The Art of Concession in General Lotto Games

Abstract

Success in adversarial environments often requires investment into additional resources in order to improve one’s competitive position. But, can intentionally decreasing one’s own competitiveness ever provide strategic benefits in such settings? In this paper, we focus on characterizing the role of “concessions” as a component of strategic decision making. Specifically, we investigate whether a player can gain an advantage by either conceding budgetary resources or conceding valuable prizes to an opponent. While one might naïvely assume that the player cannot, our work demonstrates that – perhaps surprisingly – concessions do offer strategic benefits when made correctly. In the context of General Lotto games, we first show that neither budgetary concessions nor value concessions can be advantageous to either player in a 1-vs.-1 scenario. However, in settings where two players compete against a common adversary, we find opportunities for one of the two players to improve her payoff by conceding a prize to the adversary. We provide a set of sufficient conditions under which such concessions exist.

This work is supported by UCOP Grant LFR-18-548175, ONR grant #N00014-20-1-2359, AFOSR grants #FA9550-20-1-0054 and #FA9550-21-1-0203, and the Army Research Lab through the ARL DCIST CRA #W911NF-17-2-0181.

Appendices

A Proof of Theorem 1

Proof

The proof amounts to showing that player B’s payoff is nonincreasing for any budgetary concession $x\in (0,X_B]$ such that $(X_B-x,X_C)$ is in any of the regions $\mathcal {R}_{j}$.

We first consider the scenario where $(X_B,X_C)\in \mathcal {R}_{1C}(\varPhi _B,\varPhi _C)$. Recall that, in this scenario, player A commits no budget to the battlefields in the front $\mathcal {B}_B$. Thus, player B’s payoff before the concession is $\varPhi _B$, the highest possible payoff. Furthermore, $(X_B-x,X_C)\in \mathcal {R}_{1C}(\varPhi _B,\varPhi _C)$ can only hold if $(X_B,X_C)\in \mathcal {R}_{1C}(\varPhi _B,\varPhi _C)$ as well, since the value $1-\sqrt{\varPhi _C(X_B-x)X_C/\varPhi _B}$ is increasing in x. If $(X_B-x,X_C)\in \mathcal {R}_{4}(\varPhi _B,\varPhi _C)$, then any budgetary concession $x'<x$ would be in either $\mathcal {R}_{1C}$ or $\mathcal {R}_{2C}$, since $X_B+X_C\ge 1$ in $\mathcal {R}_{4}$, while any budgetary concession $x'>x$ would be in either $\mathcal {R}_{1B}$, $\mathcal {R}_{2B}$ or $\mathcal {R}_{3}$. Thus, conceding any amount $x'<x$ would guarantee B greater payoff since $X_{A,B}=1-X_{A,C}$, and $X_{A,C}=1$ in $\mathcal {R}_{1C}$ and $X_{A,C}>X_C$ in $\mathcal {R}_{2C}$ whereas $X_{A,C}\in [\max \{0,1-X_B\},\min \{1,X_C\}]$ in $\mathcal {R}_{4}$. Further, conceding any amount $x'>x$ cannot guarantee B greater payoff since $X_{A,B}=1$ in $\mathcal {R}_{1B}$, and $X_{A,B}>X_B$ in $\mathcal {R}_{2B}$ and $\mathcal {R}_{3}$, whereas $X_{A,B}\in [\max \{0,1-X_C\},\min \{1,X_B\}]$ in $\mathcal {R}_{4}$. In all other regions, we show that player B’s payoff is strictly decreasing in x by checking the partial derivative with respect to $x\ge 0$:

If $(X_B-x,X_C)\in \mathcal {R}_{1B}(\varPhi _B,\varPhi _C)$ and $X_B-x>1$, then

$$ \frac{\partial }{\partial x} \varPhi _B\left[ 1-\frac{1}{2(X_B-x)}\right] = -\frac{\varPhi _B}{2(X_B-x)^2} < 0.$$

Else, if $(X_B-x,X_C)\in \mathcal {R}_{1B}(\varPhi _B,\varPhi _C)$ and $X_B-x\le 1$, then

$$\begin{aligned} \frac{\partial }{\partial x} \frac{\varPhi _B(X_B-x)}{2} = -\frac{\varPhi _B}{2} < 0. \end{aligned}$$

If $(X_B-x,X_C)\in \mathcal {R}_{2B}(\varPhi _B,\varPhi _C)$, then

$$ \frac{\partial }{\partial x} \frac{\varPhi _B(X_B-x)}{2\sqrt{\frac{\varPhi _B(X_B-x)X_C}{\varPhi _C}}} = -\frac{\varPhi _B}{4\sqrt{\frac{\varPhi _B(X_B-x)X_C}{\varPhi _C}}} < 0.$$

If $(X_B-x,X_C)\in \mathcal {R}_{2C}(\varPhi _B,\varPhi _C)$, then

$$ \frac{\partial }{\partial x} \varPhi _B\left[ 1-\frac{1-\sqrt{\frac{\varPhi _C(X_B-x)X_C}{\varPhi _B}}}{2(X_B-x)}\right] = -\frac{\varPhi _B\left[ 2-\sqrt{\frac{\varPhi _C(X_B-x)X_C}{\varPhi _B}}\right] }{4(X_B-x)^2} < 0, $$

which is strictly negative as the condition $1-\sqrt{\varPhi _C(X_B-x)X_C/\varPhi _B}\ge 0$ must hold in $\mathcal {R}_{2C}$. Finally, if $(X_B-x,X_C)\in \mathcal {R}_{3}(\varPhi _B,\varPhi _C)$, then

$$ \frac{\partial }{\partial x} \frac{\varPhi _B(X_B-x)}{2\frac{\sqrt{\varPhi _B(X_B-x)}}{\sqrt{\varPhi _B(X_B-x)}+\sqrt{\varPhi _CX_C}}} = -\frac{\varPhi _B}{2} -\frac{\varPhi _B\sqrt{\varPhi _CX_C}}{4\sqrt{\varPhi _B(X_B-x)}} < 0.$$

This concludes the proof.    $\square$

B Proof of Theorem 2

Before presenting the proof, we note that, in the case of battlefield concessions, we can disregard the scenario when $(X_B,X_C)\in \mathcal {R}_{4}(\varPhi _B-v,\varPhi _C)$, for any $v\in [0,\varPhi _B]$. To see why, consider a battlefield concession of value v such that $(X_B,X_C)$ is in $\mathcal {R}_{4}$, i.e., $X_B+X_C\ge 1$ and $(\varPhi _B-v)/X_B=\varPhi _C/X_C$. Observe that by conceding a battlefield of value slightly greater than v (i.e., $v'=v+\epsilon$ for $\epsilon \rightarrow 0^+$), player B obtains strictly higher payoff as $(X_B,X_C)$ now falls in region $\mathcal {R}_{1C}$ (if $X_C\ge 1$) or region $\mathcal {R}_{2C}$ (if $X_C<1$). Thus, in the following proof, we assume that any point $(X_B,X_C)$ with $X_B+X_C\ge 1$ will transit directly from $\mathcal {R}_{1B}(\varPhi _B-v,\varPhi _C)$ (if $X_B\ge 1$) or $\mathcal {R}_{2B}(\varPhi _B-v,\varPhi _C)$ (if $X_B<1$), to $\mathcal {R}_{1C}(\varPhi _B-v,\varPhi _C)$ (if $X_C\ge 1$) or $\mathcal {R}_{2C}(\varPhi _B-v,\varPhi _C)$ (if $X_C<1$), as v is increased, without first passing through $\mathcal {R}_{4}$.

Proof

The proof amounts to verifying that the conditions laid out in the claim guarantee that player B’s payoff after the battlefield concession is greater than her payoff in the nominal three-player General Lotto game. Before we continue, it is critical to note that $v^*$ as defined in the claim is precisely the value that satisfies $(\varPhi _B-v^*)/X_B=\varPhi _C/X_C$. Thus, for $(X_B,X_C)\in \mathcal {R}_{1B}\cup \mathcal {R}_{2B}$ and $X_B+X_C\ge 1$, the battlefield concession of value $v^*$ satisfies $(X_B,X_C)\in \mathcal {R}_{1C}\cup \mathcal {R}_{2C}$. We present the remainder of the proof in parts corresponding with each of the conditions in the claim.

Conditions (i): The point $(X_B,X_C)$ is nominally in the region $\mathcal {R}_{1B}(\varPhi _B,\varPhi _C)$ with $X_B\ge 1$, and, thus, player B’s nominal payoff is $\varPhi _B\cdot \left( 1-1/2X_B\right)$. Since $X_C\ge 1$, the battlefield concession of value $v^*$ satisfies $(X_B,X_C)\in \mathcal {R}_{1C}(\varPhi _B-v^*,\varPhi _C)$, and player B’s resulting payoff is $\varPhi _B-v^*$. It follows that the battlefield concession of value $v^*$ benefits player B if

$$\begin{aligned} \varPhi _B-v^* > \varPhi _B\cdot \left( 1-\frac{1}{2X_B}\right) . \end{aligned}$$

Rearranging the above inequality gives the condition in the claim.

Condition (ii): Once again, player B’s nominal payoff is $\varPhi _B\cdot \left( 1-1/2X_B\right)$. Since $X_C<1$ and $X_B+X_C\ge 1$, the battlefield concession of value $v^*$ satisfies $(X_B,X_C)\in \mathcal {R}_{2C}(\varPhi _B-v^*,\varPhi _C)$, and player B’s resulting payoff is $(\varPhi _B-v^*)\cdot [1-(1-\sqrt{\varPhi _CX_BX_C/(\varPhi _B-v^*)})/(2X_B)]$. It follows that the battlefield concession of value $v^*$ benefits player B if

$$\begin{aligned} (\varPhi _B-v^*)\cdot \left[ 1-\frac{1-\sqrt{\varPhi _CX_BX_C/(\varPhi _B-v^*)}}{2X_B}\right] > \varPhi _B\cdot \left( 1-\frac{1}{2X_B}\right) . \end{aligned}$$

Condition (iii): Observe that this condition resembles Condition (i), except that $X_B<1$. Thus, the only difference is that player B’s nominal payoff is $\varPhi _BX_B/2$. It follows that the battlefield concession of value $v^*$ benefits player B if

$$\begin{aligned} \varPhi _B-v^* > \frac{\varPhi _BX_B}{2}. \end{aligned}$$

Rearranging the above inequality gives the condition in the claim.

Condition (iv): Observe that this condition resembles Condition (iii), except that $X_C<1$. Thus, the only difference is that player B’s resulting payoff is $(\varPhi _B-v^*)\cdot [1-(1-\sqrt{\varPhi _CX_BX_C/(\varPhi _B-v^*)})/(2X_B)]$, as in Condition (ii). It follows that the battlefield concession of value $v^*$ benefits player B if

$$\begin{aligned} (\varPhi _B-v^*)\cdot \left[ 1-\frac{1-\sqrt{\varPhi _CX_BX_C/(\varPhi _B-v^*)}}{2X_B}\right] > \frac{\varPhi _BX_B}{2\sqrt{\varPhi _BX_BX_C/\varPhi _C}}. \end{aligned}$$

Condition (v): The point $(X_B,X_C)$ is nominally in the region $\mathcal {R}_{2B}(\varPhi _B,\varPhi _C)$, and, thus, player B’s nominal payoff is $\varPhi _B\cdot X_B/(2\sqrt{\varPhi _BX_BX_C/\varPhi _C})$ since $X_B<1$ must hold in $\mathcal {R}_{2B}$. Since $X_C\ge 1$, player B’s resulting payoff after the battlefield concession of value $v^*$ is $\varPhi _B-v^*$, as in Condition (i). It follows that the battlefield concession of value $v^*$ benefits player B if

$$\begin{aligned} \varPhi _B-v^* > \frac{\varPhi _BX_B}{2\sqrt{\varPhi _BX_BX_C/\varPhi _C}}. \end{aligned}$$

Rearranging the above inequality gives the condition in the claim.

Condition (vi): This condition resembles Condition (iv), except that $X_C<1$. Thus, the only difference is that player B’s resulting payoff is $(\varPhi _B-v^*)\cdot [1-(1-\sqrt{\varPhi _CX_BX_C/(\varPhi _B-v^*)})/(2X_B)]$, as in Condition (ii). It follows that the battlefield concession of value $v^*$ benefits player B if

$$\begin{aligned} (\varPhi _B-v^*)\cdot \left[ 1-\frac{1-\sqrt{\varPhi _CX_BX_C/(\varPhi _B-v^*)}}{2X_B}\right] > \frac{\varPhi _BX_B}{2\sqrt{\varPhi _BX_BX_C/\varPhi _C}}. \end{aligned}$$

This concludes the proof.    $\square$