Waging simple wars: a complete characterization of two-battlefield Blotto equilibria

Abstract

We analyze the strategic allocation of resources across two contests as in the canonical Colonel Blotto game. In the games we study, two players simultaneously allocate their forces across two fields of battle. The larger force on each battlefield wins that battle, and the payoff to a player is the sum of the values of battlefields won. We completely characterize the set of Nash equilibria of all two-battlefield Blotto games and provide the unique equilibrium payoffs. We also show how to extend our characterization to cover previously unstudied games with nonlinear resource constraints.

Appendices

Appendix 1: Proofs of Theorems 1–3

First, we prove that pairs of strategies from $\Omega ^{B}$ and $\Omega ^{E}$ provide the expected payoffs from Theorem 1. Then we prove that all pairs of strategies from $\Omega ^{B}$ and $\Omega ^{E}$ constitute a Nash equilibrium. Finally, we show that no other strategies are a part of any Nash equilibrium.

First, define two projection operators. $\Psi _{1}(S)$ is the set of all scalars that are the first dimension of some two dimensional point in the set $S$, the set of Battlefield 1 allocations given a set $S$ of two dimensional battlefield allocations.

$$\begin{aligned} \Psi _{1}(S)\equiv \{x_{1}:\;(\exists x_{2}\in \mathbb {R}\; \hbox {s.t.}\;(x_{1},x_{2})\in S)\} \end{aligned}$$

$\Psi _{2}(S)$ is defined similarly:²⁰

$$\begin{aligned} \Psi _{2}(S)\equiv \{x_{2}:\;(\exists x_{1}\in \mathbb {R}\; \hbox {s.t.}\;(x_{1},x_{2})\in S)\} \end{aligned}$$

We also define the set of all points in some $T_{i}^{b}$ ($T_{i}^{e}$):

$$\begin{aligned} T^{b}\equiv T_{1}^{b}\cup T_{2}^{b}\cup \ldots \cup T_{n}^{b}\\ T^{e}\equiv T_{1}^{e}\cup T_{2}^{e}\cup \ldots \cup T_{n}^{e} \end{aligned}$$

Recall the definitions of $h$ and $p$ from Eqs. 12 and 13. Lemma 5 gives us the intervals of battlefield allocations within the various $T_{i}$’s.

Lemma 5

$\forall i=1,2,\ldots ,n\!:$

$$\begin{aligned} \Psi _{1}(T_{i}^{b})&= \left[ h^{n-i}(E_{1}),f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) \right] ,\end{aligned}$$

(26)

$$\begin{aligned} \Psi _{2}(T_{i}^{b})&= \left[ p^{i-1}(E_{2}),f\left( h^{n-i}\left( E_{1}\right) \right) \right] . \end{aligned}$$

(27)

$\forall i=2,3,\ldots ,n-1\!:$

$$\begin{aligned}&\displaystyle \Psi _{1}(T_{i}^{e})=\left( f^{-1}\left( p^{i-2}\left( E_{2}\right) \right) ,h^{n-i}(E_{1})\right) ,\end{aligned}$$

(28)

$$\begin{aligned}&\displaystyle \Psi _{2}(T_{i}^{e})=\left( f\left( h^{n-i-1}\left( E_{1}\right) \right) ,p^{i-1}(E_{2})\right) , \end{aligned}$$

(29)

For $i=1$ we have the following:

$$\begin{aligned} \Psi _{1}(T_{1}^{e})&= [0,h^{n-1}(E_{1})),\end{aligned}$$

(30)

$$\begin{aligned} \Psi _{2}(T_{1}^{e})&= \left( f\left( h^{n-2}\left( E_{1}\right) \right) ,E_{2}\right] , \end{aligned}$$

(31)

and for $i=n:$

$$\begin{aligned} \Psi _{1}(T_{n}^{e})&= \left( f^{-1}\left( p^{n-2}\left( E_{2}\right) \right) ,E_{1}\right] ,\end{aligned}$$

(32)

$$\begin{aligned} \Psi _{2}(T_{n}^{e})&= [0,p^{n-1}(E_{2})), \end{aligned}$$

(33)

The last four equations are implied by Eqs. 28 and 29, but vary due to the closed bounds at the extremes. We will simultaneously prove Lemma 5 with our proof of Lemma 6.

First, informally we could rewrite the interval $\left[ 0,f^{-1}\left( p^{n-1}\left( E_{2}\right) \right) \right]$ as

$$\begin{aligned}{}[\Psi _{1}(T_{1}^{e}),\Psi _{1}(T_{1}^{b}),\Psi _{1}(T_{2}^{e}),\Psi _{1}(T_{2}^{b}),\ldots ,\Psi _{1}(T_{n}^{e}),\Psi _{1}(T_{n}^{b})] \end{aligned}$$

or we could rewrite the interval $\left[ 0,f\left( h^{n-1}\left( E_{1}\right) \right) \right]$ as

$$\begin{aligned}{}[\Psi _{2}(T_{n}^{e}),\Psi _{2}(T_{n}^{b}),\Psi _{2}(T_{n-1}^{e}),\Psi _{2}(T_{n-1}^{b}),\ldots ,\Psi _{2}(T_{1}^{e}),\Psi _{2}(T_{1}^{b})]. \end{aligned}$$

More formally,

Lemma 6

$\Psi _{1}(T^{e})\cup \Psi _{1}(T^{b})=\left[ 0,f^{-1}\left( p^{n-1}\left( E_{2}\right) \right) \right]$ and $\Psi _{2}(T^{e})\cup \Psi _{2}(T^{b})=\left[ 0,f\left( h^{n-1}\left( E_{1}\right) \right) \right]$ while $\Psi _{1}(T^{e})\cap \Psi _{1}(T^{b})=\{E_{1}\}$ and $\Psi _{2}(T^{e})\cap \Psi _{2}(T^{b})=\{E_{2}\}$.

Proof

Refer to Eqs. 16–19. Consider the bounds for any $e_{1}\in \Psi _{1}(T_{i}^{e})$. Its open (closed when $i=1$) infimum is $f^{-1}\left( p^{i-2}\left( E_{2}\right) \right)$. Changing the two other constraints on $T_{i}^{e}$ to equalities and solving we find that for $e_{1}\in \Psi _{1}(T_{i}^{e})$ the open (closed when $i=n$) supremum is $h^{n-i}(E_{1})$, which is the closed infimum of $b_{1}\in \Psi _{1}(T_{i}^{b})$. Similar algebra for the other relevant bounds in Lemma 5 and iteration from $i=1$ confirms Lemmas 5 and 6. $\square$

Based on this we know that $\forall i\in \{1,\ldots ,n\}$ any $e_{1}\in \Psi _{1}(T_{i}^{e})$ is strictly less than any $b_{1}\in \Psi _{1}(T_{i}^{b})$, which is strictly less than any $e_{1}\in \Psi _{1}(T_{i+1}^{e})$. The former inequality is weak when $i=n$. (Obviously, we ignore the latter when $i=n$). Also, $\forall i\in \{1,\ldots ,n\}$ any $e_{2}\in \Psi _{2}(T_{i-1}^{e})$ is strictly greater than any $b_{2}\in \Psi _{2}(T_{i}^{b})$ which is strictly greater than any $e_{2}\in \Psi _{2}(T_{i}^{e})$. The later inequality is weak when $i=1$. Obviously, we ignore the former when $i=1$. More formally we have:

Lemma 7

$$\begin{aligned}&\forall i\in \{1,\ldots ,n-1\},\;\nonumber \\&\quad \left( (e_{1}^{i}\in \Psi _{1}(T_{i}^{e}),\; e_{1}^{i+1}\in \Psi _{1}(T_{i+1}^{e}),\; b_{1}^{i}\in \Psi _{1}(T_{i}^{b}))\implies (e_{1}^{i}<b_{1}^{i}<e_{1}^{i+1})\right) \end{aligned}$$

(34)

$$\begin{aligned}&\qquad \qquad \qquad \qquad \qquad \qquad \;(e_{1}^{n}\in \Psi _{1}(T_{n}^{e}),b_{1}^{n}\in \Psi _{1}(T_{n}^{b}))\implies e_{1}^{n}\le b_{1}^{n}\end{aligned}$$

(35)

$$\begin{aligned}&\forall i\in \{2,3,\ldots ,n\},\;\nonumber \\&\quad \;\left( (e_{2}^{i-1}\in \Psi _{2}(T_{i-1}^{e}),\; e_{2}^{i}\in \Psi _{2}(T_{i}^{e}),\; b_{2}^{i}\in \Psi _{2}(T_{i}^{b})\implies (e_{2}^{i}<b_{2}^{i}<e_{2}^{i-1})\right) \qquad \end{aligned}$$

(36)

$$\begin{aligned}&\qquad \;\qquad \qquad \qquad \qquad \qquad (e_{2}^{1}\in \Psi _{2}(T_{1}^{e}),b_{2}^{1}\in \Psi _{2}(T_{1}^{e}))\implies e_{2}^{1}\le b_{2}^{1} \end{aligned}$$

(37)

Proof

Lemma 7 follows directly by examining the bounds in Lemma 5. $\square$

1.1 Proof of Theorem 1

Recall Theorem 1: When Blotto plays a strategy $\mu _{B}\in \Omega ^{B}$ and Enemy plays a strategy $\mu _{E}\in \Omega ^{E}$, Blotto’s expected payoff is $\frac{\sum _{j=0}^{n}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ and Enemy’s expected payoff is $\frac{\sum _{j=1}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}.$

Proof

Given Lemma 7, we know that against any Blotto strategy in $\Omega ^{B}$, when Enemy plays in $T_{i}^{e}$ his probability of winning on Battlefield 1 is:

$$\begin{aligned} \mu _{B}(T_{1}^{b}\cup \ldots \cup T_{i-1}^{b})=\frac{\sum _{j=n-(i-1)}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}} \end{aligned}$$

His probability of winning Battlefield 2 is:

$$\begin{aligned} \mu _{B}(T_{i+1}^{b}\cup \ldots \cup T_{n}^{b})=\frac{\sum _{j=0}^{n-i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}. \end{aligned}$$

The total expected payoff is then:

$$\begin{aligned} \frac{\sum _{j=n-(i-1)}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}+w\cdot \frac{\sum _{j=0}^{n-i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}=\frac{\sum _{j=1}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}} \end{aligned}$$

(38)

for any allocation in (any) $T_{i}^{e}$.

Similarly, against any Enemy strategy from above, when Blotto plays in $T_{i}^{b}$ his probability of winning Battlefield 1 is:

$$\begin{aligned} \mu _{E}\left( T_{1}^{e}\cup \ldots \cup T_{i}^{e}\right) =\frac{\sum _{j=0}^{i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}} \end{aligned}$$

His probability of winning Battlefield 2 is:

$$\begin{aligned} \mu _{E}\left( T_{i}^{e}\cup \ldots \cup T_{n}^{e}\right) =\frac{\sum _{j=i-1}^{n-i}w^{j}}{\sum _{j=0}^{n-1}w^{j}} \end{aligned}$$

His total expected payoff is then

$$\begin{aligned} 1\cdot \frac{\sum _{j=0}^{i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}+w\cdot \frac{\sum _{j=i-1}^{n-i}w^{j}}{\sum _{j=0}^{n-1}w^{j}}=\frac{\sum _{j=0}^{n}w^{j}}{\sum _{j=0}^{n-1}w^{j}} \end{aligned}$$

(39)

for any allocation in (any) $T_{i}^{b}$. $\square$

1.2 Proof of Theorem 2

In this section, we prove that any pair of strategies $\left\{ \mu _{B},\mu _{E}\right\}$, such that $\mu _{B}\in \Omega ^{B}$ and $\mu _{E}\in \Omega ^{E},$ in fact forms a Nash equilibrium. Before proceeding with the formal proof, we provide the intuition. Properties 1b and 1e specify that in any equilibrium Blotto and Enemy each randomize over $n$ distinct areas ($T_{1}^{b},\ldots ,T_{n}^{b}$ and $T_{1}^{e},\ldots ,T_{n}^{e}$). Blotto and Enemy’s potential equilibrium allocations on either battlefield only overlap at one point in the following sense:

$$\begin{aligned} \Psi _{1}(T^{b})\cap \Psi _{1}(T^{e})=\{E_{1}\}, \end{aligned}$$

$$\begin{aligned} \Psi _{2}(T^{b})\cap \Psi _{2}(T^{e})=\{E_{2}\}. \end{aligned}$$

Though in the figures it may appear as though all boundaries should be included in these sets, careful inspection of the boundary conditions shows that the boundaries for Enemy are open, and closed for Blotto. Therefore, they should not in fact be included in these sets.

Given that ties always go to Blotto, we calculated players’ expected payoffs in “Proof of Theorem 1 of Appendix 1.” When they both play strategies satisfying Properties 1b and 1e, Blotto achieves an expected payoff of $\frac{\sum _{j=0}^{n}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ while Enemy earns $\frac{\sum _{j=1}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$. We will show that given these payoffs, Property 2b(e) ensures that Enemy (Blotto) has no full expenditure allocation which provide a payoff strictly greater than $\frac{\sum _{j=1}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}} \left( \frac{\sum _{j=0}^{n}w^{j}}{\sum _{j=0}^{n-1}w^{j}}\right)$. Since all allocations in the players’ supports provide the same payoff, and there exist no allocations providing higher payoffs, pairs of strategies from these distributions constitute a Nash equilibrium. We now move to the formal proof.

Recall Theorem 2: Any pair of strategies $\left\{ \mu _{B},\mu _{E}\right\}$ such that $\mu _{B}\in \Omega ^{B}$ and $\mu _{E}\in \Omega ^{E}$ constitute a Nash equilibrium.

Proof

We show that there are no allocations for Enemy or Blotto that provide a strictly higher expected payoff than their payoffs from Theorem 1. Note that if either player were to have an expected payoff improving deviation from the strategies we defined, they must have a full expenditure payoff improving deviation, as the expected payoffs must be weakly increasing in strength on either battlefield. Therefore, we only need to show that there are no payoff improving full expenditure deviations. So, we check full expenditure deviations outside of any $T_{i}^{e}$ or $T_{i}^{b}$.

Consider a full expenditure Enemy deviation $e^{d}=(e_{1}^{d},e_{2}^{d}),\; e^{d}\notin T^{e}$.²¹ Given the bounds on the $T_{i}^{e}$’s and Lemma 5, $e^{d}$ must lie “between” some $T_{i}^{e}$ and $T_{i+1}^{e}$ in the following sense: $\forall e_{1}^{i}\in \Psi _{1}(T_{i}^{e}),e_{1}^{i+1}\in \Psi _{1}(T_{i+1}^{e})\quad e_{1}^{i}<e_{1}^{d}<e_{1}^{i+1}$, and similarly for $e_{2}^{d}$. Let $(e_{1},e_{2})$ be an allocation in $T_{i}^{e}$. Examine Property 2b with $x=e_{1}^{d}$. Given Lemma 7, the realized payoff to Enemy of playing $e^{d}$ against any of our Blotto strategies will be the same as if he had played $(e_{1},e_{2})$ unless Blotto plays in $T_{i}^{b}$ or $T_{i+1}^{b}$. If Blotto plays in $T_{i}^{b},$ the deviant allocation may do better ²² on Battlefield 1 (without changing the outcome on Battlefield 2). The cost is that if Blotto plays in $T_{i+1}^{b}$ the deviant strategy may do worse on Battlefield 2 (without changing the outcome on Battlefield 1). Using the notation of Property 2b, any $b_{1}$ in $j_{b}^{e_{1}^{d},i}$ will lose to $e_{1}^{d}$ (while it would have beat $e_{1}$) and any $b_{2}$ in $k_{b}^{e_{1}^{d},i}$ will beat $e_{2}^{d}$ (while it would have lost to $e_{2}$). Property 2b then says that by moving from any $(e_{1},e_{2})$ in $T_{i}^{e}$ to $(e_{1}^{d},e_{2}^{d})$, the additional probability of winning on Battlefield 1 is weakly less than the additional probability of losing on Battlefield 2 times the weight placed on that battlefield. Therefore, no full expenditure deviation $(e_{1}^{d},e_{2}^{d})$ is payoff improving, and therefore no deviation is payoff improving.

The same line of reasoning applies directly to Property 2e and full expenditure deviations by Blotto which lie “between” some $T_{i}^{b}$ and $T_{i+1}^{b}$. Specifically, Property 2e ensures that a full expenditure deviating allocation by Blotto, $(b_{1}^{d},b_{2}^{d})$, cannot be payoff improving. Simply set $b_{1}^{d}=x$ in the property and the same line of reasoning follows. Additionally, there are full expenditure deviations which do not lie “between” some $T_{i}^{b}$ and some $T_{i+1}^{b}$ (e.g., $(B,0)$ and $(0,B)$). Specifically, there are two more deviating types of full expenditure allocations: a $(b_{1}^{d},b_{2}^{d})$ where $\forall (b_{1},b_{2})\in T_{1}^{b}\quad b_{1}^{d}<b_{1}\quad and\quad f\left( b_{1}^{d}\right) =b_{2}^{d}>b_{2}$ or a $(b_{1}^{\#},b_{2}^{\#})$ where $\forall (b_{1},b_{2})\in T_{n}^{b}\quad b_{1}^{\#}>b_{1}\quad and\quad f\left( b_{1}^{\#}\right) =b_{2}^{\#}<b_{2}$. In the former, Blotto increases allocations to Battlefield 2 at the expense of Battlefield 1, relative to $T_{1}^{b}$. However, in $T_{1}^{b}$, Blotto is guaranteeing victory on Battlefield 2, so this can not be payoff improving. Similar logic applies to the later type of allocations. Given Lemma 5, and Lemma 7, there are no other types of full expenditure deviations.

Thus, if Blotto plays $\mu _{B}\in \Omega ^{B}$ and Enemy plays $\mu _{E}\in \Omega ^{E}$, they would both be playing best responses to the other’s strategy. Therefore any such pair $\left\{ \mu _{B},\mu _{E}\right\}$ constitutes a Nash equilibrium. $\square$

1.3 Proof of Theorem 3

We now prove that there are no uncharacterized strategies which could be part of a Nash equilibrium. Before proceeding, we will need two lemmas that hold for any two-player constant-sum game.

Consider a two-player constant-sum game where Player 1 chooses a strategy $x\in X$ and Player 2 chooses a strategy $y\in Y$. Let $f^{i}(x,y)$ denote the expected payoffs to Player $i$ when Player 1 plays $x$ and Player 2 plays $y$. Also, let $\Xi ^{i}$ denote the set of player $i$ strategies that are a part of some Nash equilibrium.

Definition 8

A game is said to feature constant payoffs if, for each player, the expected payoff is the same in each Nash equilibrium.

Lemma 9

If a Nash equilibrium exists for a two-player constant-sum game, that game features constant payoffs. In other words, $\exists \left\{ c_{i}\right\} _{i=1}^{2}$ such that in every Nash equilibrium $\{x_{j},y_{j}\}$, $f^{i}(x_{j},y_{j})=c_{i}$ for all $i=1,2$.

For a formal proof of the preceding and proceding lemmas see (Vorob’ev (1977), pp. 1–10)

Lemma 10

Equilibrium Interchangeability If a Nash equilibrium exists to a two-player constant-sum game, then every strategy that a player uses in any Nash equilibrium forms a Nash equilibrium with any opponent strategy from any (other) Nash equilibrium. In other words: for all $x^{*}\in \Xi ^{1}$ and all $y^{*}\in \Xi ^{2}$, $\left\{ x^{*},y^{*}\right\}$ constitutes a Nash equilibrium.

Equilibrium Interchangeability and constant payoffs are useful when proving the completeness of our characterization.²³ Equilibrium Interchangeability allows us to consider equilibrium strategies for Blotto and Enemy separately. Unlike with most multiple equilibria games, there is no need to worry about pairing with a particular opponent equilibrium strategy. If we discover just one Nash equilibrium (pair of strategies), all the remaining equilibria are simply the cross of all the Blotto strategies that form an equilibrium with the one known Enemy strategy, and all the Enemy strategies that form an equilibrium with the one known Blotto strategy.

Our proof of the completeness of our characterization proceeds as follows. We first prove that all Enemy strategies that are a part of some Nash equilibrium are in $\Omega ^{E}.$ Then we prove that all Blotto strategies that are a part of some equilibrium are in $\Omega ^{B}.$ Therefore, the set of all Nash equilibria is the set of pairs of strategies from $\Omega ^{E}$ and $\Omega ^{B}$ by Equilibrium Interchangeability.

In the proof, we make use of the strategy $\mu _{B}^{*}$ where in each $T_{i}^{b}$ Blotto plays the allocations $\left( h^{n-i}(E_{1}),f\left( h^{n-i}(E_{1})\right) \right)$ and $\left( f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ,p^{i-1}\left( E_{2}\right) \right)$ with probability $\frac{w^{n-i}}{2\cdot {{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}}$ each. These points correspond to $z_{1}-z_{6}$ in Fig. 11 for a game in region 3. Blotto never plays any other allocations. Recall that if $E_{2}=f\left( h^{n-i}(E_{1})\right)$ each of the $T_{i}^{b}$’s contain only one allocation, $\left( h^{n-i}(E_{1}),f\left( h^{n-i}(E_{1})\right) \right) =\left( f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ,p^{i-1}\left( E_{2}\right) \right) .$ In this case this single allocation is played with probability $\frac{w^{n-i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}.$

In each $T_{i}^{b},$ $\left( h^{n-i}(E_{1}),f\left( h^{n-i}(E_{1})\right) \right)$ and $\left( f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ,p^{i-1}\left( E_{2}\right) \right)$ are the intersections of Blotto’s resource constraint with the two other bounds on $T_{i}^{b}$. So, Property 1b holds as $2\cdot \frac{w^{n-i}}{2\cdot {{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}}=\frac{w^{n-i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}=\mu _{B}^{*}\left( T_{i}^{b}\right)$. Now consider Property 2b. As Blotto is playing $\left( f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ,p^{i-1}\left( E_{2}\right) \right)$ in $T_{i}^{b}$ with probability $\frac{w^{n-i}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$, $\mu _{B}^{*}(j_{b}^{x,i})\le \frac{w^{n-i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}-\frac{w^{n-i}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}=\frac{w^{n-i}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}},$ $\forall i<1,2,\ldots ,n-1,\quad \forall x\in [h^{n-i}(E_{1}),f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ].$ As $g$ is strictly decreasing, the minimum of $\mu _{B}^{*}(k_{b}^{x,i})$ for $x\in [h^{n-i}(E_{1}),f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ]$ occurs when $x=h^{n-i}(E_{1}).$ Then $\forall x\in [h^{n-i}(E_{1}),f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ],$ $\mu _{B}^{*}(k_{b}^{x,i})\ge \frac{w^{n-(i+1)}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$ as $g\left( h^{n-i}(E_{1})\right) =f\left( h^{n-i-1}(E_{1})\right)$ and $\left( h^{n-(i+1)}(E_{1}),f\left( h^{n-(i+1)}(E_{1})\right) \right)$ is played with probability $\frac{w^{n-(i+1)}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$ in $T_{i+1}^{b}.$ Therefore $\mu _{B}^{*}(k_{b}^{x,i})\cdot w\ge \frac{w^{n-i}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}\ge \mu _{B}^{*}(j_{b}^{x,i})$ and Property 2b holds, $\mu _{B}^{*}\in \Omega ^{B}.$

Fig. 11

An example

Lemma 11

Any Enemy strategy which is a part of some Nash equilibrium is in $\Omega ^{E}$.

Proof

We prove Lemma 11 by contradiction. Suppose there exists a Nash equilibrium Enemy strategy that is not in $\Omega ^{E}$. Such a strategy must then either violate Property 1e or satisfy Property 1e and violate Property 2e. In proving that all our strategies were indeed part of a Nash equilibrium, we have already shown how a violation of Property 2e alone would provide Blotto with a payoff improving deviation, so we rule out that possibility. The only other way Lemma 11 could be false is if there were a Nash equilibrium Enemy strategy which violated property 1e. We divide deviations from property 1e into three possible cases. Figure 12 provides a graphical reference (in Region 3) to aid the reader.

Fig. 12

Enemy deviations in region 3

Intuitively, Deviation 1 represents Enemy placing some mass on allocations which, relative to some $T_{i}^{e}$, always send less to one battlefield, without increasing the allocation to the other. Deviation 2 represents Enemy placing mass on allocations which, relative to any $T_{i}^{e}$ always send less to one battlefield, but increase the allocation to the other. Deviation 3 has him playing an “incorrect” mass on some $T_{i}^{e}$ (i.e., $\mu _{E}(T_{i}^{e})\ne \frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$).

Deviation 1: Enemy could play over an area that sends less to both battlefields than some $(e_{1},e_{2})$ in some $T_{i}^{e}$. Formally, this would have Enemy play a $\mu _{E}^{d_{1}}$ such that the following three statements hold for some feasible set of Enemy allocations $S$:²⁴

$$\begin{aligned}&\displaystyle S\cap T^{e}=\emptyset \end{aligned}$$

(40)

$$\begin{aligned}&\displaystyle \mu _{E}^{d_{1}}(S)>0\end{aligned}$$

(41)

$$\begin{aligned}&\displaystyle \forall (e_{1}^{d},e_{2}^{d})\in S,\quad \exists (e_{1},e_{2})\in T^{e}\quad \hbox {s.t.}\quad e_{1}^{d}\le e_{1}\quad and\quad e_{2}^{d}\le e_{2} \end{aligned}$$

(42)

Condition 40 implies that at least one of the two inequalities in 42 always holds strictly.

Suppose Deviation 1 holds for some $S$ and consider an allocation $(e_{1}^{d},e_{2}^{d})\in S$. Without loss of generality, suppose $\exists \left( e_{1},e_{2}\right) \in T_{i}^{e}$ such that $e_{1}^{d}\le e_{1}$ and $e_{2}^{d}<e_{2}.$ Either $e_{2}^{d}<e_{2}^{\#}$ for all $e_{2}^{\#}\in \Psi _{2}\left( T_{i}^{e}\right)$ or not. If so, $(e_{1}^{d},e_{2}^{d})$ cannot be a best response to $\mu _{B}^{*}$ which has Blotto playing Battlefield 2 allocations equal to the open lower bound of $e_{2}^{\#}\in \Psi _{2}\left( T_{i}^{e}\right)$ with positive probability. Enemy could increase his expected payoff by playing $(e_{1},e_{2}).$ Alternatively, there could exist an $e_{2}^{\#}\in \Psi _{2}\left( T_{i}^{e}\right)$ for which $e_{2}^{d}\ge e_{2}^{\#}.$ Since we know $e_{2}^{d}<e_{2}\in \Psi _{2}\left( T_{i}^{e}\right)$ we must have that $e_{2}^{d}\in \Psi _{2}\left( T_{i}^{e}\right) .$ Given the bounds of $T_{i}^{e}$ and Eq. 40, it must be the case that $e_{1}^{d}\le f^{-1}\left( p^{i-2}\left( E_{2}\right) \right)$ the open lower bound of $\Psi _{1}\left( T_{i}^{e}\right) ,$ a Battlefield 1 allocation Blotto plays with positive probability in $\mu _{B}^{*}$. Therefore $(e_{1}^{d},e_{2}^{d})$ cannot be a best response to $\mu _{B}^{*}$ which has Blotto playing the open lower bounds of $e_{1}^{\#}\in \Psi _{1}\left( T_{i}^{e}\right)$ with positive probability. Enemy could increase his payoff by playing $\left( e_{1},e_{2}\right) .$ Given equilibrium interchangeability and the fact that no allocation $(e_{1}^{d},e_{2}^{d})\in S$ could be a best response to $\mu _{B}^{*},$ Deviation 1 cannot happen in any Nash equilibrium.

This only leaves two possible types of deviations by Enemy: He could play with mass other than $\frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$ over some $T_{i}^{e}$ (Deviation 3) and/or he could play with mass over a region $S$ where $\forall (e_{1}^{d},e_{2}^{d})\in S,\quad \forall (e_{1},e_{2})\in T^{e}$ :

$$\begin{aligned}&\displaystyle e_{1}^{d}>e_{1}\; or\; e_{2}^{d}>e_{2}\\&\displaystyle S\cap T^{e}=\emptyset \end{aligned}$$

(Deviation 2). Given the bounds of the $T_{i}^{e}$’s it is easy to show that any such region $S$ must be within the set of points $D_{i}^{e}$, indexed by $i=1,2,\ldots ,n-1$, where

$$\begin{aligned} D_{i}^{e}\equiv \left\{ (e_{1},e_{2}):\;(e_{1}\ge h^{n-i}(E_{1}),\; e_{2}\ge p^{i}(E_{2}),\; and\; e_{2}\le g\left( e_{1}\right) )\right\} \end{aligned}$$

Given Lemma 5 there are no other allocations for which Deviation 2 holds.

We simultaneously prove that neither of the latter two deviations is possible. Consider a $T_{i}^{e}$ and $D_{i}^{e}$ and a deviating Enemy strategy $\mu _{E}^{d}$ that forms an Nash equilibrium with any $\mu _{B}\in \Omega ^{B}$. Consider some $i\in \left\{ 1,2,3,\ldots ,n-1\right\}$. Assume that

$$\begin{aligned} \forall k=1,2,\ldots ,i-1,\quad \mu _{E}^{d}(D_{k}^{e})=0\; and\;\mu _{E}^{d}(T_{k}^{e})=\frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}} \end{aligned}$$

(43)

In other words, there has not “yet” been a Deviation 2 or Deviation 3.

Suppose the mass over $\mu _{E}^{d}(T_{i}^{e})<\frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$. Given Lemma 7, Eq. 43 and the fact that we’ve ruled out Deviation 1, when Blotto plays $\left( h^{n-i}(E_{1}),f\left( h^{n-i}(E_{1})\right) \right)$ (which he does with probability $\frac{w^{n-i}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$ in strategy $\mu _{B}^{*}$) he wins Battlefield 1 with probability $\mu _{E}^{d}\left( T_{1}^{e}\cup \ldots \cup T_{i}^{e}\right) <\frac{\sum _{j=0}^{i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ but still wins Battlefield 2 with probability $1-\mu _{E}^{d}\left( T_{1}^{e}\cup \ldots \cup T_{i-1}^{e}\right) =\frac{\sum _{j=i-1}^{n-i}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ for a total expected payoff strictly less than $\frac{\sum _{j=0}^{n}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$, which is Blotto’s constant expected payoff in all equilibria, a contradiction. Therefore, $\mu _{E}^{d}(T_{i}^{e})\ge \frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}.$

Similarly, if $\mu _{E}^{d}\left( T_{i}^{e}\right) \!>\!\frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}},$ then when Blotto plays $\left( h^{n-i}(E_{1}),f\left( h^{n-i}(E_{1})\right) \right)$, he wins Battlefield 1 with probability $\mu _{E}^{d}\left( T_{1}^{e}\cup \ldots \cup T_{i}^{e}\right) >\frac{\sum _{j=0}^{i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ but still wins Battlefield 2 with probability $1-\mu _{E}^{d}\left( T_{1}^{e}\cup \ldots \cup T_{i-1}^{e}\right) =\frac{\sum _{j=i-1}^{n-i}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ for a total expected payoff strictly greater than his constant equilibrium payoff, $\frac{\sum _{j=0}^{n}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$, a contradiction. Therefore, $\mu _{E}^{d}(T_{i}^{e})=\frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$.

Now suppose $\mu _{E}^{d}(D_{i}^{e})>0$. Now, when Blotto plays $\left( f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ,p^{i-1}\left( E_{2}\right) \right)$ (which he does with probability $\frac{w^{n-i}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$ in strategy $\mu _{B}^{*}$) he expects to win on Battlefield 1 with probability $\mu _{E}^{d}\left( T_{1}^{e}\cup \ldots \cup T_{i}^{e}\cup D_{i}^{e}\right) >\frac{\sum _{j=0}^{i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ and expects to win on Battlefield 2 with probability $1-\mu _{E}^{d}\left( T_{1}^{e}\cup \ldots \cup T_{i-1}^{e}\right) =\frac{\sum _{j=i-1}^{n-i}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$. Therefore his total expected payoff is strictly greater than $\frac{\sum _{j=0}^{n}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$, his constant equilibrium payoff, another contradiction. Therefore, $\mu _{E}^{d}(D_{i}^{e})$ must equal zero.

As the above analysis holds for all $i=1,2,\ldots ,n-1$, the mass over all such $T_{i}^{e}$ and $D_{i}^{e}$ must equal $\frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$ and $0$, respectively. The remaining mass of $\frac{w^{n-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$ must then be distributed over the only region left, $T_{n}^{e}$. Therefore, $\mu _{E}^{d}$ satisfies Property 1e. We’ve already discussed why it must also satisfy Property 2e. Therefore, $\mu _{E}^{d}\in \Omega ^{E},$ and there can be no Enemy strategies which are a part of some Nash equilibrium which do not satisfy our characterization. $\square$

We have ruled out any potential Enemy strategies that deviate from our characterization of possible Nash equilibrium Enemy strategies. The proof that we have characterized the complete set of Blotto Nash equilibrium strategies proceeds much the same way as the proof of completeness for Enemy’s strategies. However, due to the open boundaries of the $T_{i}^{e}$’s, the proof requires additional consideration.

Lemma 12

Any Blotto strategy which is a part of some Nash equilibrium is in $\Omega ^{B}.$

Proof

Suppose not. Then there exists at least one Blotto strategy which is a part of some Nash equilibrium that does not satisfy properties 1b and 2b. Call such a strategy $\mu _{B}^{d}$ We’ve already shown how a strategy that satisfied property 1b, but violated property 2b would give Enemy an allocation offering a payoff higher than his constant equilibrium payoff. So, any uncharacterized Blotto strategy which is part of some Nash equilibrium must violate property 1b. Property 1b specifies that Blotto must only play in his $T_{i}^{b}$’s and provides the probability of play in each. There are two ways Blotto could violate this property: He could sometimes play outside of $T^{b}$, or his $\mu _{B}$ could assign an incorrect probability to some $T_{i}^{b}$. We break the former down into two separate deviations. The first type of deviation we consider (Deviation 1) are where Blotto mixes over allocations that send weakly less to both battlefields than some non-deviating allocation. Formally, a strategy, $\mu _{B}^{d_{1}}$, exhibits Deviation 1 if the following conditions hold for some $S$ :

$$\begin{aligned}&\displaystyle \mu _{B}^{d_{1}}(S)>0\end{aligned}$$

(44)

$$\begin{aligned}&\displaystyle S\cap T^{b}=\emptyset \end{aligned}$$

(45)

$$\begin{aligned}&\displaystyle \forall (b_{1}^{d},b_{2}^{d})\in S,\quad \exists (b_{1},b_{2})\in T^{b}\quad \hbox {s.t.}\quad b_{1}^{d}\le b_{1}\quad and\quad b_{2}^{d}\le b_{2} \end{aligned}$$

(46)

The second condition implies that one of the two inequalities in the third condition always holds strictly.

The next type of deviation we consider (Deviation 2) are the remaining feasible allocations outside of $T^{b}$. Specifically, these are the allocations that are outside $T^{b}$, and, relative to any allocation in $T^{b},$ send strictly more to one battlefield. Formally a strategy, $\mu _{B}^{d_{2}}$, exhibits Deviation 2 if the following conditions hold for some $S$:

$$\begin{aligned}&\displaystyle \mu _{B}^{d_{2}}(S)>0\end{aligned}$$

(47)

$$\begin{aligned}&\displaystyle S\cap T^{b}=\emptyset \end{aligned}$$

(48)

$$\begin{aligned}&\displaystyle \forall (b_{1}^{d},b_{2}^{d})\in S,(b_{1},b_{2})\in T^{b}\quad either\quad b_{1}^{d}>b_{1}\quad or\quad b_{2}^{d}>b_{2} \end{aligned}$$

(49)

Consider the following sets of allocations:

$$\begin{aligned} D_{0}^{b}&\equiv \{(b_{1},b_{2}):\;(b_{1}\ge 0),(b_{2}>f\left( h^{n-1}\left( E_{1}\right) \right) ),(b_{2}\le f\left( b_{1}\right) )\}\end{aligned}$$

(50)

$$\begin{aligned} D_{i}^{b}&\equiv \{(b_{1},b_{2}):\;(b_{1}>f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ),(b_{2}>f\left( h^{n-i-1}\left( E_{1}\right) \right) ),\nonumber \\&(b_{2}\le f\left( b_{1}\right) )\}\;\forall i=1,\ldots ,n-1\end{aligned}$$

(51)

$$\begin{aligned} D_{n}^{b}&\equiv \{(b_{1},b_{2}):\;(b_{1}>f^{-1}\left( p^{n-1}\left( E_{2}\right) \right) ),(b_{2}\ge 0),(b_{2}\le f\left( b_{1}\right) )\} \end{aligned}$$

(52)

Note that Eqs. 50 and 52 are implied by 51 with the exception of the weak inequalities at the zero bounds. A strategy, $\mu _{B}^{d_{2}}$, satisfying Deviation 2 must allocate some mass over at least one of the $D_{i}^{b}$’s as, given Lemma 5, these are the only regions where conditions 47–49 hold. See Fig. 13 for a graph of the $D_{i}$’s.

Fig. 13

Blotto deviations in region 3

The last type of deviation we consider (Deviation 3) is simply where Blotto plays inappropriate mass over one of his $T_{i}^{b}$’s. Formally, a strategy $\mu _{B}^{d_{3}}$ exhibits Deviation 3 if the following condition holds for at least one of Blotto’s $T_{i}^{b}$’s:

$$\begin{aligned} \mu _{B}^{d_{3}}\left( T_{i}^{b}\right) \ne \frac{w^{n-i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}. \end{aligned}$$

Because of Lemma 10 (Equilibrium Interchangeability), any Blotto strategy from any Nash equilibrium must form a Nash equilibrium with any Enemy strategy from $\Omega ^{E}$. Specifically, we consider the following sequence of Enemy strategies: For any $k=1,2,\ldots$ let $\mu _{E}^{k}$ be the strategy where in each $T_{i}^{e}$ Enemy plays points

$$\begin{aligned} \left( g^{-1}\left( p^{i-1}\left( E_{2}-i\frac{\epsilon }{k}\right) \right) ,p^{i-1}\left( E_{2}-i\frac{\epsilon }{k}\right) \right) \end{aligned}$$

and

$$\begin{aligned} \left( h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{k}\right) ,g\left( h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{k}\right) \right) \right) \end{aligned}$$

with probability $\frac{w^{i-1}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}$. This implies that no other allocations are in the support of $\mu _{E}^{k}$ since $2\cdot \sum _{i=1}^{n}\frac{w^{i-1}}{2\cdot {{\sum \nolimits _{j=0}^{n-1}}}w^{j}}=1.$ For $\epsilon$ sufficiently small $\mu _{E}^{1}$ (and any other $\mu _{E}^{k}$) satisfies properties 1e and 2e.

Intuitively, this is a sequence of strategies that has Enemy randomizing over allocations on his resource constraint, arbitrarily close to the corners in his $T_{i}^{e}$’s. In other words, for any $T_{i}^{e}$ we will be able to find some $\mu _{E}^{k}$ where Enemy plays arbitrarily close to each intersection of his resource constraint and either of the other two bounds for $T_{i}^{e}$ with strictly positive probability.

To see that any $\mu _{E}^{k}$ is in fact in $\Omega ^{E},$ consider the following. So long as $\epsilon$ is sufficiently small, the points from $\mu _{E}^{1}$, $\left( g^{-1}\left( p^{i-1}\left( E_{2}-i\frac{\epsilon }{1}\right) \right) ,p^{i-1}\left( E_{2}-i\frac{\epsilon }{1}\right) \right)$ and $\left( h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{1}\right) ,g\left( h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{1}\right) \right) \right)$, will be in $T_{i}^{e}.$ These points are a small (note the $i\epsilon$ or $\left( n+1-i\right) \epsilon$ terms) distance from the intersection with their respective boundary of $T_{i}^{e}.$ As they are a small distance toward the interior, Property 1e holds (for sufficiently small $\epsilon$). Since these points will simply get closer to the boundary as $k$ increases, (yet never touch it) Property 1e will hold for any $\mu _{E}^{k}.$

The $\frac{\epsilon }{k}$ terms are multiplied by $i$ and $\left( n+1-i\right)$ to ensure that property 2e holds. Note that full expenditure deviating play by Blotto of $\left( f^{-1}\left( p^{i-1}\left( E_{2}-i\frac{\epsilon }{k}\right) \right) , p^{i-1}\left( E_{2}-i\frac{\epsilon }{k}\right) \right)$ will exactly match the $p^{i-1}\left( E_{2}-i\frac{\epsilon }{k}\right)$ Enemy sometimes plays on Battlefield 2, but will not increase his payoff on Battlefield 1 as $f^{-1}\left( p^{i-1}\left( E_{2}-i\frac{\epsilon }{k}\right) \right)$ is strictly less than $f^{-1}\left( p^{i}\left( E_{2}-\left( i+1\right) \frac{\epsilon }{k}\right) \right)$ (from $T_{i+1}^{b}$). So, Blotto’s expected payoff will not increase relative to play in $T_{i}^{b}.$ Similar analysis on other potential full expenditure Blotto deviations ensures Property 2e holds.

Now we are ready to start considering Blotto’s potential deviations. Deviation 1 has Blotto mix over allocations which send weakly less to both battlefields than some allocation in some $T_{i}^{b}.$ Since these deviating allocations are not themselves in $T_{i}^{b}$ they must send strictly less to at least one battlefield. Suppose these conditions hold for some set of allocations $S$ and let $b^{d}=\left( b_{1}^{d},b_{2}^{d}\right) \in S$ satisfy Deviation 1 relative to a $\left( b_{1}^{i},b_{2}^{i}\right) \in T_{i}^{b}.$ Suppose $b^{d}$ sends strictly less to Battlefield 1, or, by Lemma 5, $b_{1}^{d}<h^{n-i}(E_{1})$ (and $b_{2}^{d}\le f\left( h^{n-i}\left( E_{1}\right) \right)$). There exists some $k^{*}$ where $b_{1}^{d}<h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{k^{*}}\right) <h^{n-i}\left( E_{1}\right)$. Now, we know $\left( b_{1}^{d},b_{2}^{d}\right)$ cannot be a best response to $\mu _{E}^{k^{*}}.$ Blotto could play $\left( h^{n-i}(E_{1}),f\left( h^{n-i}(E_{1})\right) \right)$, but he plays $\left( b_{1}^{d},b_{2}^{d}\right)$ which strictly lowers his probability of winning on Battlefield 1 when Enemy plays $\mu _{E}^{k^{*}}$, and it does so without increasing Blotto’s probability of winning on Battlefield 2. Therefore $\left( b_{1}^{d},b_{2}^{d}\right)$ provides a strictly lower payoff and cannot be a best response. Similar logic applies to $\left( b_{1}^{d},b_{2}^{d}\right)$ that send strictly less to Battlefield 2 (or where $b_{1}^{d}\le f^{-1}\left( p^{i-1}\left( E_{2}\right) \right)$ and $b_{2}^{d}<p^{i-1}(E_{2})$). Therefore, all allocations which could be randomized over in Deviation 1 are not best responses to some $\mu _{E}^{k}.$ Therefore, no equilibrium Blotto strategy exhibits Deviation 1.

Consider a deviating Blotto strategy $\mu _{B}^{d}$ that forms a Nash equilibrium with any $\mu _{E}\in \Omega ^{E}$. All allocations in $D_{0}^{b}$ and $D_{n}^{b}$ are not best responses to certain Nash equilibrium Enemy strategies. For instance, take an allocation $\left( b_{1}^{d},b_{2}^{d}\right) \in D_{0}^{b}.$ Blotto is increasing his Battlefield 2 allocation while reducing his Battlefield 1 allocation relative to $T_{1}^{b}.$ However, in $T_{1}^{b}$ Blotto was already guaranteeing victory on Battlefield 2 so this cannot be payoff improving. Clearly $b_{1}^{d}<h^{n-1}(E_{1}).$ We can find some $k^{*}\in \mathbb {N}$ such that $b_{1}^{d}<h^{n-1}\left( E_{1}-\left( n+1-1\right) \frac{\epsilon }{k^{*}}\right) <h^{n-1}(E_{1}).$ As Enemy plays $h^{n-1}\left( E_{1}-\left( n+1-1\right) \frac{\epsilon }{k^{*}}\right)$ on Battlefield 1 with positive probability in $\mu _{E}^{k^{*}},$ $\left( b_{1}^{d},b_{2}^{d}\right)$ must provide Blotto with a strictly lower payoff than $\left( h^{n-1}(E_{1}),f\left( h^{n-1}(E_{1})\right) \right) ,$ which also guarantees victory on Battlefield 2. Therefore, $\mu _{B}^{d}\left( D_{0}^{b}\right) =0$. Similar logic implies that $\mu _{B}^{d}\left( D_{n}^{b}\right) =0.$

We now simultaneously prove that neither Deviation 2 nor Deviation 3 is possible in a Nash equilibrium. Consider some $i\in \left\{ 1,2,3,\ldots ,n-1\right\} .$ Assume that

$$\begin{aligned} \forall j=1,2,\ldots ,i-1,\quad \mu _{B}^{d}(D_{j}^{b})=0\;\quad and\;\quad \mu _{B}^{d}(T_{j}^{b})=\frac{w^{n-j}}{{{\sum \nolimits _{l=0}^{n-1}}}w^{l}} \end{aligned}$$

(53)

In other words, there has not “yet” been a Deviation 2 or Deviation 3.

Consider possible versions of Deviation 3 for $T_{i}^{b}$. First suppose, $\mu _{B}^{d}\left( T_{i}^{b}\right) >\frac{w^{n-i}}{{\sum \nolimits _{j=0}^{{n-1}}}w^{j}}$. Given Eq. 53 and the fact that we’ve already ruled out Deviation 1, when Enemy plays in $T_{i}^{e}$ his probability of winning on Battlefield 1 is $\mu _{B}^{d}\left( T_{1}^{b}\!\cup \!\ldots \!\cup \! T_{i-1}^{b}\right) \!=\!\frac{\sum _{j=n-(i-1)}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$.²⁵ However, his probability of winning on Battlefield 2 is $1-\mu _{B}^{d}(T_{1}^{b}\cup \ldots \cup T_{i}^{b})<\frac{\sum _{j=0}^{n-i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}.$ Therefore, Enemy’s total expected payoff is then strictly less than $\frac{\sum _{j=1}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$ which is his constant equilibrium payoff, a contradiction. Similar logic applies to the case where $\mu _{B}^{d}\left( T_{i}^{b}\right) <w^{n-i}$ and implies that his expected payoff would be greater than $\frac{\sum _{j=1}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}},$ his constant equilibrium payoff, a contradiction. Therefore, $\mu _{B}^{d}\left( T_{i}^{b}\right) =\frac{w^{n-i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}.$

Now consider a possible Deviation 2. Specifically, $\mu _{B}^{d}\left( D_{i}^{b}\right) >0.$ Note that all Battlefield 2 allocations in $D_{i}^{b}$ are strictly greater than $f\left( h^{n-i-1}\left( E_{1}\right) \right) .$ Define $S_{i}\left( \delta \right) =\left\{ \left( b_{1},b_{2}\right) :\;\left( b_{1},b_{2}\right) \in D_{i}^{b}\; and\; b_{2}\ge f\left( h^{n-i-1}\left( E_{1}\right) \right) +\delta \right\} .$ We are then assured that $\exists \delta >0$ sufficiently small that $\mu _{B}^{d}\left( S_{i}\left( \delta \right) \right) >0$. Then we are also assured that $\exists k^{*}\in \mathbb {N}$ such that $f\left( h^{n-i-1}\left( E_{1}\right) \right) +\delta >g\left( h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{k^{*}}\right) \right)$.²⁶ Note that Enemy plays $\left( h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{k^{*}}\right) ,g\left( h^{n-i}\left( E_{1}-\left( n+1-i\right) \frac{\epsilon }{k^{*}}\right) \right) \right)$ with positive probability in $\mu _{E}^{k^{*}}.$ When he does so the probability that he wins on Battlefield 1 is $\mu _{B}(T_{1}^{b}\cup \ldots \cup T_{i-1}^{b})=\frac{\sum _{j=n-(i-1)}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}$, but the probability he wins on Battlefield 2 is weakly less than $1-\mu _{B}(T_{1}^{b}\cup \ldots \cup T_{i-1}^{b}\cup S_{i}(\delta ))<\frac{\sum _{j=0}^{n-i-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}}.$ Therefore his expected payoff is strictly less than $\frac{\sum _{j=1}^{n-1}w^{j}}{\sum _{j=0}^{n-1}w^{j}},$ his constant equilibrium payoff, a contradiction.

This analysis holds for all $i=1,\ldots ,n-1$. Therefore, given Lemma 5 and the definitions of the $T_{i}^{b}$’s, $D_{i}^{b}$’s, and Deviation 1, we have now determined the mass over sets containing all feasible Blotto allocations other than those in $T_{n}^{b}$ (see Fig. 13 for a graphical aide). Thus, $\mu _{B}^{d}\left( T_{n}^{b}\right) =\frac{1}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}},$ the remaining mass. Therefore, $\mu _{B}^{d}$ does not violate Property 1b. Since we’ve already shown it can’t violate Property 2b, $\mu _{B}^{d}\in \Omega ^{B}.$ $\square$

Recall Theorem 3: The complete set of Nash Equilibria of any two-battlefield Colonel Blotto game is the set of pairs $\left\{ \mu _{B},\mu _{E}\right\}$ such that $\mu _{B}\in \Omega ^{B}$ and $\mu _{E}\in \Omega ^{E}.$

Proof

The theorem follows directly from Lemmas 10, 11, and 12. $\square$

Appendix 2: Method of equilibrium strategy construction

1.1 Blotto construction

Here we will demonstrate how to construct an equilibrium Blotto strategy from any continuous CDF whose range is $\left[ 0,1\right]$ over the domain of $b_{1}$ within $T_{1}^{b}$.

Let $z_{1}\left( b_{1}\right)$ be any continuous, strictly increasing function over the domain $\left[ h^{n-1}\left( E_{1}\right) ,f^{-1}\left( E_{2}\right) \right]$ such that:

$$\begin{aligned} z_{1}\left( b_{1}\right)&= 0\;\quad if\; b_{1}\le h^{n-1}\left( E_{1}\right) \end{aligned}$$

(54)

$$\begin{aligned} z_{1}\left( b_{1}\right)&= 1\;\quad if\; b_{1}\ge f^{-1}\left( E_{2}\right) . \end{aligned}$$

(55)

So, $z_{1}$ is a continuous, strictly increasing function with a range of $\left[ 0,1\right]$ over the domain of $b_{1}$ within $T_{1}^{b}.$ We let this represent a CDF of $b_{1}$ values that Blotto might play within $T_{1}^{b}.$ Now $\forall i=2,\ldots ,n$ we iteratively define:

$$\begin{aligned} z_{n}\left( h^{-(n-1)}\left( b_{1}\right) \right) \equiv \ldots z_{i}\left( h^{-(i-1)}\left( b_{1}\right) \right) \ldots \equiv z_{2}\left( h^{-\left( 2-1\right) }\left( b_{1}\right) \right) \equiv z_{1}\left( b_{1}\right) \qquad \end{aligned}$$

(56)

Each $z_{i}$ represents a CDF over $b_{1}$ values that Blotto might play within $T_{i}^{b}.$ Note that by Eq. 26 the lower and upper bounds of $b_{1}$ values in $\Omega \left( T_{i}^{b}\right)$ are $h^{n-i}(E_{1})$ and $f^{-1}\left( p^{i-1}(E_{2})\right)$ respectively. Then, by Eqs. 54, and 56:

$$\begin{aligned} z_{i}\left( h^{n-i}\left( E_{1}\right) \right) =z_{i}\left( h^{-\left( i-1\right) }\left( h^{n-1}\left( E_{1}\right) \right) \right) =z_{1}\left( h^{n-1}\left( E_{1}\right) \right) =0. \end{aligned}$$

Note that by definition we have that $f^{-1}\left( p^{j}\left( x\right) \right) =h^{-j}\left( f^{-1}\left( x\right) \right) .$ Therefore, by Eqs. 55, and 56:

$$\begin{aligned} z_{i}\left( f^{-1}\left( p^{i-1}(E_{2})\right) \right) =z_{i}\left( h^{-\left( i-1\right) }\left( f^{-1}\left( E_{2}\right) \right) \right) =z_{1}\left( f^{-1}\left( E_{2}\right) \right) =1. \end{aligned}$$

So, given this construction method, $z_{i}$ equals zero at the lower bound of $b_{1}$ values in $\Psi _{1}\left( T_{i}^{b}\right)$ and equals 1 for the upper bound of $b_{1}$ values in $\Psi _{1}\left( T_{i}^{b}\right) .$ Also note that all $z_{i}$ are strictly increasing functions over $\Psi _{1}\left( T_{i}^{b}\right)$ as $h^{-\left( i-1\right) }$ and $z_{1}$ are strictly increasing functions. Therefore, each $z_{i}$ is a CDF over $b_{1}$ values that Blotto might play within $\Psi _{1}\left( T_{i}^{b}\right) .$

Proposition 13

For any $z_{1},$ the following mixed strategy, $\mu _{B}^{*},$ is in $\Omega ^{B}:$ Blotto chooses $i\in \{1,2,\ldots ,n\}$ with the probability he chooses any particular $i$ being given by

$$\begin{aligned} \frac{w^{n-i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}. \end{aligned}$$

He then randomly chooses a $b_{1}$ within $\Psi _{1}\left( T_{i}^{b}\right)$ according to the CDF $z_{i}.$ He then allocates $\left( b_{1},f\left( b_{1}\right) \right)$.

Proof

This strategy trivially satisfies Property 1b. We now show that such a strategy satisfies Property 2b. For some $i\in 1,2,\ldots ,n-1$ consider an $x\in [h^{n-i}(E_{1}),f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ]=\Psi _{1}\left( T_{i}^{b}\right) .$ By construction, and Eq. 20 $\mu _{B}^{*}\left( j_{b}^{x,i}\right) =z_{i}\left( x\right) \cdot \frac{w^{n-i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}.$ Note that since under this strategy Blotto always fully expends his resources, the set of allocations such that $b_{2}\ge g\left( x\right)$ is equivalent to the set of allocations such that $b_{1}\le f^{-1}\left( g\left( x\right) \right) .$ Therefore, given Eq. 21 $\mu _{B}^{*}\left( k_{b}^{x,i}\right) =z_{i+1}\left( f^{-1}\left( g\left( x\right) \right) \right) \cdot \frac{w^{n-i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}.$ Note that $h^{-1}\left( x\right) =f^{-1}\left( g\left( x\right) \right) .$ Therefore, by construction $z_{i+1}\left( f^{-1}\left( g\left( x\right) \right) \right) =z_{i}\left( x\right)$, and $\mu _{B}^{*}\left( j_{b}^{x,i}\right) =\mu _{B}^{*}\left( k_{b}^{x,i}\right) \cdot w.$ Thus Property 2b is satisfied. $\square$

1.2 Enemy construction

Here we will demonstrate how to construct an equilibrium Enemy strategy from any continuous CDF whose range is $\left[ 0,1\right]$ over $\Psi _{1}\left( T_{1}^{e}\right) .$

Let $v_{1}\left( e_{1}\right)$ be any continuous, strictly increasing function such that:

$$\begin{aligned} v_{1}\left( e_{1}\right) =0\;\quad if\; e_{1}\le 0 \end{aligned}$$

(57)

$$\begin{aligned} v_{1}\left( e_{1}\right) =1\;\quad if\; e_{1}\ge h^{n-1}\left( E_{1}\right) . \end{aligned}$$

(58)

So, $v_{1}$ is a continuous, strictly increasing function with a range of $\left[ 0,1\right]$ over the domain of $e_{1}$ within $T_{1}^{e}.$ We let this represent a CDF of $e_{1}$ values that Enemy might play within $T_{1}^{e}.$ Now $\forall i=2,\ldots ,n$ we iteratively define:

$$\begin{aligned} v_{n}\left( h^{-(n-1)}\left( e_{1}\right) \right) \equiv \ldots v_{i}\left( h^{-(i-1)}\left( e_{1}\right) \right) \ldots \equiv v_{2}\left( h^{-\left( 2-1\right) }\left( e_{1}\right) \right) \equiv v_{1}\left( e_{1}\right) \end{aligned}$$

(59)

Note that by Eqs. 28, 30, and 32 the lower and upper bounds of $e_{1}$ values within $T_{i}^{e}$ are $f^{-1}\left( p^{i-2}(E_{2})\right)$ and $h^{n-i}(E_{1})$ respectively. Then, by Eqs. 57, and 59:

$$\begin{aligned} v_{i}\left( h^{n-i}\left( E_{1}\right) \right) =v_{i}\left( h^{-\left( i-1\right) }\left( h^{n-1}\left( E_{1}\right) \right) \right) =v_{1}\left( h^{n-1}\left( E_{1}\right) \right) =1. \end{aligned}$$

Note that by definition we have that $f^{-1}\left( p^{j}\left( x\right) \right) =h^{-j}\left( f^{-1}\left( x\right) \right) .$ Therefore, by Eqs. 58, and 59:

$$\begin{aligned} v_{i}\left( f^{-1}\left( p^{i-2}(E_{2})\right) \right)&= v_{i}\left( h^{-\left( i-2\right) }\left( f^{-1}\left( E_{2}\right) \right) \right) =v_{1}\left( h\left( f^{-1}\left( E_{2}\right) \right) \right) \\&= v_{1}\left( g^{-1}\left( f\left( f^{-1}\left( E_{2}\right) \right) \right) \right) =v_{1}\left( 0\right) =0. \end{aligned}$$

So, given this construction method, $v_{i}$ equals zero at the lower bound of $e_{1}$values in $\Psi _{1}\left( T_{i}^{e}\right)$ and equals 1 for the upper bound of $e_{1}$ values in $\Psi _{1}\left( T_{i}^{e}\right) .$ Also note that all $v_{i}$ are strictly increasing over $\Psi _{1}\left( T_{i}^{e}\right)$ as $h^{-\left( i-1\right) }$ and $v_{1}$ are strictly increasing functions. Therefore, each $v_{i}$ is a CDF over $e_{1}$ values that Enemy might play within $\Psi _{1}\left( T_{i}^{e}\right) .$

Proposition 14

For any $v_{1},$ the following mixed strategy, $\mu _{E}^{*},$ is in $\Omega ^{e}:$ Enemy chooses $i\in \{1,2,\ldots ,n\}$ with the probability he chooses any particular $i$ being given by

$$\begin{aligned} \frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}. \end{aligned}$$

He then randomly chooses a $e_{1}$ within $\Psi _{1}\left( T_{i}^{e}\right)$ according to the CDF $v_{i}.$ He then allocates $\left( e_{1},g\left( e_{1}\right) \right)$.

Proof

This strategy trivially satisfies Property 1e. We now show that such a strategy satisfies Property 2e. For some $i\in 1,2,\ldots ,n-1$ consider an $x\in \left( f^{-1}\left( p^{i-1}\left( E_{2}\right) \right) ,h^{n-i-1}\left( E_{1}\right) \right) =\Psi _{1}\left( T_{i+1}^{e}\right)$. ²⁷ By construction, and Eq. 22 $\mu _{E}^{*}\left( j_{e}^{x,i}\right) =v_{i+1}\left( x\right) \cdot \frac{w^{i}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}.$ Note that since under this strategy Enemy always fully expends his resources, the restriction that $e_{2}\ge f\left( x\right)$ is equivalent to the restriction that $e_{1}\le g^{-1}\left( f\left( x\right) \right) .$ Therefore, given Eq. 23 $\mu _{E}^{*}\left( k_{e}^{x,i}\right) =v_{i}\left( g^{-1}\left( f\left( x\right) \right) \right) \cdot \frac{w^{i-1}}{{{\sum \nolimits _{j=0}^{n-1}}}w^{j}}.$ Note that $h\left( x\right) =g^{-1}\left( f\left( x\right) \right) .$ Therefore, by construction

$$\begin{aligned} v_{i+1}\left( x\right) =v_{i+1}\left( h^{-1}\left( h\left( x\right) \right) \right) =v_{i}\left( h\left( x\right) \right) =v_{i}\left( g^{-1}\left( f\left( x\right) \right) \right) , \end{aligned}$$

and $\mu _{E}^{*}\left( j_{e}^{x,i}\right) =\mu _{E}^{*}\left( k_{e}^{x,i}\right) \cdot w.$ Thus Property 2e is satisfied and $\mu _{E}^{*}\in \Omega ^{e}$. $\square$