac{Y_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}^{2}}\right) \left( \frac{Y_{W}}{Z_{W}}\right) + \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{X_{S}}{Z_{S}^{2}}\right) \left( \frac{Y_{E}}{Z_{E}}\right) - 1 \end{aligned}$$

Following Lemma 1, we can substitute $X_j = Y_j$, which makes these two sets of derivatives identical. Hence writing in terms of $X_j$, our first derivatives are

$$\begin{aligned} \frac{\partial U_X}{\partial X_N}&= \frac{X_N}{4 X_N^2} \frac{X_S}{2 X_S} + \frac{X_N}{4 X_N^2} \frac{X_S}{2 X_S} \frac{X_E}{2 X_E} - \frac{X_N}{4 X_N^2} \frac{X_S}{2 X_S} \frac{X_W}{2 X_W} - 1 \\ \frac{\partial U_X}{\partial X_E}&= \frac{X_N}{2 X_N} \frac{X_S}{2 X_S} \frac{X_E}{4 X_E^2} - 1 \\ \frac{\partial U_X}{\partial X_W}&= \frac{X_N}{2 X_N} \frac{X_S}{2 X_S} \frac{X_W}{4 X_W^2} - 1 \\ \frac{\partial U_X}{\partial X_S}&= \frac{X_N}{2 X_N} \frac{X_S}{4 X_S^2} - \frac{X_N}{2 X_N} \frac{X_S}{4 X_S^2} \frac{X_E}{2 X_E} + \frac{X_N}{2 X_N} \frac{X_S}{4 X_S^2} \frac{X_W}{2 X_W} - 1 \end{aligned}$$

This simplifies to

$$\begin{aligned} \frac{1}{4X_N} \frac{1}{2} + \frac{1}{4X_N} \frac{1}{4} - \frac{1}{4X_N} \frac{1}{2} \frac{1}{2} - 1&= 0\\ \frac{1}{2} \frac{1}{2} \frac{1}{4X_E} - 1&= 0 \\ \frac{1}{2} \frac{1}{2} \frac{1}{4X_W} - 1&= 0\\ \frac{1}{2} \frac{1}{4X_S} - \frac{1}{4} \frac{1}{4X_S} + \frac{1}{4} \frac{1}{4X_S} - 1&= 0 \end{aligned}$$

Solving this system of equations gives us $X_N = X_S = \frac{1}{8}$, and $X_E = X_W = \frac{1}{16}$. By Lemma 1, this also gives us $Y_N = Y_S = \frac{1}{8}$, and $Y_E = Y_W = \frac{1}{16}$ in equilibrium.

To determine the nature of this solution, we require a full set of second derivatives, evaluated at $X_N = X_S = Y_N = Y_S = \frac{1}{8}, X_E = X_W = Y_E = Y_W = \frac{1}{16}$. This gives us the following matrix of second derivatives

$$\begin{aligned} \left[ \begin{array}{llll} \frac{\partial ^2 U_X}{\partial X_N^2} &{} \quad \frac{\partial ^2 U_X}{\partial X_N \partial X_E} &{} \quad \frac{\partial ^2 U_X}{\partial X_N \partial X_W} &{} \quad \frac{\partial ^2 U_X}{\partial X_N \partial X_S} \\ \frac{\partial ^2 U_X}{\partial X_E \partial X_N} &{} \quad \frac{\partial ^2 U_X}{\partial X_E^2} &{} \quad \frac{\partial ^2 U_X}{\partial X_E \partial X_W} &{} \quad \frac{\partial ^2 U_X}{\partial X_E \partial X_S} \\ \frac{\partial ^2 U_X}{\partial X_W \partial X_N} &{} \quad \frac{\partial ^2 U_X}{\partial X_W \partial X_E} &{} \quad \frac{\partial ^2 U_X}{\partial X_W^2} &{} \quad \frac{\partial ^2 U_X}{\partial X_W \partial X_S} \\ \frac{\partial ^2 U_X}{\partial X_S \partial X_N} &{} \quad \frac{\partial ^2 U_X}{\partial X_S \partial X_E} &{} \quad \frac{\partial ^2 U_X}{\partial X_S \partial X_W} &{} \quad \frac{\partial ^2 U_X}{\partial X_S^2} \\ \end{array} \right] = \left[ \begin{array}{llll} -8 &{} \quad 4 &{} \quad -4 &{} \quad 0 \\ 4 &{} \quad -16 &{} \quad 0 &{} \quad -4 \\ -4 &{} \quad 0 &{} \quad -16 &{} \quad 4 \\ 0 &{} \quad -4 &{} \quad 4 &{} \quad -8 \\ \end{array} \right] \end{aligned}$$

We need to find the characteristic polynomial of the determinant

$$\begin{aligned} \left| \begin{array}{llll} -8-\lambda &{} \quad 4 &{} \quad -4 &{} \quad 0 \\ 4 &{} \quad -16-\lambda &{} \quad 0 &{} \quad -4 \\ -4 &{} \quad 0 &{} \quad -16-\lambda &{} \quad 4 \\ 0 &{} \quad -4 &{} \quad 4 &{} \quad -8-\lambda \\ \end{array} \right| \end{aligned}$$

This matrix has a characteristic polynomial of $\lambda ^4 + 48 \lambda ^3 + 784 \lambda ^2 + 4864 \lambda + 7936$, which has roots of approximately $(-2.4805, -10.8345, -15.0041, -19.6809)$, and thus is negative definite. As the Hessian matrix is negative definite, the solution to first order equations must be a local maximum, and as it is the only equilibrium, this local maximum is the only possible interior solution.

In order to find a global maximum, we now have to consider the boundary cases. Negative bids are not allowed to begin with, and clearly placing bids greater than the value of the prize cannot result in positive values, so we will restrict all possible solutions to the $[0,1]^4$ hypercube. We will now consider boundary conditions.

First note that a bid of 1 placed on more than one cell will yield negative payoffs. Hence a bid of 1 must be restricted to only one cell and on this boundary point all other bids must be zero for non-negative profits. However this will never lead to a winning path and hence can also be ruled out.

We can eliminate boundary solutions involving bids of zero, because if there were a solution with zero bids, it would be profitable for a player to deviate from this solution by instead investing an arbitrarily small amount in place of those zero bids. This would ensure winning those areas, thus increasing the probability of overall victory. Increasing the probability of victory for an arbitrarily small increased expenditure would increase the deviating player’s expected value. Because a player could gain by deviating, the boundary cases involving zero bids cannot be equilibrium solutions. Thus we have only one possible pure strategy equilibrium solution, which we need now verify to be an equilibrium solution.

In order to do so, we must also check the possibility that one player could unilaterally deviate from the candidate interior solution, and thus gain a greater expected payoff. Clearly, they cannot deviate to placing three or four zero bids, as this would eliminate the possibility of forming a winning path. Thus, we will only need to examine deviations involving one or two zero bids.

A single zero bid cannot be on North or South. By conceding either North or South, the player will have no benefit from either East or West, and thus will place a bid of zero on that cell as well. Thus, the player making a single zero bid must do so on either East or West. We will take the case of Player X placing a zero bid on East without loss of generality.

Given that $X_E = 0$, Player X will have an expected payoff of

$$\begin{aligned} U_X \left( \cdot \right)&= \left( \frac{X_{N}}{\frac{1}{8} + X_{N}}\right) \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right) + \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right) \left( \frac{X_{W}}{\frac{1}{16} + X_{W}}\right) \\&- \left( \frac{X_{N}}{\frac{1}{8} + X_{N}}\right) \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right) \left( \frac{X_{W}}{\frac{1}{16} + X_{W}}\right) - X_N - X_S - X_W \end{aligned}$$

This yields first order conditions

$$\begin{aligned}&\left( \frac{\frac{1}{8}}{(\frac{1}{8} + X_{N})^{2}}\right) \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right) \left( \frac{\frac{1}{16}}{\frac{1}{16} + X_{W}}\right) = 1 \\&\left( \frac{\frac{1}{8}}{\frac{1}{8} + X_{N}}\right) \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right) \left( \frac{\frac{1}{16}}{(\frac{1}{16} + X_{W})^{2}}\right) = 1 \\&\left( \frac{X_{N}}{\frac{1}{8} + X_{N}}\right) \left( \frac{\frac{1}{8}}{(\frac{1}{8} + X_{S})^{2}}\right) \left( \frac{X_{W}}{\frac{1}{16} + X_{W}}\right) + \left( \frac{\frac{1}{8}}{\frac{1}{8} + X_{N}}\right) \left( \frac{\frac{1}{8}}{(\frac{1}{8} + X_{S})^{2}}\right) \\&\qquad \times \left( \frac{X_{W}}{\frac{1}{16} + X_{W}}\right) + \left( \frac{X_{N}}{\frac{1}{8} + X_{N}}\right) \left( \frac{\frac{1}{8}}{(\frac{1}{8} + X_{S})^{2}}\right) \left( \frac{\frac{1}{16}}{\frac{1}{16} + X_{W}}\right) = 1 \end{aligned}$$

The first two conditions yield that $\frac{1}{8} + X_N = \frac{1}{16} + X_W$, however further symbolic substitution does not give useful results. Thus, we must use numeric approximation to find the best deviation. This gives us an optimal solution of $X_N \approx 0.0407$, $X_S \approx 0.1740$, $X_W \approx 0.1032$, giving Player X an expected value of approximately 0.0985. This is worse than the expected value of 0.125 from the symmetric bids, and thus no player will unilaterally deviate to a single zero bid.

There are two distinct possibilities for two zero bid cases. One possibility is to place positive bids on a winning path for the deviating player alone, while the other is to place positive bids on North and South (positive bids on East and West do not form a winning set, and thus cannot have positive expected value).

Without loss of generality, we will consider the case of Player X placing positive bids on the South and West cells for the case of having positive bids on a winning path for only the deviating player. This leaves an expected payoff of

$$\begin{aligned} U_X \left( \cdot \right) = \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right) \left( \frac{X_{W}}{\frac{1}{16} + X_{W}}\right) - X_S - X_W \end{aligned}$$

which yields first order conditions of

$$\begin{aligned} \left( \frac{\frac{1}{8}}{(\frac{1}{8} + X_{S})^{2}}\right) \left( \frac{X_{W}}{\frac{1}{16} + X_{W}}\right)&= 1\\ \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right) \left( \frac{\frac{1}{16}}{(\frac{1}{16} + X_{W})^{2}}\right)&= 1 \end{aligned}$$

Again, symbolic manipulation is insufficient in this case, and we instead rely on numeric approximation to find the optimal deviation. At $X_S \approx 0.1640,\, X_W \approx 0.1258$, we have an approximate expected value of 0.0893, worse than the 0.125 from symmetric bids.

When the positive bids are place on North and South, we have an expected payoff of

$$\begin{aligned} U_X \left( \cdot \right) = \left( \frac{X_{N}}{\frac{1}{8} + X_{N}}\right) \left( \frac{X_{N}}{\frac{1}{8} + X_{N}}\right) - X_N - X_S \end{aligned}$$

which yields first order conditions

$$\begin{aligned} \left( \frac{\frac{1}{8}}{(\frac{1}{8} + X_{N})^{2}}\right) \left( \frac{X_{S}}{\frac{1}{8} + X_{S}}\right)&= 1\\ \left( \frac{X_{N}}{\frac{1}{8} + X_{N}}\right) \left( \frac{\frac{1}{8}}{(\frac{1}{8} + X_{S})^{2}}\right)&= 1 \end{aligned}$$

Combining these, we see that $X_N = X_S$, and thus that $\frac{1}{8} X_N = (\frac{1}{8} + X_N)^3$. This has two solutions over the unit interval, with $X_N = \frac{1}{8}$ yielding an expected value of zero, and $X_N = \frac{1}{8} (\sqrt{5} - 2)$ yielding a negative expected value. Thus, no deviation is beneficial, and so we have an equilibrium at $X_N = X_S = \frac{1}{8}$, $X_E = X_W = \frac{1}{16}$.

Case 2: Four cells sequentially

We now will solve the sequential cases. Combining these with the simultaneous cases gives us Proposition 1, while comparing the sequential cases proves Proposition 2. In order to obtain these results, we must work backwards from the terminal nodes. If only one cell remains to be contested, either the cell is irrelevant as we already have a winner, or the winner of this one cell will win the contest. Let the remaining cell be East without loss of generality, we have $U_X \left( \cdot |R_1, R_2, R_3,R_4 \right) = \left( \frac{X_{E}}{Z_{E}}\right) - X_E$, and $U_Y \left( \cdot |R_1, R_2, R_3,R_4 \right) = \left( \frac{Y_{E}}{Z_{E}}\right) - Y_E$. Taking derivatives gives us $\left( \frac{X_{E}}{Z_{E}^{2}}\right) = 1, \left( \frac{Y_{E}}{Z_{E}^{2}}\right) = 1$, so $X_E = Y_E = \frac{1}{4}$, giving $U_X \left( \cdot | \cdot ,R_4 \right) = U_Y \left( \cdot | \cdot ,R_4 \right) = \frac{1}{4}$.

Now we can work backwards to the previous stage. If there are two cells remaining, there are three possibilities. If the contest has already been won, they are both irrelevant, and we are done. However, if only one of the two remaining cells is relevant, we ignore the irrelevant cell. For simplicity assume that that irrelevant cell appears in Round 3. Then the fourth round will play out as described above for East and this reduces to the above. Finally, there is the possibility that both are relevant, with one player needing to win both and the other needing to win one of the two. We will assume without loss of generality that North and East remain to be contested sequentially, with Player X requiring both to win. Then $U_X \left( \cdot | \cdot ,R_3,R_4 \right) = \left( \frac{X_{N}}{Z_{N}}\right) \frac{1}{4} - X_N$, $U_Y \left( \cdot | \cdot ,R_3,R_4 \right) = \left( \frac{Y_{N}}{Z_{N}}\right) + \left( \frac{X_{N}}{Z_{N}}\right) \frac{1}{4} - Y_N$ gives the expected utility for each player in the North round. Taking derivatives gives us

$$\begin{aligned} \left( \frac{Y_{N}}{Z_{N}^{2}}\right) \frac{1}{4}&= 1 \\ \left( \frac{X_{N}}{Z_{N}^{2}}\right) - \left( \frac{X_{N}}{Z_{N}^{2}}\right) \frac{1}{4}&= \left( \frac{X_{N}}{Z_{N}^{2}}\right) \frac{3}{4} = 1 \end{aligned}$$

Solving this gives $X_N = \frac{3}{64}$, while $Y_N = \frac{9}{64}$, and thus expected payoffs $U_X \left( \cdot | \cdot ,R_3,R_4 \right) = \frac{1}{64}$, $U_Y \left( \cdot | \cdot ,R_3,R_4 \right) = \frac{43}{64}$. These two solutions will be used extensively in the subcases below. This is as far backwards as we can work without having the specifics of the complementarity, so we now must break into subcases.

Subcase 2a: North or South in $R_1$

After the first round, one of the remaining cells will become irrelevant. The winner of the first round will need to win one of the remaining two relevant cells, which is a subgame with an expected payoff of $\frac{43}{64}$, and the loser of the first round will need to win both remaining relevant cells, a subgame with an expected value of $\frac{1}{64}$, as shown above. Thus, if North is the first cell, we have $U_X = \left( \frac{X_{N}}{Z_{N}}\right) \frac{43}{64} + \left( \frac{Y_{N}}{Z_{N}}\right) \frac{1}{64} - X_N$. Taking the derivative with respect to $X_N$ gives $\left( \frac{X_{N}}{Z_{N}^{2}}\right) \frac{43}{64} - \left( \frac{X_{N}}{Z_{N}^{2}}\right) \frac{1}{64} = 1$. From Lemma 1, $X_N = Y_N$, so solving gives us $X_N = Y_N = \frac{21}{128}$ and $U_X = U_Y = \frac{23}{128}$.

Subcase 2b: East and West in $R_1$ and $R_2$

Without loss of generality, we consider the cases where East is the first round, results for West as opening round are symmetric. If player X wins the East, winning the West means he will need either the North or South, while losing the West means that the South is irrelevant, and the North will determine the overall winner. These have expected payoffs of $\frac{43}{64}$ and $\frac{1}{4}$ respectively for player X and $\frac{1}{64}$ and $\frac{1}{4}$ for player Y. Thus, if we consider the subgames in which player X wins the East, the expected payoff functions for the second round are

$$\begin{aligned} U_X \left( \cdot | \cdot ,R_2,R_3,R_4 \right)&= \left( \frac{X_{W}}{Z_{W}}\right) \frac{43}{64} + \left( \frac{Y_{W}}{Z_{W}}\right) \frac{1}{4} - X_W \\ U_Y \left( \cdot | \cdot ,R_2,R_3,R_4 \right)&= \left( \frac{Y_{W}}{Z_{W}}\right) \frac{1}{4} + \left( \frac{X_{W}}{Z_{W}}\right) \frac{1}{64} - Y_W \\ \end{aligned}$$

This gives us first order conditions of

$$\begin{aligned} \left( \frac{Y_{W}}{Z_{W}^{2}}\right) \frac{27}{64} = \left( \frac{X_{W}}{Z_{W}^{2}}\right) \frac{15}{64} = 1 \end{aligned}$$

and so $15 X_W = 27 Y_W$. Thus $X_W = \frac{27}{15} Y_W$, which when substituted into the first order conditions gives us $X_W = \frac{(15)(27^2)}{(42^2)(64)}$, $Y_W = \frac{(15^2)(27)}{(42^2)(64)}$. Using these payoffs in the expected payoff functions gives $U_X = \frac{47907}{112896}$ and $U_Y = \frac{5139}{112896}$. These payoffs will be reversed if player Y wins the East.

Thus, in the initial round, the expected payoff function for player X is $\left( \frac{X_{E}}{Z_{E}}\right) \frac{47907}{112896} + \left( \frac{Y_{E}}{Z_{E}}\right) \frac{5139}{112896} - X_E$, giving a first order condition of $\left( \frac{Y_{E}}{Z_{E}^{2}}\right) \frac{42768}{112896} = 1$, with $X_E = Y_E$ due to symmetry. Thus, the optimal strategy is $X_E = Y_E = \frac{42768}{451584}$, which yields expected payoffs of $\frac{63486}{451584}$.

Subcase 2c: East or West in $R_1$ and North or South in $R_2$

Without loss of generality, we consider the subcases where East is contested in the first round. If player X wins the East, winning the North means victory, while losing North means she must win both West and South, for an expected payoff of $\frac{1}{64}$ for X and $\frac{43}{64}$ for Y. Thus the expected payoff functions in the second round in the subgame where player X won the East are

$$\begin{aligned} U_X \left( \cdot | \cdot ,R_2,R_3,R_4 \right)&= \left( \frac{X_{N}}{Z_{N}}\right) + \left( \frac{Y_{N}}{Z_{N}}\right) \frac{1}{64} - X_N \\ U_Y \left( \cdot | \cdot ,R_2,R_3,R_4 \right)&= \left( \frac{Y_{N}}{Z_{N}}\right) \frac{43}{64} - Y_N \end{aligned}$$

This gives the first order conditions of

$$\begin{aligned} \left( \frac{Y_{N}}{Z_{N}^{2}}\right) \frac{63}{64} = \left( \frac{X_{N}}{Z_{N}^{2}}\right) \frac{43}{64} = 1 \end{aligned}$$

Thus we have $43 X_N = 63 Y_N$, which with our first order condition gives $X_N = \frac{(43)(63^2)}{(64)(106^2)}$ and $Y_N = \frac{(43^2)(63)}{(64)(106^2)}$. Plugging these into the expected payoff equations give approximations of $U_X \approx 0.2373,\, U_Y \approx 0.1106$.

If player Y wins the East, winning either the South or North and West wins. This is the same as in the E–W–N–S case, which gives decimal approximations of $U_X \approx 0.0455$ and $U_Y \approx 0.4245$.

Thus for the first round, we have expected payoff functions of

$$\begin{aligned} U_X&= \left( \frac{X_{E}}{Z_{E}}\right) .2373 + \left( \frac{Y_{E}}{Z_{E}}\right) .0456 - X_E \\ U_Y&= \left( \frac{Y_{E}}{Z_{E}}\right) .4245 + \left( \frac{X_{E}}{Z_{E}}\right) .1106 - Y_E \end{aligned}$$

These yield first order conditions of

$$\begin{aligned} \left( \frac{Y_{E}}{Z_{E}^{2}}\right) (.2373 - .0455)&= \left( \frac{X_{E}}{Z_{E}^{2}}\right) (.4245 - .1106) = 1 \\ \left( \frac{Y_{E}}{Z_{E}^{2}}\right) .1918&= \left( \frac{X_{E}}{Z_{E}^{2}}\right) .3139 = 1 \\ Y_E .1918&= X_E .3139 \\ Y_E&\approx X_E 1.6365 \end{aligned}$$

Thus we find $X_E \approx 0.0452,\, Y_E \approx 0.0739$. Placing these in our expected payoff function gives $U_X \approx 0.0731,\, U_Y \approx 0.2315$.

Combining cases 1 and 2 gives us Proposition 1, while comparing the expected payoffs for each player across case 2 gives us Proposition 2. $\square$

Appendix 2: The other sequential contests

To explain the other types of sequential contests here we will go through the details of one case. The other cases are similar and details of these cases can be found in the working paper version.¹²

We will provide an analysis of the Hex contest involving two rounds where two cells are contested in each round. This case will require a solution to the expected payoffs of each player if only 2 cells remain, they are being contested simultaneously, and neither cell has been rendered irrelevant by the first round. In such a case, one player requires winning both cells for victory, while the other player requires winning either cell for victory. To illustrate, we will take take the subgame where player X needs both the North and East cells. In this subgame, player X’s expected payoff is $U_X(\cdot | \cdot , R_2) = \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{X_{E}}{Z_{E}}\right) - X_{N} - X_{E}$. As player Y wins this subgame as long as player X does not win both the North and East, which occurs with probability $1 - \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{X_{E}}{Z_{E}}\right)$, player Y has an expected payoff of $U_Y(\cdot | \cdot , R_2) = (1 - \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{X_{E}}{Z_{E}}\right) ) - Y_{N} - Y_{E}$. Taking derivatives, we obtain the following set of first order conditions

$$\begin{aligned} \frac{\partial U_X}{\partial X_N}&= \left( \frac{Y_{N}}{Z_{N}^{2}}\right) \left( \frac{X_{E}}{Z_{E}}\right) - 1 = 0 \\ \frac{\partial U_X}{\partial X_E}&= \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{Y_{E}}{Z_{E}^{2}}\right) - 1 = 0 \\ \frac{\partial U_Y}{\partial Y_N}&= \left( \frac{X_{N}}{Z_{N}^{2}}\right) \left( \frac{X_{E}}{Z_{E}}\right) - 1 = 0 \\ \frac{\partial U_Y}{\partial Y_E}&= \left( \frac{X_{N}}{Z_{N}}\right) \left( \frac{X_{E}}{Z_{E}^{2}}\right) - 1 = 0 \end{aligned}$$

These may be solved simultaneously to obtain $X_{N} = Y_{N} = X_{E} = Y_{E} = \frac{1}{8}$, so player X has an expected payoff of $0$, while player Y has an expected payoff of $\frac{1}{2}$. Thus, if both cells in the two cell subgame are relevant, the player who needs to win both has an expected payoff of $0$, while the player that needs to win at least one of two cells has an expected payoff of $\frac{1}{2}$. If only one of the two cells is relevant, both players have an expected value in the subgame of $\frac{1}{4}$, as seen in Appendix 1.

As we now have the expected values for the players in the final round, we can now solve for the first round. However, we will need to break this case into three subcases, those where East and West comprise the first round, where North and South comprise the first round, and all other combinations of two cells comprising the first round. The results for the other cases are summarized in the final table.

Case 3: East and west as the first round If player X wins both the East and West in the initial round, she will complete a winning set with either North or South, and thus X will have an expected payoff of $\frac{1}{2}$, while Y will have an expected payoff of 0 in the ensuing second round. If players X and Y split East and West, only one of North and South will matter, and thus each player has an expected payoff of $\frac{1}{4}$ in the second round. Thus, player X has an overall expected payoff of:

$$\begin{aligned} \left( \frac{X_{E}}{Z_{E}}\right) \left( \frac{X_{W}}{Z_{W}}\right) \frac{1}{2} + \left( \frac{X_{E}}{Z_{E}}\right) \left( \frac{Y_{W}}{Z_{W}}\right) \frac{1}{4} + \left( \frac{Y_{E}}{Z_{E}}\right) \left( \frac{X_{W}}{Z_{W}}\right) \frac{1}{4} - X_E - X_W \end{aligned}$$

As East and West are symmetric, as well as players X and Y, by logic similar to Lemma 1, we know that $X_E = X_W = Y_E = Y_W$. Thus, taking the derivative with respect to $X_E$ gives us the following

$$\begin{aligned}&\left( \frac{Y_{E}}{Z_{E}^{2}}\right) \left( \frac{X_{W}}{Z_{W}}\right) \frac{1}{2} + \left( \frac{Y_{E}}{Z_{E}^{2}}\right) \left( \frac{Y_{W}}{Z_{W}}\right) \frac{1}{4} - \left( \frac{Y_{E}}{Z_{E}^{2}}\right) \left( \frac{X_{W}}{Z_{W}}\right) \frac{1}{4} - 1\\&\quad = \left( \frac{Y_{E}}{Z_{E}^{2}}\right) \left( \frac{Y_{E}}{Z_{E}}\right) \frac{1}{2} - 1 = \left( \frac{Y_{E}}{Z_{E}^{2}}\right) \frac{1}{4} - 1 = \frac{1}{4 X_E} \frac{1}{4} - 1 \end{aligned}$$

Setting this equal to zero gives us a solution $X_E = X_W = Y_E = Y_W = \frac{1}{16}$, and using this yields expected payoffs for each player of $\frac{1}{8}$.