# The Asymmetric Colonel Blotto Game

This paper explores the Nash equilibria of a variant of the Colonel Blotto game, which we call the Asymmetric Colonel Blotto game. In the Colonel Blotto game, two players simultaneously distribute forces across n battlefields. Within each battlefield, the player that allocates the higher level of force wins. The payoff of the game is the proportion of wins on the individual battlefields. In the asymmetric version, the levels of force distributed to the battlefields must be nondecreasing. In this paper, we find a family of Nash equilibria for the case with three battlefields and equal levels of force and prove the uniqueness of the marginal distributions. We also find the unique equilibrium payoff for all possible levels of force in the case with two battlefields, and obtain partial results for the unique equilibrium payoff for asymmetric levels of force in the case with three battlefields.

## Authors

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## 1. Introduction

In this section we discuss the background and origins of the Asymmetric Colonel Blotto game.

The Colonel Blotto game, which originates with Borel in [Bor53], is a constant-sum game involving two players, and , and independent battlefields. distributes a total of units of force among the battlefields, and distributes a total of units of force among the battlefields, in such a way that each player allocates a nonnegative amount of force to each battlefield. The player who sends the higher level of force to a particular battlefield wins that battlefield. The payoff for the whole game is the proportion of the wins on the individual battlefields.

Roberson in [Rob06] characterizes the unique equilibrium payoffs for all (symmetric and asymmetric) configurations of the players’ aggregate levels of force, and characterizes the complete set of equilibrium univariate marginal distributions for most of these configurations for the Colonel Blotto game.

A possible variant of the Colonel Blotto game, which has not been studied before, is the Asymmetric Colonel Blotto game, where the forces distributed among the battlefields must be in non-decreasing order.

The Asymmetric Colonel Blotto game is a constant-sum game involving two players, and , and independent battlefields. distributes units of force among the battlefields in a nondecreasing manner and distributes units of force among the battlefields in a non-decreasing manner. Each player distributes forces without knowing the opponent’s distribution. The player who provides the higher amount of force to a battlefield wins that battlefield. If both players deploy the same amount of force to a battlefield, we declare that battlefield to be a draw, and the payoff of that battlefield is equally distributed among the two players.111As we show in Theorem 3.5

, Nash equilibria of games with equal levels of force do not contain atoms, so the probability that the two players place equal force on some battlefield is 0. Thus we may, if we choose, use a different tie-breaking rule without altering the result in this case.

The payoff for each player is the proportion of battlefields won.

In this paper, we study the Nash equilibria and equilibrium payoffs of Asymmetric Colonel Blotto games.

In Section 3, we find a family of equilibria for the game with three battlefields and equal levels of force, and we prove the uniqueness of the marginal distribution functions. We also prove that in any equilibrium strategies for a game with equal levels of force and at least three battlefields, there are no atoms in the marginal distributions.

In Section 4, we find the unique equilibrium payoffs of all cases of the Asymmetric Colonel Blotto game involving only two battlefields, and in Section 5 we find the unique equilibrium payoffs in the case of three battlefields in certain cases. We conclude with Section 6, where we discuss the difficulties in extending our work to the case of battlefields.

## 2. The model

In this section we introduce the model and related concepts. The definitions in this section are adaptations from those in [Rob06] to the asymmetric version.

### 2.1. Players

Two players, and , simultaneously allocate their forces and across battlefields in a nondecreasing manner. Each player distributes forces without knowing the opponent’s distribution. The player who provides the higher level of force to a battlefield wins that battlefield, gaining a payoff of . If both players deploy the same level of force to a battlefield, that battlefield is a draw and both players gain a payoff of . The payoff for each player is the proportion of battlefields won, or equivalently, the sum of the payoffs across all the battlefields.222

That the payoff for each player is the sum of the payoffs across all the battlefields means that two different joint distributions are equivalent if they have the same marginal distributions. Hence, this definition makes it possible to separate a joint distribution into the marginal distributions and a

-copula later in this paper.

Player sends units to the th battlefield. For player , the set of feasible allocations of force across the battlefields in the Asymmetric Colonel Blotto game is denoted by :

 Bi={x∈Rn ∣∣ ∣∣ n∑j=1xji=Xi,0≤x1≤x2≤⋯≤xn}.
###### Definition 2.1.

Given an

-variate cumulative distribution function

, for every such that for all , the -volume of the -box is,

 VH([x,y])=Δnynxnn−1Δyn−1xn−1…Δ2y2x2Δ1y1x1H(t),

where

 ΔkykxkH(t)=H(t1,…,tk−1,yk,tk+1,…,tn)−H(t1,…,tk−1,xk,tk+1,…,tn).

Intuitively, the -volume of a -box just measures the probability that a point within that -box will be chosen given the cumulative distribution function .

###### Definition 2.2.

The support of an -variate cumulative distribution function is the complement of the union of all open sets of with -volume zero. Intuitively, the support of a mixed strategy is just the closure of the set of pure strategies that might be chosen.

### 2.2. Strategies

A mixed strategy, or a distribution of force, for player is an -variate cumulative distribution function (cdf) with support in the set of feasible allocations of force . This means that if player chooses strategy , then the probability that is . has marginal cumulative distribution functions , one univariate marginal cumulative distribution function for each battle field . is the probability that . Equivalently, , where the th argument is , and the rest of the arguments are , the player’s entire allocation of force. We write .

In the case where the mixed strategy is the combination of finite pure strategies, the mixed strategy where units of force are distributed the battlefields respectively with probability is denoted by

 Pi={((i1j,i2j,…,inj),pj)}.

Here and .

### 2.3. The Asymmetric Colonel Blotto Game

The Asymmetric Colonel Blotto game with battlefields, denoted by

 ACB(XA,XB,n),

is a one-shot game in which players simultaneously and independently announce distributions of force subject to their budget constraints and , for each , and such that for . Each battlefield, providing a payoff of , is won by the player that provides the higher allocation of force on that battlefield (and declared a draw if both players allocate the same level of force to a battlefield, each gaining a payoff of ), and players’ payoffs equal the sum of the payoffs over all the battlefields.

### 2.4. Nash equilibrium

Mixed strategies and form a Nash equilibrium if and only if neither player can increase payoff by changing to a different strategy.

Since this particular game is two-player and constant-sum, it has the interesting property that the equilibrium payoff is always unique:

###### Theorem 2.1.

The Nash equilibrium payoff for both players of any two-player and constant-sum game is unique.

###### Proof.

Suppose and is a pair of Nash equilibrium strategies. Let be the payoff for player . For any pair of Nash equilibrium strategies and , let be the payoff for player .

Let us consider the payoff for both players when player plays strategy and player plays strategy . Call the payoff for player and the payoff for player . Since is a strategy in a Nash equilibrium, . Similarly, . So . Since we are considering a constant sum game, . Hence, . Similarly, we must have . So . Similarly . ∎

## 3. Optimal univariate marginal distributions for three battlefields

In this section we use copulas to separate the joint distributions of players into the marginal distributions and suitable copula. We also find and prove the unique univariate marginal distribution for .

Let us first introduce the concept of copulas:

###### Definition 3.1.

Let denote the unit interval . An -copula is a function from to such that

1. For all , if at least one coordinate of is ; and if all coordinates of are except , then .

2. For every such that for all , the -volume of the -box satisfies

 VC([x,y])≥0.

The crucial property of -copulas that we need is the following theorem of Sklar:

###### Theorem 3.1 (Sklar, [Skl59]).

Let be an -variate distribution function with univariate marginal distribution functions . Then there exists an -copula such that for all ,

 H(x1,…,xn)=C(F1(x1),…,Fn(xn)). (1)

Conversely, if is an -copula and are univariate distribution functions, then the function defined by equation 1 is an -variate distribution function with univariate marginal distribution functions .

The proof of this theorem can be found in [SS83].

This theorem establishes the equivalence between a joint distribution on the one hand, and a combination of a complete set of marginal distributions and a -copula on the other hand. We will now show that the univariate marginal distribution functions and the -copula are separate components of the players’ best responses.

###### Proposition 3.2.

In the game , suppose that the opponent’s strategy is fixed as the distribution , and that . Then, in order for player to maximize payoff under the constraint that the support of the chosen strategy must be in , player must solve an optimization problem. Given that there are no atoms in Nash equilibrium strategies (Theorem 3.5), we can write the Lagrangian for this optimization problem as

 (2)

where the set of univariate marginal distribution functions satisfy the constraint that there exists an -copula such that the support of the -variate distribution

 C(F1i(x1),…,Fni(xn))

is contained in .333Here we only maximize over the set of that satisfy the constraint, not all of them.

###### Proof.

The payoff for player given the opponent’s marginal distribution functions is the sum of the payoffs across all the battlefields:

 n∑j=1∫∞01nFj−i(x)dFji.

Here, the integral is the Riemann-Stieltjes integral, so the integrand is for . We also use the Riemann-Stieltjes integral for other integrals later in the paper.

 maxPin∑j=1∫∞01nFj−i(x)dFji.

That is contained in implies that the sum of the levels of force across all battlefields is :

 n∑j=1∫∞0x dFji=Xi.

Hence, the Lagrangian is

 maxPi[n∑j=1∫∞01nFj−i(x)dFji−λi[n∑j=1∫∞0x dFji−Xi]]=max{Fji}nj=1λin∑j=1[∫∞0[1nλiFj−i(x)−x]dFji]+λiXi.

Finally, from Theorem 3.1 the -variate distribution function is equivalent to the set of univariate marginal distribution functions combined with an appropriate -copula, , so the result follows directly. ∎

###### Theorem 3.3.

The unique Nash equilibrium univariate marginal distribution functions of the game are for each player to allocate forces according to the following univariate distribution functions:

 F1(u) =⎧⎨⎩3u0≤u≤13113

The expected payoff for both players is .

This means that any equilibrium strategies must have the marginal distributions described above, and that any joint distribution with support in with such marginal distributions is an equilibrium strategy.

Intuitively, it is easy to see why this particular set of marginal distributions might guarantee a Nash equilibrium. Since the distribution density is the same among the three battlefields, the payoff of a pure strategy remains constant at when it changes inside the region , , and . A player can only hope to increase payoff above that given by by moving below the lower bound of the marginal distribution in some battlefield and staying inside the bounds of the marginal distributions in the other battlefields. However, this is impossible: cannot be negative; any attempt to bring below would result in being above the upper bound ; , as the biggest of the 3, cannot be below . (The rigorous proof of this can be found in Lemma 3.6.)

Before we give the formal proof of this theorem, let us first examine some joint distributions that satisfy the conditions in Theorem 3.3.

Consider the -variate distribution function that uniformly places mass on each of the three sides of the equilateral triangle with vertices , , and to (Depicted in Figure 0(b)). Clearly its marginal distributions are those described in Theorem 3.3.

Similarly, as in Figure 0(c), divide the original equilateral triangle into three smaller equilateral triangles with side lengths of the original, and let

be the strategy that uniformly distribute on the sides of the smaller triangles. Clearly

has the same marginal distributions as , and is thus a joint distribution as described in Theorem 3.3. As shown in Figure 0(d), we can continue this process on the smaller triangles (or only on some of the smaller triangles), and thus we obtain an countably-infinite family of joint distributions with marginal distributions as described in Theorem 3.3. Furthermore, given any two such suitable joint distributions, their weighted average is also a suitable joint distribution, and thus we obtain a continuum of suitable joint distributions.

Given these joint distributions that have marginal distributions as characterized by Theorem 3.3, we have the following theorem:

###### Theorem 3.4.

For the unique set of equilibrium univariate marginal distribution functions characterized in Theorem 3.3, there exists a -copula such that the support of the -variate distribution function

 C(F1i(x1),F2i(x2),F3i(x3))

is contained in .

###### Proof.

Consider the -variate distribution function that uniformly places mass on each of the three sides of the equilateral triangle with vertices , , and to (depicted in Figure 0(b)). Clearly its marginal distributions are those described in Theorem 3.3, and its support is in . Hence, according to Sklar’s theorem (Theorem 3.1), for the unique set of equilibrium univariate marginal distribution functions characterized in Theorem 3.3, there exists a -copula such that the support of the -variate distribution function is contained in .

Before we provide the formal proof of Theorem 3.3, we first seek to provide some intuition for the outline of the proof, which takes inspiration from the proofs in [Rob06] and [BKdV96].

From equation 2 in Proposition 3.2, we know that in an Asymmetric Colonel Blotto game , each player’s Lagrangian can be written as

subject to the constraint that there exists an -copula, , such that the support of the -variate distribution is contained in . If there exists a suitable -copula, then, for different , is independent. So equation 3 is the maximization of three independent sums, hence the sum of three independent maximizations:

 max{Fji}3j=1λi3∑j=1[∫∞0[13λiFj−i(x)−x]dFji]+λiXi=3∑j=1maxFjiλi∫∞0[13λiFj−i(x)−x]dFji+λiXi.

Hence we have reduced the maximization problem over a joint distribution to separate maximization problems over univariate distributions, which can be easily solved.

Note that each separate maximization problem has the same form as that of an all-pay auction. An all-pay auction is an auction where several players simultaneously call out a bid for a prize, and all bidders pay regardless of who wins the prize; the prize is awarded to the highest bidder. In an all-pay auction with two bidders, let represent bidder ’s distribution of the bid, and represent the value of the auction for bidder . Each bidder ’s problem is

 maxFi∫∞0[viF−i(x)−x]dFi.

In the separate maximization problems for the Asymmetric Colonel Blotto game, the quantity acts as the value for the all-pay auctions. Lemma 3.13 establishes the uniqueness of the Lagrange multipliers, hence the uniqueness of the value .

A potential issue that arises is whether the constraint that the strategy must be in leads to equilibria outside those characterized by Theorem 3.3. From Sklar’s Theorem (Theorem 3.1), we know that the joint distribution is equivalent to a set of marginal distributions , together with a suitable -copula . So if a suitable -copula exists, the constraint that be in places no restraint on the set of potential univariate marginal distribution functions, ; instead, this constraint and the set of univariate marginal distributions places a restraint on the set of feasible -copulas. Since Theorem 3.4 establishes the existence of suitable -copula, this is not an issue.

On the other hand, the restriction on the -copula implies that the set of equilibrium -variate distributions for the game forms a strict subset of the set of all -variate distribution functions with univariate marginal distribution functions characterized by Theorem 3.3.

The proof of Theorem 3.3 under the assumption that suitable -copula exists is contained in the results that fill up the rest of this section. The proof takes inspiration from the proofs found in [Rob06] and [BKdV96].

First, for the form of the Lagrangian in Proposition 3.2 to be accurate, we need show that there are no atoms in any Nash equilibrium strategies. The following theorem proves this in the more general case of equal levels of force for both players and any number of battlefields where :

###### Theorem 3.5.

If , then Nash equilibrium strategies for cannot contain atoms.

###### Proof.

Suppose that we have an equilibrium strategy with an atom in battlefield on .

Let be any pure strategy in the support of that contains playing on battlefield . The general idea of this proof will be to find a pure strategy that does strictly better against than , thus reaching a contradiction that cannot be an equilibrium strategy as we supposed.

Let denote the possibility of choosing on battlefield in . If is any point that is not an atom and is greater than (or greater than in the case of ), then consider the pure strategy that plays lower on battlefield and plays higher on battlefield and all the battlefields after that. We can always find sufficiently small positive and such that there is no atom between and in . So the payoff of against minus the payoff of against is at least

 1nfk(aj)−δ

for any . Hence, does strictly better than against .

Therefore, for to be an equilibrium strategy, all such that are not atoms must be equal to (or if ). So every pure strategy in the support of containing the atom on battlefield must be of the following form: a series of zeros in the first few battlefields (possibly none), an atom, the same level of force in the next few battlefields (also possibly none), another atom, the same level of force (as in the previous atom) in the next few battlefields, and so forth.

Now, one of the following statements must be true:

1. All in the support of containing the atom on battlefield is played with probability .

2. There exists some in the support of containing the atom on battlefield that is played with a positive probability, hence every is an atom on battlefield .

Suppose that statement is true. For to be played with some positive probability, there must be a continuum of such . Hence there must also be a continuum of atoms, which is clearly impossible. So statement must be true. Let be such a pure strategy in the support of where every is an atom on battlefield . Some casework is needed here:

1. All the are the same. Then they must all be . In any pure strategy where player plays on the first battlefield, he must also play on all the other battlefields. So for all . Hence,

 f1(1n)

Consider the pure strategy that plays on battlefield and plays on all the other battlefields. We can find a sufficiently small positive such that there are no atoms between and on battlefield . The payoff of against minus the payoff of against is at least

 1n⋅(n∑k=2fk(1n)−f1(1n))−δ

for any . So does strictly better against than .

2. All the fall into exactly two values, and . () Suppose contains battlefields with level of force and then battlefields with level of force .i

1. If , then . Given any pure strategy in the support of that plays on battlefield , it must also play on all the battlefields after that, and play on the battlefields to . So where . Hence,

 ∑k≠m+1fk(ck)>fm+1(d1=cm+1).

Consider the pure strategy that plays on battlefield and plays on battlefield for all . We can find a sufficiently small positive such that there are no atoms between and on battlefield . The payoff of against minus the payoff of against is at least:

 1n⋅⎛⎝∑k≠m+1fk(ck)−fm+1(d1)⎞⎠−δ

for any . So does strictly better against than .

2. If , then at least one of the following two must be true:

1.  fm+1(cm+1)<∑k≠m+1fk(ck).
2.  n∑k=m+1fk(ck)>m∑k=1fk(ck).

Similar to the arguments above, if the first one is true, then we can construct a by playing lower on battlefield and higher on all the other battlefields; if the second one is true, then we can construct a by playing higher on battlefield and all the battlefields after that, and playing lower on battlefields to . In either case, does strictly better than against .

3. All the fall into at least three different values. So from these values we can choose two different values that are not zero. Then we apply the proof in item 2b and obtain the needed pure strategy .

In all the cases, a contradiction is reached, showing that cannot be an equilibrium strategy. ∎

In the following discussions, let be any joint distribution characterized in Theorem 3.3, and let be any equilibrium strategy. Our goal is to prove that is an equilibrium strategy, and that and have the same marginal distributions.

###### Lemma 3.6.

Suppose is any pure strategy in . Then the payoff of against is if , , and ; and the payoff is less than otherwise.

###### Proof.

Suppose plays the mixed strategy and plays the pure strategy , where and . Then, let be the payoff for . So

 W(a,b,c)=13(F1(a)+F2(b)+F3(c))

Our goal is to find the maximum value of in and to show that it is no greater than .

Clearly, , so . And

• If , then , so and . And

 W(a,b,c) =13(3a+0+1) <13(3⋅16+0+1) =12.

So .

• If , then . Since , .

 W(a,b,c) ≤13(3a−12+3b+3c−1) =(a+b+c)−12 =12.

Equality holds if and only if , which is equivalent to . In this case , , and .

Otherwise, equality does not hold, and the payoff is less than .

###### Lemma 3.7.

Any joint strategy as characterized in Theorem 3.3 is a Nash equilibrium strategy.

###### Proof.

We know that the game is symmetrical and has constant sum , and since Lemma 3.6 indicates that gives a payoff of at least against any pure strategy, so must be an equilibrium strategy. ∎

Let , , , , , . Clearly, is just the upper bound of on battlefield , and is the lower bound.

for and all .

###### Proof.

This is self-evident from the representation of in Theorem 3.3:

 F1(u) ={3u0≤u≤13113

###### Lemma 3.9.

If , then . If , then . Or, in other words, does not place any strategy outside .

###### Proof.

Since is a two player symmetric constant sum game, every pure strategy in the support of , an equilibrium strategy, must give the unique equilibrium payoff, , when played against another equilibrium strategy, . From Lemma 3.6 we know that a pure strategy only gives payoff against when plays a level of force between and on battlefield for all . So cannot play any strategy outside that range. ∎

and .

###### Proof.

Theorem 3.5 implies that is continuous. This, together with Lemma 3.9, gives the desired result. ∎

Let us recall player ’s optimization problem for (equation 2 in Proposition 3.2):

 (3)

where the set of univariate marginal distribution functions satisfy the constraint that there exists a -copula such that the support of the -variate distribution

 C(F1i(x1),F2i(x2),F3i(x3))

is contained in .

From the lemmas above, we can add some further restrictions to it. From Lemma 3.9, we know that must be played within for every battlefield . From Lemma 3.7, we know that is an equilibrium strategy, so must be a best response against and vice versa. Since Theorem 3.4 establishes the existence of suitable -copula, we can disregard that restriction for now and focus on the rest.

For different , is independent. So equation 3 is the maximization of three independent sums, hence the sum of three independent maximizations:

 max{Fji}3j=1λi3∑j=1[∫∞0[13λiFj−i(x)−x]dFji]+λiXi=3∑j=1maxFjiλi∫∞0[13λiFj−i(x)−x]dFji+λiXi.

The term is just a constant, so we could throw that away. Thus the problem for player becomes:

 maxFjiλi∫∞0[13λiFj−i(x)−x]dFji

for all battlefields , under the constraint that is a best response against , is a best response against , and is played within . Let us set . Since we assume the existence of a suitable -copula, the different can be considered independent and the different maximizations for different battlefields can also be considered independent. Hence, and form an equilibrium for all .

Let . This is the payoff for player by playing when player plays in the maximization problem for battlefield .

###### Lemma 3.11.

is constant for all .

###### Proof.

Since is an equilibrium strategy against , every strategy in the support of gives a constant payoff against . Since the support of is , the result directly follows. ∎