[1,2]Jan-Tino Brethouwer

1]Institute of Applied Mathematics, Delft University of Technology, Delft,
2628 CD, The Netherlands

2]Faculty of Military Sciences, Netherlands Defence Academy, Breda, 4818 CR,
The Netherlands

# General Lotto Games with Scouts: Information versus Strength

<EMAIL_ADDRESS>Bart van Ginkel  Roy Lindelauf  [  [

###### Abstract

We introduce General Lotto games with Scouts: a General Lotto game with
asymmetric information. There are two players, Red and Blue, who both allocate
resources to a field. However, scouting capabilities afford Blue to gain
information, with some probability, on the number of Red’s resources before
allocating his own. We derive optimal strategies for this game in the case of
a single field. In addition we provide upper and lower bounds of the value of
the game in a multi-stage case with multiple battlefields. We devise several
ways to characterise the influence of information versus strength. We conclude
by drawing qualitative insights from these characterisations and the game
values, and draw parallels with military practice.

###### keywords:

General Lotto, Colonel Blotto, Game Theory, Asymmetric information

##  1 Introduction

The Colonel Blotto game in its most basic form was introduced by [1]. The main
idea is that two opposing Colonels simultaneously allocate a fixed amount of
resources over a finite number of fields. Each field is won by whichever
player sends the most resources there, and the goal is to win as many fields
as possible. The Colonel Blotto game and its variations and generalisations
have been used to model various strategic situations, ranging from division of
troops and the placement of missiles ([2]) to budget allocation during
elections ([3, 4]).

Borel introduced two versions of the Colonel Blotto game in his original
paper: the discrete Colonel Blotto game, where the number of resources sent to
a field has to be an integer, and the continuous Colonel Blotto game, where
the number of resources can be any non-negative number. The first solutions to
the continuous Colonel Blotto game were found by [5]. They solved the game
completely for 2 fields. They also provided solutions for 3 or more fields,
under the condition that both sides have an equal number of resources. The
continuous Blotto game was eventually completely solved by [6], allowing for
any number of fields and asymmetric resource numbers between players. The
discrete version turned out to be quite difficult to solve, as even now it is
considered very computationally intensive to find solutions. There is a
substantial literature on the computational complexity of the discrete Blotto
game, see [7] and the references therein. Even so, [8] showed that the
equilibrium payoff for two-battlefield games is generally the same in the
continuous version as in the discrete version.

[9] introduced a more computable variant of Colonel Blotto games, which he
called General Lotto games. These games relax the budget constraint on the
number of resources allocated to a field, only requiring it to hold in
expectation. Hart showed how these games could be used to find optimal
strategies for symmetric discrete Colonel Blotto games. While General Lotto
games were introduced as a tool to study Colonel Blotto games, they have shown
to be interesting games in their own right.

In standard General Lotto games, the resources allocated to each of the fields
are revealed simultaneously. In practice, however, this need not always be the
case. Intelligence, surveillance and reconnaissance efforts might unveil
adversary troop allocations beforehand. A reconnaissance drone or satellite
image could identify troops en route to the field, a spy might uncover the
number of missiles in a specific location, or a leaked document could detail
the budget allocated to a certain state during elections. Knowledge of such
events significantly influences the strategies of both players.

In this paper, we present a new variation of the General Lotto game that
accommodates such information. We consider the following variation of a
General Lotto game. Consider two players, Blue and Red. First, Red allocates a
number of resources to the field, as done in a standard General Lotto game.
Next, Blue may obtain knowledge of Red’s resource allocation, i.e., with some
probability Blue learns the exact number of resources Red has allocated. Blue
can now choose to match this number and win the corresponding field, or
allocate no resources to the field. Since it is a General Lotto game, the
number of resources allocated by the players has to match their budget in
expectation.

Our research is inspired by the real-world scenario of a scouting drone
spotting enemy troops on their way to a field, which is a problem that was
provided to us by the Dutch ministy of Defense. This setting has since then
only grown in relevance, with the war in Ukraine using surveillance drones on
a large scale, making troop movements more transparent than ever ([10]). We
abstracted this to a problem of information versus strength with fixed
budgets.

The goal of this paper is to study the effect of information asymmetry on
resource allocation. Furthermore, we want to gain insights in optimal
configurations of different types of resources: resources that yield strength
(troops) or information (drones).
We focus on two main research questions. First, we assume that the amount of
information and strength is predetermined. With the information that is
provided, what are the optimal strategies? Specifically, how does Blue use his
information efficiently and how does Red change her strategy to deal with Blue
possibly gaining information? We provide several insights and guidelines to
answer to these questions.

The second research question is a generalisation of the first: What if you are
allowed to choose your own configuration of information and strength under a
given budget? This is the so-called weapons mix problem ([11]). You are about
to face a series of battles, and you have to assemble your troops. Given is a
fixed budget to divide between scouting drones which provide information, and
weaponized unmanned vehicles which provide firepower. What is the mix of
information and strength that maximises expected wins?

#### Relevant literature

General Lotto games are a valuable framework for exploring information and
asymmetry in resource allocation. There are many aspects of the game where
there can exist asymmetry: players may have different field valuations,
resources could vary in strength between players, or a player might have a
strategic advantage, such as in our game.

Several extensions of General Lotto Games have been developed to investigate
the effects of these asymmetries. For example, in [12] the players have
asymmetric valuations for the fields, which are known to both players. Another
approach is in [13], where the valuation of fields is the same for both
players, but only one player knows the exact value, and the other player only
knows its distribution.

[14] provided solutions for General Lotto games with favouritism, where some
of the resources are pre-allocated beforehand, and remaining resources have
differing strengths per field and player. The number of resources pre-
allocated for each field is fixed. [15] builds upon this idea, by including
pre-allocation in the strategy. They extend General Lotto games to two-stage
games, where in the first phase a fixed budget of resources must be pre-
allocated over fields, but the player now has freedom to choose how to divide
these resources. In the second stage, these resources are revealed, and the
General Lotto game with favouritism is played.

An alternate extension of General Lotto games are Winner-Take-All games, as
studied by [16]. Where General Lotto games have two players competing, Winner-
Take-All games allow for any number of players. Each player is tasked with
selecting a distribution following predefined constraints. Subsequently,
scores are sampled from these distributions, and the player achieving the
highest score wins. Notably, two player Winner-Take-All games with constraints
on the expectation of the distribution are equivalent to General Lotto games.

#### Main contributions and results

We first study General Lotto games with Scouts on a single field. We show that
based on the amount of resources and the probability that the field is
revealed, the analysis of this game can be split into three different cases.
We provide optimal strategies for all three. We also study a multi-stage
version of the game with multiple fields, which all have their own value and
own probability to be revealed. For this version, we provide upper and lower
bounds on the value of the game. These bounds are tight in some settings. We
devise several ways to compare the value of information and resources. We
conclude with several qualitative insights.

#### Overview of the paper

We first introduce the game model and its notation in Section 2. We focus on
single field General Lotto games with Scouts in Section 3, where we provide
optimal strategies for both players. In Section 4 we move on to a multi-stage
version with multiple fields and provide upper and lower bounds on the value
of this game. In Section 5, we introduce methods to measure information versus
strength. We discuss the insights gained about the value of information in
Section 6. We conclude and discuss future work in Section 7.

##  2 Model, notation and assumptions

As described in the introduction, General Lotto games as introduced by [9] are
a relaxed version of Colonel Blotto games, in the sense that the resource
constraint should only hold in expectation, rather than with probability $1$.
We want to study a version of the General Lotto game where one player receives
information on the resource configuration chosen by the other player. This act
of receiving information is symbolised by the idea of sending a scout. The
game is therefore denoted as $GL\text{-}S$, meaning General Lotto with Scouts.
In this section we first formally introduce the General Lotto game and its
optimal strategies. Afterwards we generalise it to General Lotto with Scouts
on a single field, which exhibits insightful properties. In later sections we
analyse a multi-stage version with multiple fields, which will be introduced
in the relevant section.

###  2.1 General Lotto

Before introducing General Lotto with Scouts, we first define the General
Lotto game. This game involves two players, Red and Blue, competing on a
single field. Blue is endowed with resource budget $B>0$, and Red possesses a
resource budget $R>0$. Red chooses as her strategy a (distribution of a)
nonnegative random variable X with $\mathbb{E}(X)=R$, Blue chooses as his
strategy a (distribution of a) nonnegative random variable $Y$ with $E(Y)=B$.

To determine the winner, two numbers, $r$ and $b$, representing the amount of
resources allocated to the field by Red and Blue respectively, are sampled
from their strategies. The player with the highest amount of allocated
resources wins the game, with ties being resolved in favor of Blue. I.e. the
game is constant-sum with the following pay-off for Blue:

$\displaystyle P(b,r)=\mathbbm{1}(b\geq r).$

The way in which ties are broken is chosen to avoid technical complications
and does not influence the value of the game. We define the value of the game
as the probability Blue wins:

$\displaystyle V=P(X\leq Y)$

This game was solved by [9]. He proved the value of the game and the optimal
strategies were as in Theorem 1.

######  Theorem 1 (Hart).

Let $R\geq B>0$. The value $V$ of the General Lotto game is

$\displaystyle V=\frac{B}{2R}$

and the unique optimal strategies are $X^{*}$ for Red and $Y^{*}$ for Blue,
where

$\displaystyle X^{*}$ $\displaystyle\sim\mathbb{U}[0,2R]$ $\displaystyle
Y^{*}$ $\displaystyle\sim\mathbb{U}[0,2R]\text{ with prob. }\frac{B}{R}\text{,
else }0$

###  2.2 General Lotto with Scouts

We now consider General Lotto games with Scouts, denoted as
$GL\text{-}S(B,R,u)$. Similar to the standard General Lotto game, $GL-S$
involves the two players Blue and Red, with resource budgets $B,R>0$,
competing on a single field. However, in this variant, Blue obtains
information insight into Red’s resource allocation with a probability
$u\in[0,1]$. This influences how both players approach the game.

To describe the feasible strategies, we begin with an assumption and an
observation.

  * •

First, we assume that both players choose their strategies before knowing
whether the number of Red resources will be revealed. This means that the
strategy of Blue consists of two parts: actions in case the field is revealed,
and actions for when it is not. The expectation over these two together has to
match Blue’s total resources $B$.

  * •

Second, note that if the number of resources allocated by Red is revealed,
Blue either allocates the exact same amount of resources (Call) or allocates
no resources at all (Fold). It is clear that any other course of action is
less beneficial for Blue, i.e. is dominated by this strategy.

A strategy for Red, denoted by $\sigma_{R}=(X)$, is a random variable $X$ on
$[0,\infty)$ that satisfies:

$\displaystyle\mathbb{E}(X)\leq R.$

The space of all Reds strategies is denoted by $\Sigma_{R}$. We denote the
number of resources allocated by Red by $T_{R}(X)$.

A strategy for Blue, denoted by $\sigma_{B}=(t,Z)$, consists of a function
$t:[0,\infty)\to[0,1]$ and a random variable $Z$ on $[0,\infty)$. Here $t(x)$
is defined as the probability with which Blue calls if it is revealed that Red
has allocated $x$ resources and $Z$ determines the number of resources that
Blue allocates if the field is not revealed. We denote the number of resources
allocated by Blue by $T_{B}(t,Z,X)$. Now we see that

$\displaystyle T_{B}(t,Z,X)=\begin{cases}X\text{ with prob. $t(X)$, else
0\hskip 14.22636pt}&\text{if the field is revealed}\\\ Z,&\text{if the field
is not revealed }\end{cases}$

Given that Red plays strategy $\sigma_{R}=(X)$ and Blue plays strategy
$\sigma_{B}=(t,Z)$, we see that

$\displaystyle\mathbb{E}T_{B}(t,Z,X)=u\mathbb{E}\left(t(X)X\right)+(1-u)\mathbb{E}(Z).$

Since the Blue resource constraint must hold for any strategy of Red, we
require that

$\displaystyle
u\mathbb{E}\left(t(X)X\right)+(1-u)\mathbb{E}(Z)=\mathbb{E}T_{B}(t,Z,X)\leq B$

for all random variables $X$ on $[0,\infty)$ that satisfy $\mathbb{E}(X)\leq
R$. The space of all such strategies of $B$ is denoted by $\Sigma_{B}$.

The value of the game is the expected payoff to Blue. Given that Blue plays
strategy $\sigma_{B}=(t,Z)$ and Red plays strategy $\sigma_{R}=(X)$, the value
of the game $V$ is given by

$\displaystyle
V=P(\sigma_{B},\sigma_{R})=u\mathbb{E}(t(X))+(1-u)\mathbb{P}(Z\geq X).$

Blue want to find a strategy which maximises the value, and Red want to find a
strategy which minimises the value.

##  3 General Lotto with Scouts: Single Field

In the last section we introduced the $GL\text{-}S$ game. In this section we
find its value and optimal strategies. The main result is given in Theorem 2
at the end of the section.

Based on the resource ratio $B/R$, we split the analysis into the following
three cases:

  * •

Case 1: $GL\text{-}S$ with $1\leq B/R$

  * •

Case 2: $GL\text{-}S$ with $u\leq B/R\leq 1$

  * •

Case 3: $GL\text{-}S$ with $B/R\leq u$

The cases have different optimal strategies and are therefore treated
separately. At the end of the section, the results are combined into Theorem
2. Every time we propose two strategies and show that they both guarantee the
same payoff $V$. Thus these strategies must be optimal and the value of the
game is $V$.

###  3.1 Case 1: $GL\text{-}S$ with $1\leq B/R$

We introduce two new parameters $q(u)$ and $C(u)$:

$\displaystyle q(u)$ $\displaystyle=\frac{1-u}{\left(\frac{B}{R}\right)-u}$
$\displaystyle C(u)$
$\displaystyle=\frac{R}{q(u)}=R\left(\frac{\frac{B}{R}-u}{1-u}\right)=B+u\frac{B-R}{1-u}$

Since $B\geq R$, it follows that $0\leq q(u)\leq 1$ and $C(u)\geq B\geq R$.

We show that $\sigma_{R}^{*}=(X^{*})$ and $\sigma_{R}^{*}=(t^{*},Z^{*})$ are
optimal strategies, where:

$\displaystyle X^{*}$ $\displaystyle\sim\mathbb{U}[0,2C(u)]\text{ with prob.
}q(u)\text{, else }0$ $\displaystyle t^{*}$ $\displaystyle=1\quad\forall x$
$\displaystyle Z^{*}$ $\displaystyle\sim\mathbb{U}[0,2C(u)]$

$\mathbb{U}[a,b]$ denotes the uniform distribution with support $[a,b]$. This
strategy consists of Blue always replicating Reds allocation if the field is
revealed, or Blue plays a uniform distribution if the field is not revealed.
Red plays the same uniform distribution with probability $q(u)$, and 0 with
probability $1-q(u)$.

######  Lemma 1.

These strategies are optimal strategies in the case $\frac{B}{R}\geq 1$, and
the value of the game equals $V=1-\frac{(1-u)^{2}}{2(\frac{B}{R}-u)}$.

###### Proof.

First we show that the resource constraints are satisfied:

$\displaystyle\mathbb{E}T_{R}(X^{*})$ $\displaystyle=$
$\displaystyle\mathbb{E}(X^{*})=q(u)\cdot C(u)+(1-q(u))\cdot
0=q(u)\cdot\frac{R}{q(u)}=R$ $\displaystyle\mathbb{E}T_{B}(t^{*},Z^{*},X)$
$\displaystyle=$ $\displaystyle
u\mathbb{E}\left(t^{*}(X)X\right)+(1-u)\mathbb{E}(Z^{*})=u\cdot R+(1-u)\cdot
C(u)=B,$

where $X$ is an arbitrary non-negative random variable such that
$\mathbb{E}X=R$.
Now it suffices to prove the following two claims.

Claim 1: given that Red plays $\sigma_{R}^{*}$, the payoff is at most $V$.
If Red plays $\sigma_{R}^{*}$, the payoff for any $\sigma_{B}$ is:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$
$\displaystyle=u\mathbb{E}(t(X^{*}))+(1-u)\mathbb{E}\left(\mathbbm{1}[Z\geq
X^{*}]\right)$

We can rewrite $\mathbb{E}\left(\mathbbm{1}[Z\geq X^{*}]\right)$ by
conditioning on $Z$:

$\mathbb{E}\left(\mathbbm{1}[Z\geq
X^{*}]\right)=\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[Z\geq
X^{*}]|Z\right)\right)\\\
=1-q(u)+q(u)\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[Z\geq
U]|Z\right)\right)$

Where $U\sim\mathbb{U}[0,2C(u)]$. We can use this to rewrite as follows:

$\displaystyle\mathbb{E}\left(\mathbbm{1}[Z\geq U]|Z\right)$
$\displaystyle=\left(\frac{Z}{2C(u)}\right)\mathbbm{1}[2C(u)\geq
Z]+1\cdot\mathbbm{1}[2C(u)<Z]$
$\displaystyle=\left(\frac{Z}{2C(u)}\right)\left(1-\mathbbm{1}[2C(u)<Z]\right)+1\cdot\mathbbm{1}[2C(u)<Z]$
$\displaystyle=\left(\frac{Z}{2C(u)}\right)-\left(\frac{Z}{2C(u)}-1\right)\cdot\mathbbm{1}[2C(u)<Z]$

Plugging this back in gives:

$\displaystyle\mathbb{E}\left(\mathbbm{1}[Z\geq X^{*}]\right)$
$\displaystyle=1-q(u)+q(u)\left(\mathbb{E}\left(\frac{Z}{2C(u)}\right)-\mathbb{E}\left(\left(\frac{Z}{2C(u)}-1\right)\cdot\mathbbm{1}[2C(u)<Z]\right)\right)$
$\displaystyle=1-q(u)+q(u)\left(\frac{\mathbb{E}(Z)}{2C(u)}-\mathbb{E}\left(\left(\frac{Z}{2C(u)}-1\right)\cdot\mathbbm{1}[2C(u)<Z]\right)\right)$
$\displaystyle=1-q(u)\left(1-\frac{\mathbb{E}(Z)}{2C(u)}\right)-q(u)\mathbb{E}\left(\left(\frac{Z}{2C(u)}-1\right)\cdot\mathbbm{1}[2C(u)<Z]\right)$
$\displaystyle\leq 1-q(u)\left(1-\frac{\mathbb{E}(Z)}{2C(u)}\right)$

Using this equation, we can rewrite the payoff as:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$ $\displaystyle\leq
u\mathbb{E}(t(X^{*}))+(1-u)\left(1-q(u)\left(1-\frac{\mathbb{E}(Z)}{2C(u)}\right)\right)$
$\displaystyle=u\mathbb{E}(t(X^{*}))+(1-u)(1-q(u))+\frac{q(u)}{2C(u)}\left(B-u\mathbb{E}\left(t(X^{*})X^{*}\right)\right)$
$\displaystyle=u\mathbb{E}\left(t(X^{*})\left(1-\frac{q(u)X^{*}}{2C(u)}\right)\right)+(1-u)(1-q(u))+\frac{q(u)B}{2C(u)}$

Since $0\leq\left(1-\frac{q(u)X^{*}}{2C(u)}\right)$, this expression is
maximised by taking $t(x)=1\forall x$. This yields

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$ $\displaystyle\leq
u\mathbb{E}\left(1-\frac{q(u)X^{*}}{2C(u)}\right)+(1-u)(1-q(u))+\frac{q(u)B}{2C(u)}$
$\displaystyle=1+u-u-q(u)(1-u)+\frac{q(u)}{2C(u)}\left(B-uR\right)$
$\displaystyle=1-q(u)(1-u)+\frac{q(u)}{2}\frac{B-uR}{C(u)}$
$\displaystyle=1-\frac{1}{2}q(u)(1-u)=1-\frac{(1-u)^{2}}{2(\frac{B}{R}-u)}=V$

This proves Claim 1.

Claim 2: given Blue plays $\sigma_{B}^{*}$, the payoff is at least $V$.
If Blue uses the strategy $\sigma_{B}^{*}$, the payoff for any $\sigma_{R}$
is:

$\displaystyle P(\sigma_{B}^{*},\sigma_{R})$
$\displaystyle=u\mathbb{E}(t^{*}(X))+(1-u)\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]\right)$
$\displaystyle=u\mathbb{E}(1)+(1-u)\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]\right)$ $\displaystyle=u+(1-u)\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]\right)$

We can rewrite $\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq X]\right)$ by
conditioning on $X$:

$\displaystyle\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq X]\right)$
$\displaystyle=\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]|X\right)\right)$

Where $Z^{*}\sim\mathbb{U}[0,2C(u)]$. We can use this to rewrite as follows:

$\displaystyle\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]|X\right)\right)$
$\displaystyle=\mathbb{E}\left(\left(1-\frac{X}{2C(u)}\right)\mathbbm{1}[2C(u)\geq
X]+0\cdot\mathbbm{1}[2C(u)<X]\right)$
$\displaystyle=\mathbb{E}\left(\left(1-\frac{X}{2C(u)}\right)\left(1-\mathbbm{1}[2C(u)<X]\right)\right)$
$\displaystyle=\mathbb{E}\left(1-\frac{X}{2C(u)}\right)+\mathbb{E}\left(\mathbbm{1}[2C(u)<X]\left(\frac{X}{2C(u)}-1\right)\right)$
$\displaystyle=1-\frac{R}{2C(u)}+\mathbb{E}\left(\mathbbm{1}[2C(u)<X]\left(\frac{X}{2C(u)}-1\right)\right)$
$\displaystyle\geq 1-\frac{R}{2C(u)}=1-\frac{q(u)}{2}$

The last step follows from the fact that the expectation on the RHS is non-
negative. Plugging this into the original equation for payoff gives:

$\displaystyle P(\sigma_{B}^{*},\sigma_{R})$ $\displaystyle\geq
u+(1-u)\left(1-\frac{q(u)}{2}\right)=1-(1-u)\frac{q(u)}{2}$
$\displaystyle=1-\frac{(1-u)^{2}}{2(\frac{B}{R}-u)}=V$

This proves Claim 2 and therefore completes the proof. ∎

###  3.2 Case 2: $GL\text{-}S$ with $u\leq B/R\leq 1$

This is the most interesting of the three scenario’s. Blue is outnumbered, but
can use information to be competitive. We introduce the following parameter:

$\displaystyle p(u)=\frac{\frac{B}{R}-u}{1-u}=\frac{1}{q(u)}$

Since $u\leq\frac{B}{R}\leq 1$, it follows that $0\leq p(u)\leq 1$.

We show that $\sigma_{R}^{*}=(X^{*})$ and $\sigma_{R}^{*}=(t^{*},Z^{*})$ are
optimal strategies, where:

$\displaystyle X^{*}$ $\displaystyle\sim\mathbb{U}[0,2R]$ $\displaystyle
t^{*}$ $\displaystyle=1\quad\forall x$ $\displaystyle Z^{*}$
$\displaystyle\sim\mathbb{U}[0,2R]\text{ with prob. }p(u)\text{, else }0$

This strategy consists of Blue always replicating Reds allocation if the field
is revealed, or Blue plays a uniform distribution with probability $p(u)$ if
the field is not revealed, and 0 with probability $1-p(u)$. Red always plays
the same uniform distribution, which is equivalent to her strategy in General
Lotto games without scouts.

######  Lemma 2.

These strategies are optimal in the case $u\leq\frac{B}{R}\leq 1$, and the
value of the game equals $V=\frac{1}{2}\left(u+\frac{B}{R}\right)$.

###### Proof.

First we show that the resource constraints are satisfied:

$\displaystyle\mathbb{E}T_{R}(X^{*})$ $\displaystyle=$
$\displaystyle\mathbb{E}(X^{*})=R$
$\displaystyle\mathbb{E}T_{B}(t^{*},Z^{*},X)$ $\displaystyle=$ $\displaystyle
u\mathbb{E}\left(t^{*}(X)X\right)+(1-u)\mathbb{E}(Z^{*})=uR+(1-u)p(u)R=B,$

where $X$ is an arbitrary non-negative random variable such that
$\mathbb{E}X=R$.
Now it suffices to prove the following two claims.

Claim 1: given that Red plays $\sigma_{R}^{*}$, the payoff is at most $V$.
If Red plays $\sigma_{R}^{*}$, the payoff for any $\sigma_{B}$ is:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$
$\displaystyle=u\mathbb{E}(t(X^{*}))+(1-u)\mathbb{E}\left(\mathbbm{1}[Z\geq
X^{*}]\right)$

We can rewrite $\mathbb{E}\left(\mathbbm{1}[Z\geq X^{*}]\right)$ by
conditioning on $Z$:

$\displaystyle\mathbb{E}\left(\mathbbm{1}[Z\geq X^{*}]\right)$
$\displaystyle=\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[Z\geq
X^{*}]|Z\right)\right)$

Where $X^{*}\sim\mathbb{U}[0,2R]$. We can use this to rewrite as follows:

$\displaystyle\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[Z\geq
X^{*}]|Z\right)\right)$
$\displaystyle=\mathbb{E}\left(\left(\frac{Z}{2R}\right)\mathbbm{1}[2R\geq
Z]+1\cdot\mathbbm{1}[2R<Z]\right)$
$\displaystyle=\mathbb{E}\left(\left(\frac{Z}{2R}\right)\left(1-\mathbbm{1}[2R<Z]\right)+1\cdot\mathbbm{1}[2R<Z]\right)$
$\displaystyle=\mathbb{E}\left(\frac{Z}{2R}\right)-\mathbb{E}\left(\left(\frac{Z}{2R}-1\right)\cdot\mathbbm{1}[2R<Z]\right)$
$\displaystyle=\frac{\mathbb{E}(Z)}{2R}-\mathbb{E}\left(\left(\frac{Z}{2R}-1\right)\cdot\mathbbm{1}[2R<Z]\right)$
$\displaystyle\leq\frac{\mathbb{E}(Z)}{2R}$

Using this equation, we can rewrite the payoff as:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$ $\displaystyle\leq
u\mathbb{E}(t(X^{*}))+(1-u)\left(\frac{\mathbb{E}(Z)}{2R}\right)$

If we rewrite the expectation requirement, we get:

$\displaystyle(1-u)\mathbb{E}(Z)=B-u\mathbb{E}\left(t(X)X\right)$

Plugging this in gives:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$ $\displaystyle\leq
u\mathbb{E}(t(X^{*}))+\frac{1}{2R}\left(B-u\mathbb{E}\left(t(X^{*})X^{*}\right)\right)$
$\displaystyle=\frac{B}{2R}+u\mathbb{E}\left(t(X^{*})\left(1-\frac{X^{*}}{2R}\right)\right)$

Since $0\leq\left(1-\frac{X^{*}}{2R}\right)$, the payoff is optimised by
taking $t(\cdot)$ as large as possible. It it therefore maximised by taking
$t(x)=1\forall x$, which gives:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$
$\displaystyle\leq\frac{B}{2R}+u\mathbb{E}\left(1\left(1-\frac{X^{*}}{2R}\right)\right)=\frac{B}{2R}+u\left(1-\frac{R}{2R}\right)$
$\displaystyle=\frac{1}{2}\left(\frac{B}{R}+u\right)=V$

This proves Claim 1.

Claim 2: given that Blue plays $\sigma_{B}^{*}$, the payoff is at least $V$.
If Blue uses the strategy $\sigma_{B}^{*}$, the payoff for any $\sigma_{R}$
is:

$\displaystyle P(\sigma_{B}^{*},\sigma_{R})$
$\displaystyle=u\mathbb{E}(t^{*}(X))+(1-u)\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]\right)$
$\displaystyle=u\mathbb{E}(1)+(1-u)\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]\right)$ $\displaystyle=u+(1-u)\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]\right)$

We can rewrite $\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq X]\right)$ by
conditioning on $X$:

$\displaystyle\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq X]\right)$
$\displaystyle=\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]|X\right)\right)$
$\displaystyle=\mathbb{E}\left(\mathbb{E}\left((1-p(u))\cdot
0+p(u)\mathbbm{1}[U\geq X]|X\right)\right)$
$\displaystyle=p(u)\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[U\geq
X]|X\right)\right)$

Where $U\sim\mathbb{U}[0,2R]$. We can use this to rewrite as follows:

$\displaystyle\mathbb{E}\left(\mathbb{E}\left(\mathbbm{1}[U\geq
X]|X\right)\right)$
$\displaystyle=\mathbb{E}\left(\left(1-\frac{X}{2R}\right)\mathbbm{1}[2R\geq
X]+0\cdot\mathbbm{1}[2R<X]\right)$
$\displaystyle=\mathbb{E}\left(\left(1-\frac{X}{2R}\right)\left(1-\mathbbm{1}[2R<X]\right)\right)$
$\displaystyle=\mathbb{E}\left(1-\frac{X}{2R}\right)+\mathbb{E}\left(\mathbbm{1}[2R<X]\left(\frac{X}{2R}-1\right)\right)$
$\displaystyle=\left(1-\frac{R}{2R}\right)+\mathbb{E}\left(\mathbbm{1}[2R<X]\left(\frac{X}{2R}-1\right)\right)$
$\displaystyle=\frac{1}{2}+\mathbb{E}\left(\mathbbm{1}[2R<X]\left(\frac{X}{2R}-1\right)\right)$
$\displaystyle\geq\frac{1}{2}$

Here the last step follows from the fact that the expectation on the RHS is
non-negative. Plugging this into the original equation for the game payoff
yields

$\displaystyle P(\sigma_{B}^{*},\sigma_{R})$ $\displaystyle\geq
u+(1-u)\left(\frac{p(u)}{2}\right)=u+\frac{\frac{B}{R}-u}{2}$
$\displaystyle=\frac{1}{2}\left(u+\frac{B}{R}\right)=V$

This proves Claim 2 and therefore completes the proof. ∎

###  3.3 Case 3: $GL\text{-}S$ with $B/R\leq u$

In this case Blue is outnumbered, but has a significant amount of information.
The optimal strategies of this case are surprising, as Red will do away with
any randomness in her strategy.

We show that $\sigma_{R}^{*}=(X^{*})$ and $\sigma_{R}^{*}=(t^{*},Z^{*})$ are
optimal strategies, where:

$\displaystyle X^{*}$ $\displaystyle=R$ $\displaystyle t^{*}$
$\displaystyle=\frac{B}{uR}\quad\forall x$ $\displaystyle Z^{*}$
$\displaystyle=0$

This strategy consists of Blue replicating Reds allocation with probability
$\frac{B}{uR}$ whenever the field is revealed, independently of the number of
resources spotted. If the field is not revealed, Blue allocates no resources.
Red always allocates $R$ resources, effectively removing any random element
from her strategy, and thus removing the advantage Blue has with his scouts.

######  Lemma 3.

These strategies are optimal strategies in the case $u\geq\frac{B}{R}$, and
the value of the game is $V=\frac{B}{R}$:

###### Proof.

First we show that the resource constraints are satisfied:

$\displaystyle\mathbb{E}T_{R}(X^{*})$ $\displaystyle=$
$\displaystyle\mathbb{E}(X^{*})=R$
$\displaystyle\mathbb{E}T_{B}(t^{*},Z^{*},X)$ $\displaystyle=$ $\displaystyle
u\mathbb{E}\left(t^{*}(X)X\right)+(1-u)\mathbb{E}(Z^{*})u\left(\frac{B}{uR}\right)\mathbb{E}\left(X\right)+(1-u)\cdot
0=B,$

where $X$ is an arbitrary non-negative random variable such that
$\mathbb{E}X=R$.
Now it suffices to prove the following two claims.

Claim 1: given that Red plays $\sigma_{R}^{*}$, the payoff is at most $V$.
If Red uses the strategy $\sigma_{R}^{*}$, the payoff for any $\sigma_{B}$ is:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$
$\displaystyle=u\mathbb{E}(t(X^{*}))+(1-u)\mathbb{E}\left(\mathbbm{1}[Z\geq
X^{*}]\right)$
$\displaystyle=u\mathbb{E}(t(X^{*}))+(1-u)\mathbb{E}\left(\mathbbm{1}[Z\geq
R]\right)$ $\displaystyle=\frac{1}{R}\left(u\mathbb{E}(R\cdot
t(X^{*}))+(1-u)\mathbb{E}\left(R\cdot\mathbbm{1}[Z\geq R]\right)\right)$

Note that:

$\displaystyle\mathbb{E}(R\cdot t(X^{*}))=\mathbb{E}(t(X^{*})X^{*})$

and

$\displaystyle\mathbb{E}\left(R\cdot\mathbbm{1}[Z\geq
R]\right)\leq\mathbb{E}\left(Z\cdot\mathbbm{1}[Z\geq
R]\right)\leq\mathbb{E}(Z)$

Plugging these into the equation for the payoff gives:

$\displaystyle P(\sigma_{B},\sigma_{R}^{*})$
$\displaystyle\leq\frac{1}{R}\left(u\mathbb{E}\left(t(X^{*})X^{*}\right)+(1-u)\mathbb{E}\left(Z\right)\right)$
$\displaystyle=\frac{1}{R}\left(B\right)=\frac{B}{R}=V,$

where the last step follows from the expectation requirement for Blue.
This proves Claim 1.

Claim 2: given that Blue plays $\sigma_{B}^{*}$, the payoff is at least $V$.
If Blue uses the strategy $\sigma_{B}^{*}$, the payoff for any $\sigma_{R}$
is:

$\displaystyle P(\sigma_{B}^{*},\sigma_{R})$
$\displaystyle=u\mathbb{E}(t^{*}(X))+(1-u)\mathbb{E}\left(\mathbbm{1}[Z^{*}\geq
X]\right)$ $\displaystyle=u\left(\frac{B}{uR}\right)+(1-u)\cdot P[X=0]$
$\displaystyle\geq\frac{B}{R}=V.$

This proves Claim 2 and therefore completes the proof. ∎

As the payoff in this case no longer depends on the $u$, it follows that
increasing the $u$ past $\frac{B}{R}$ no longer increases payoff.

###  3.4 Solution of $GL\text{-}S$

We combine the results of Lemma 1, 2 and 3 to form our main result:

######  Theorem 2.

In $GL\text{-}S$, it is optimal for Blue to play
$\sigma_{B}^{*}=(t^{*},Z^{*})$ and optimal for Red to play
$\sigma_{R}^{*}=(X^{*})$, where $X^{*},t^{*}$ and $Z^{*}$ are chosen according
to Table 1.

| $X^{*}$ | $t^{*}(\cdot)$ | $Z^{*}$
---|---|---|---
$\frac{B}{R}\leq u$ | R | $\frac{B}{uR}$ | 0
$u\leq\frac{B}{R}\leq 1$ | $\mathbb{U}[0,2R]$ | 1 | $\mathbb{U}[0,2R]\text{ w.p. }p(u),\text{ else }0$
$1\leq\frac{B}{R}$ | $\mathbb{U}[0,2C(u)]\text{ w.p. }q(u),\text{ else }0$ | 1 | $\mathbb{U}[0,2C(u)]$
Table 1: Optimal strategy of $GL\text{-}S$.

The value $V$ of $GL\text{-}S$ is therefore:

$\displaystyle
V=P(\sigma_{B}^{*},\sigma_{R}^{*})=\begin{cases}\frac{B}{R},&\text{if
}\frac{B}{R}\leq u,\\\ \frac{u+\frac{B}{R}}{2},&\text{if }u\leq\frac{B}{R}\leq
1\\\ 1-\frac{(1-u)^{2}}{2(\frac{B}{R}-u)},&\text{if
}1\leq\frac{B}{R},\end{cases}$

The value of $GL\text{-}S$ depends both $u$ and the ratio $B/R$. We illustrate
this function in three different figures. The detection chance $u$ is fixed in
Figure 1, the ratio $B/R$ is fixed in Figure 2 and the value of the game is
portrayed as a heatmap in Figure 3.

Figure 1: The value of $GL\text{-}S$ for fixed detection probability $u$ as a
function of $B/R$. Figure 2: The value of $GL\text{-}S$ for fixed $B/R$ ratios
as a function of the detection probability $u$. Figure 3: A heatmap of the
value of $GL\text{-}S$, with the resource ratio $B/R$ on the x-axis and the
detection probability $u$ on the y-axis.

Some interesting insights follow from this optimal solution to the
$GL\text{-}S$ game. The first is that $t(\cdot)$ is always maximised, meaning
that Blue calls as many fields as he can, given his budget. Calling a field is
always a more efficient way of using resources, then using resources on a
field that was not revealed.

The second insight follows from the question whether Red should distribute her
resources more or less evenly compared to a standard Lotto game. The answer is
both, but in different cases. If Red heavily outnumbers Blue, i.e.
$\tfrac{B}{R}<u$, she plays as evenly as possible, sending exactly $R$
resources every time. However, if Blue outnumbers Red, Red plays riskier than
in the standard General Lotto game, allowing higher number of resources per
round, but also sending 0 resources more often. This also means that, as long
as $u<1$, Red always has a chance to win, even if heavily outnumbered.

######  Remark 1 (Comparison with known results when $u=0$).

By setting the detection probability $u$ in $GL\text{-}S$ to $0$, we retrieve
a General Lotto game of which the value should be consistent with earlier
work. The value of this game is $\tfrac{B}{2R}$ whenever $B\leq R$ and
$1-\frac{R}{2B}$ otherwise. This is consistent with Corollary 1 of Kovenock
and Roberson [12]. Also the optimal strategies found there coincide with the
optimal strategies that we use. It is also consistent with [9] and [17] who
showed this result earlier.

######  Remark 2 (Comparison with known results when $u=1$).

By setting $u=1$ we obtain a game that is in some respects similar to the pre-
allocation game in Chandan et al. [15]. Namely, when $u=1$ Blue always sees
what Red does and in that sense Red ‘pre-allocates’. There are two major
differences. The first is that in our case Red does not get the opportunity to
allocate any additional resources and in that sense forcibly pre-allocates all
her resources. The second difference is that the pre-allocation in Chandan et
al. [15] is performed ‘Blotto-style’, i.e. such that the sum of the pre-
allocated resources equals a fixed constant. In our case it should only in
expectation equal the total resources.

##  4 General Lotto with Scouts: Multistage

In this section we consider a multistage version $GL\text{-}MS$ of
$GL\text{-}S$ where Blue and Red first divide their total resources over the
battlefield and then play $GL\text{-}S$ on each battlefield. The latter step
was solved in Section 3, so we know the value of the games on the individual
battle fields. Therefore, this multistage version of the game can be
considered a Colonel Blotto game with a complicated payoff function given by
those game values. First we introduce the game and discuss under which
conditions the game value exists. Subsequently, we provide strategies for Blue
and Red, which provide upper and lower bounds on the value of the multistage
game. Lastly, we specify under which conditions these bounds coincide, in
which case these strategies must be optimal.

###  4.1 The multistage game

We start out with the same two players with total resources $B,R>0$,
respectively. Next, let there be $n$ fields, where $n$ is an integer larger
than one. Each field is endowed with a worth $w_{i}>0$ and a detection
probability $u_{i}\in[0,1]$. We assume that $\sum_{i}w_{i}=1$. We denote the
Multistage General Lotto game with Scouts with these parameters by
$GL\text{-}MS(B,R,\\{u_{i}\\},\\{w_{i}\\})$.

The game is played in two stages.

  * i)

Red chooses a distribution $\underline{R}=(R_{1},..,R_{n})$ of her total
resources over the fields such that $\sum_{i}R_{i}=R$. Simultaneously Blue
chooses a distribution $\underline{B}=(B_{1},..,B_{n})$ of his total resources
over the fields such that $\sum_{i}B_{i}=B$.

  * ii)

On each field an $GL\text{-}S(B_{i},R_{i},u_{i})$ game is played with worth
$w_{i}$.

The total payoff for Blue is the sum of the individual payoffs.

Recall from Section 3 that given the resources $B_{i},R_{i}$ at field $i$ and
given its worth $w_{i}$ and the detection probability $u_{i}$, the payoff at
field $i$ equals $\phi_{i}(B_{i}/R_{i})$ where

$\phi_{i}(x)=w_{i}\begin{cases}x&x\leq u_{i}\\\ \frac{1}{2}(x+u_{i})&u_{i}\leq
x\leq 1\\\ 1-\frac{(1-u)^{2}}{2(x-u)}&x\geq 1\end{cases}$

Note that the field worth $w_{i}$ was not present before, but is included in
$\phi_{i}$ here. Since for $n=1$ there is just one field with worth $1$, this
is consistent with before.

Therefore the first stage of the game has payoff function

$H(\underline{B},\underline{R})=\sum_{i=1}^{n}\phi_{i}\left(\frac{B_{i}}{R_{i}}\right).$

Our goal is to solve this game with the constraints

$\sum_{i=1}^{n}B_{i}=B,\hskip 28.45274pt\sum_{i=1}^{n}R_{i}=R.$

###  4.2 Existence of Game Value

There are three things that make it difficult to derive the game value. First,
even though $H$ is concave in $B_{1},..,B_{n}$, it is not convex in
$R_{1},..,R_{n}$. Second, $H$ is not everywhere differentiable, because of the
sharp turn in the $\phi_{i}$’s at $x=u_{i}$. Third, $H$ ill defined at
$B_{i}=R_{i}=0$ for any $i$ and cannot be continuously extended there.

Without the singularity at (0,0), the payoff function is continuous and we can
guarantee the existence of the game value. Allocating zero resources to a
field is both theoretically and in practice unlikely to be an optimal
strategy. It is possible to remove this singularity from the game in the
following way: we fix $\epsilon>0$ and require that $B_{i},R_{i}\geq\epsilon$
for all $i$. This creates a new, slightly more restricted game. As a result of
this restriction $H$ is continuous on the domain of strategies while the
domain is still compact. By a classical result of [18], there must exists a
Nash equilibrium for the corresponding game.

Now that we have discussed existence, we want to study the exact value of the
game. To bound the value of the game, we provide strategies for Red And Blue
and prove that these strategies guarantee certain upper and lower bounds.
Whenever those bounds coincide, we know the exact value of the game.

To find a good guess for optimal strategies for Red and blue, we explore the
case where $H$ is continuously differentiable, even though it is not. If the
$\phi$’s were $C^{1}$, then any Nash equilibrium would have to satisfy the
equations in Lemma 4, with the provided optimal strategies. We use these
strategies to provide bounds on the original game.

######  Lemma 4.

Let $\phi_{i}:[0,\infty)\rightarrow\mathbb{R}$, $i=1,..,n$, be $C^{1}$ and
increasing and let $B,R>0$. Suppose that $B_{1},..,B_{n}$, $R_{1},..,R_{n}$
satisfy

$\sum_{i=1}^{n}\phi_{i}\left(\frac{B_{i}}{R_{i}}\right)=\inf_{R^{\prime}_{1},..,R^{\prime}_{n}}\max_{B^{\prime}_{1},..,B^{\prime}_{n}}\sum_{i=1}^{n}\phi_{i}\left(\frac{B^{\prime}_{i}}{R^{\prime}_{i}}\right)=\sup_{B^{\prime}_{1},..,B^{\prime}_{n}}\min_{R^{\prime}_{1},..,R^{\prime}_{n}}\sum_{i=1}^{n}\phi_{i}\left(\frac{B^{\prime}_{i}}{R^{\prime}_{i}}\right)$
(1)

given

$\sum_{i=1}^{n}B_{i}=B,\hskip 28.45274pt\sum_{i=1}^{n}R_{i}=R,\hskip
28.45274ptB_{i},R_{i}\geq 0.$

Then $B_{i}=c_{i}B$ and $R_{i}=c_{i}R$, where

$c_{i}=\frac{\phi_{i}^{\prime}(B/R)}{\sum_{j=1}^{n}\phi_{j}^{\prime}(B/R)}.$

###### Proof.

Suppose that $B_{1},..,B_{n}$, $R_{1},..,R_{n}$, form a solution. Then in
particular $B_{1},..,B_{n}$ maximise

$f:(x_{1},..,x_{n})\longmapsto\sum_{i=1}^{n}\phi_{i}\left(\frac{x_{i}}{R_{i}}\right)$

given $0=g(x_{1},..,x_{n}):=x_{1}+..+x_{n}-B$. Using the theory of Lagrange
multipliers, we see that then for all $i$:

$\lambda=\left.\frac{\partial}{\partial
x_{i}}f(x_{1},..,x_{n})\right|_{B_{1},..,B_{n}}=\frac{1}{R_{i}}\phi_{i}^{\prime}\left(\frac{B_{i}}{R_{i}}\right),$
(2)

where $\lambda\in\mathbb{R}$ is a fixed constant. Reasoning similarly starting
from the RHS of (1), we see that there is a Lagrange multiplier $\mu$ such
that

$\mu=\frac{-B_{i}}{R_{i}^{2}}\phi_{i}^{\prime}\left(\frac{B_{i}}{R_{i}}\right).$
(3)

Plugging (2) into (3), we see that

$\mu=\frac{-B_{i}}{R_{i}^{2}}\phi_{i}^{\prime}\left(\frac{B_{i}}{R_{i}}\right)=\frac{-B_{i}}{R_{i}}\frac{1}{R_{i}}\phi_{i}^{\prime}\left(\frac{B_{i}}{R_{i}}\right)=\frac{-B_{i}}{R_{i}}\lambda,$

so

$\frac{B_{i}}{R_{i}}=\frac{-\mu}{\lambda}:=c$

is constant. Using the boundary conditions, we obtain

$B=\sum_{i=1}^{n}B_{i}=\sum_{i=1}^{n}cR_{i}=c\sum_{i=1}^{n}R_{i}=cR,$

so $c=\tfrac{B}{R}$, so in particular $B_{i}=\tfrac{B}{R}R_{i}$. Plugging this
back into (2) and rearranging yields

$R_{i}=\frac{1}{\lambda}\phi_{i}^{\prime}\left(\frac{B}{R}\right).$

Since $R_{1},..,R_{n}$ must sum to $R$, this implies that

$\lambda=\frac{1}{R}\sum_{j=1}^{n}\phi_{j}^{\prime}\left(\frac{B}{R}\right),$

so $R_{i}=c_{i}R$ with $c_{i}$ defined as in the lemma statement. Using now
that $B_{i}=\tfrac{B}{R}R_{i}$, this gives what we wanted. ∎

###  4.3 Upper bound on the game value

To obtain an upper bound on the game value, we let Red use the strategy that
is suggested by Lemma 4.

######  Lemma 5 (Upper bound).

The game value is bounded from above by

$\sum_{i=1}^{n}\phi_{i}\left(\frac{B}{R}\right).$

###### Proof.

First we assume that $\tfrac{B}{R}\neq u_{i}$ for all $i$. This implies that
each $\phi_{i}$ is differentiable at $B/R$, so we can set $R_{i}$ according to
Lemma 4:

$R_{i}=R\frac{\phi_{i}^{\prime}(B/R)}{\sum_{j=1}^{n}\phi_{j}^{\prime}(B/R)}.$

Now let $B_{1},..,B_{n}$ be an arbitrary strategy of Lotto. Then, using a
Taylor expansion for each $i$ of $\phi_{i}$ around $B/R$ and the concavity of
the $\phi_{i}$’s, we obtain the following bound for the payoff:

$\displaystyle
P(R_{1},..,R_{n},B_{1},..,B_{n})=\sum_{i=1}^{n}\phi_{i}\left(\frac{B_{i}}{R_{i}}\right)\leq\sum_{i=1}^{n}\left[\phi_{i}\left(\frac{B}{R}\right)+\phi_{i}^{\prime}\left(\frac{B}{R}\right)\left(\frac{B_{i}}{R_{i}}-\frac{B}{R}\right)\right].$

Now substituting the expression for $R_{i}$ we see that

$\displaystyle\sum_{i=1}^{n}\phi_{i}^{\prime}\left(\frac{B}{R}\right)\left(\frac{B_{i}}{R_{i}}-\frac{B}{R}\right)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n}\phi_{i}^{\prime}\left(\frac{B}{R}\right)\frac{B_{i}\sum_{j}\phi_{j}^{\prime}\left(\frac{B}{R}\right)}{R\phi_{i}^{\prime}\left(\frac{B}{R}\right)}-\frac{B}{R}\sum_{i=1}^{n}\phi_{i}^{\prime}\left(\frac{B}{R}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{R}\sum_{i=1}^{n}B_{i}\sum_{j=1}^{n}\phi_{j}^{\prime}\left(\frac{B}{R}\right)-\frac{B}{R}\sum_{i=1}^{n}\phi_{i}^{\prime}\left(\frac{B}{R}\right)=0,$

where in the last equation we used that the $B_{i}$’s sum to $B$. This implies
that

$H(B_{1},..,B_{n},R_{1},..,R_{n})\leq\sum_{i=1}^{n}\phi_{i}\left(\frac{B}{R}\right),$

which is what we wanted.

To deal with the case where $u_{i}=B/R$ for some (or possibly multiple
$u_{i}$), change all such $u_{i}$ by at most $\epsilon$ to $u_{i}^{*}$. Now
note that there is a constant $L$ such that each $\phi_{i}$ only changes by at
most $L\epsilon$ (with respect to the uniform norm). This implies that the
game value can change by at most $nL\epsilon$. Similarly the bound in this
lemma only changes by at most a constant times $\epsilon$. Since this holds
for all $\epsilon$, the bound of this lemma must also hold for $\epsilon=0$. ∎

Now we work out what this means in the case $\tfrac{B}{R}<1$. We assume WLOG
that $u_{1}\leq u_{2}..\leq u_{n}$. Suppose there is $0\leq k\leq n$ such that
$u_{k}<\tfrac{B}{R}<u_{k+1}$ (where we set $u_{0}=0$ and $u_{n+1}=1$). In this
case we obtain that

$\phi_{i}^{\prime}\left(\frac{B}{R}\right)=\begin{cases}\frac{w_{i}}{2}&i\leq
k\\\ w_{i}&i>k\end{cases}$

Denoting $w_{(k)}=\sum_{i\leq k}w_{i}$, this implies that

$\sum_{i=1}^{n}\phi_{i}^{\prime}\left(\frac{B}{R}\right)=1-\frac{w_{(k)}}{2}.$

The strategy for Red from Lemma 4 is therefore given by

$R_{i}=\begin{cases}R\frac{w_{i}}{2-w_{(k)}}&i\leq k\\\
R\frac{2w_{i}}{2-w_{(k)}}&i>k\end{cases}$

The upper bound from Lemma 5 now equals

$\displaystyle\sum_{i=1}^{n}\phi_{i}\left(\frac{B}{R}\right)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{k}\frac{w_{i}}{2}\left(u_{i}+\frac{B}{R}\right)+\sum_{i=k+1}^{n}w_{i}\frac{B}{R}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{i=1}^{k}w_{i}u_{i}+\frac{B}{R}\left(1-\frac{w_{(k)}}{2}\right).$

###  4.4 Lower bound on the game value

As noted before, $GL\text{-}MS$ does not generally have a Nash equilibrium
because of the non-convexity of $H$. This non-convexity is in turn caused by
the non-convexity of $\phi_{i}(1/\cdot)$ for each $i$. To obtain a lower bound
on the game value, we devise a different version of the game where this
convexity does hold and find the optimal solution for that game.

####  4.4.1 New payoff function

To devise the new game, we take the following steps:

  * i)

We take the functions $\phi_{i}$ and consider
$\psi_{i}:x\mapsto\phi_{i}(1/x)$.

  * ii)

We then define $\psi_{i}^{\dagger}$ as the lower convex envelope of $\psi_{i}$
(i.e. the largest convex function that is dominated by $\psi_{i}$).

  * iii)

We transform $\psi_{i}^{\dagger}$ back in the same way to define
$\phi_{i}^{\dagger}:x\mapsto\psi_{i}^{\dagger}(1/x)$. This function is
dominated by $\phi_{i}$ and is still concave. However, by construction
$\phi_{i}^{\dagger}(1/\cdot)$ is now convex.

(a) $\psi_{i}$ and $\psi_{i}^{\dagger}$ for $u_{i}=\frac{2}{5}$

(b) $\psi_{i}$ and $\psi_{i}^{\dagger}$ for $u_{i}=\frac{7}{10}$

Figure 4: The function $\psi_{i}$ and its convex envelope $\psi_{i}^{\dagger}$
for two different $u_{i}$, both with $w_{i}=1$.

We first follow this recipe for $u_{i}>0$. In that case we get

$\frac{1}{w_{i}}\psi_{i}(x)=\begin{cases}1-\frac{x(1-u_{i})^{2}}{2(1-u_{i}x)}&x\leq
1\\\ \frac{u_{i}}{2}-\frac{1}{2x}&1\leq x\leq\frac{1}{u_{i}}\\\
\frac{1}{x}&x\geq\frac{1}{u_{i}}\end{cases}.$

To understand how to obtain the convex envelope $\psi_{i}^{\dagger}$, the
graph of $\psi_{i}$ is shown in Figure 4 for two different $u_{i}$, along with
their convex envelopes. As can be observed from the graph, depending on
$u_{i}$, there are two different versions of the convex envelope. The formulas
are straightforward to derive.

  * i)

For $u_{i}\leq 2-\sqrt{2}$ there is a linear part, then $u/2+1/(2x)$, another
linear part and finally $1/x$. The equations are straightforward to derive and
are given by

$\frac{1}{w_{i}}\psi_{i}^{\dagger}(x)=\begin{cases}1-\frac{x}{2\alpha_{i}^{2}}&x\leq\alpha_{i}\\\
\frac{u}{2}+\frac{1}{2x}&\alpha_{i}\leq x\leq\beta_{i}\\\
\frac{2\gamma_{i}-x}{\gamma_{i}^{2}}&\beta_{i}\leq x\leq\gamma_{i}\\\
\frac{1}{x}&x\geq\gamma_{i},\end{cases}$

where

$\alpha_{i}=\frac{2}{2-u_{i}},\hskip
28.45274pt\beta_{i}=\frac{2(\sqrt{2}-1)}{u_{i}},\hskip
28.45274pt\gamma_{i}=\frac{2(2-\sqrt{2})}{u_{i}}.$

Note that for all $i$

$1\leq\alpha_{i}\leq\beta_{i}\leq\frac{1}{u_{i}}\leq\gamma_{i}.$

  * ii)

For $u_{i}\geq 2-\sqrt{2}$ we get a linear part, followed by $1/x$:

$\frac{1}{w_{i}}\psi_{i}^{\dagger}(x)=\begin{cases}1-\frac{x}{4}&x\leq 2\\\
\frac{1}{x}&x\geq 2\end{cases}.$

(a) $\phi_{i}$ and $\phi_{i}^{\dagger}$ for $u_{i}=\frac{2}{5}$

(b) $\phi_{i}$ and $\phi_{i}^{\dagger}$ for $u_{i}=\frac{7}{10}$

Figure 5: The function $\phi_{i}$ and the transformed convex envelope
$\phi_{i}^{\dagger}$ for two different $u_{i}$, both with $w_{i}=1$.

Now we transform this to $\phi_{i}^{\dagger}$ by again considering
$\psi_{i}^{\dagger}(1/x)$. The two different versions are shown in Figure 5.
The formulas are as follows.

  * i)

For $u_{i}\leq 2-\sqrt{2}$

$\frac{1}{w_{i}}\phi_{i}^{\dagger}(x)=\begin{cases}x&x\leq\frac{1}{\gamma_{i}}\\\
\frac{2\gamma_{i}x-1}{\gamma_{i}^{2}x}&\frac{1}{\gamma_{i}}\leq
x\leq\frac{1}{\beta_{i}}\\\ \frac{u}{2}+\frac{x}{2}&\frac{1}{\beta_{i}}\leq
x\leq\frac{1}{\alpha_{i}}\\\
1-\frac{1}{2\alpha_{i}^{2}x}&x\geq\frac{1}{\alpha_{i}}\\\ \end{cases}$

  * ii)

For $u_{i}\geq 2-\sqrt{2}$

$\frac{1}{w_{i}}\phi_{i}^{\dagger}(x)=\begin{cases}x&x\leq\frac{1}{2}\\\
1-\frac{1}{4x}&x\geq\frac{1}{2}\end{cases}.$

This was all in case $u_{i}>0$. For $u_{i}=0$, we see analogously that

$\frac{1}{w_{i}}\phi_{i}(x)=\begin{cases}\frac{x}{2}&x\leq 1\\\
1-\frac{1}{2x}&x\geq 1\end{cases}$

so

$\frac{1}{w_{i}}\psi_{i}(x)=\begin{cases}1-\frac{x}{2}&x\leq 1\\\
\frac{1}{2x}&x\geq 1.\end{cases}$

Note that this function is already convex. Therefore
$\psi_{i}^{\dagger}=\psi_{i}$ and $\phi_{i}^{\dagger}=\phi_{i}$.

By construction the $\phi_{i}^{\dagger}$ are now reciprocally concave, i.e.
$\phi_{i}^{\dagger}$ is concave and $\phi_{i}^{\dagger}(1/\cdot)$ is convex.
We use these to define a new payoff function

$H^{\dagger}(B_{1},..,B_{n},R_{1},..,R_{n})=\sum_{i=1}^{n}\phi_{i}^{\dagger}\left(\frac{B_{i}}{R_{i}}\right).$

Note that $H^{\dagger}$ is concave in $B_{1},..,B_{n}$ and convex in
$R_{1},..,R_{n}$. Also note that setting $\phi_{i}(B_{i}/0)=1$ and
$B_{i}\geq\epsilon>0$ ensures continuity of $H^{\dagger}$ everywhere.

####  4.4.2 Value of the dominated game

We now define a new game with payoff $H^{\dagger}$, which has the required
convex-concavity. First fix $B,R>0$ and set

$c_{i}=\frac{(\phi_{i}^{\dagger})^{\prime}(B/R)}{\sum_{j=1}^{n}(\phi_{j}^{\dagger})^{\prime}(B/R)}.$

We fix $\epsilon$ so that $0<\epsilon<B\min_{i}c_{i}$ and require that
$B_{i}\geq\epsilon$ for all $i$. Note that this extra restriction can only
lower the value for Blue, so the value of the resulting game is still a lower
bound for the original game.

All the $\phi_{i}^{\dagger}$’s are $C^{1}$, so we can apply Lemma 4 to
conclude that the unique Nash equilibrium is obtained by setting
$B_{i}=c_{i}B$ and $R_{i}=c_{i}R$. Moreover, the game value is
$\sum_{i}\phi_{i}^{\dagger}(B/R)$. Note that by construction of $\epsilon$,
the $B_{i}$ all satisfy the restriction $B_{i}\geq\epsilon$.

To compute the $c_{i}$’s more explicitly, one needs the derivative of
$\phi_{i}^{\dagger}$, which is given below.

  * i)

For $u_{i}=0$

$\frac{1}{w_{i}}(\phi_{i}^{\dagger})^{\prime}(x)=\begin{cases}\frac{1}{2}&x\leq
1\\\ \frac{1}{2x^{2}}&x\geq 1\end{cases}$

  * ii)

For $0\leq u_{i}\leq 2-\sqrt{2}$

$\frac{1}{w_{i}}(\phi_{i}^{\dagger})^{\prime}(x)=\begin{cases}1&x\leq\frac{1}{\gamma_{i}}\\\
\frac{1}{\gamma_{i}^{2}x^{2}}&\frac{1}{\gamma_{i}}\leq
x\leq\frac{1}{\beta_{i}}\\\ \frac{1}{2}&\frac{1}{\beta_{i}}\leq
x\leq\frac{1}{\alpha_{i}}\\\
\frac{1}{2\alpha_{i}^{2}x^{2}}&x\geq\frac{1}{\alpha_{i}}\end{cases}$

  * iii)

For $u_{i}\geq 2-\sqrt{2}$

$\frac{1}{w_{i}}(\phi_{i}^{\dagger})^{\prime}=\begin{cases}1&x\leq\frac{1}{2}\\\
\frac{1}{4x^{2}}&x\geq\frac{1}{2}\end{cases}$

###  4.5 Value of the original game

So far we have obtained an upper bound and a lower bound for the game. Now
note that in some cases, the $c_{i}$’s as computed in Section 4.4.2 coincide
with the ones that were used in Section 4.3. Namely when for each $i$ one of
the following conditions holds:

  * i)

$u_{i}=0$

  * ii)

$0\leq u_{i}\leq 2-\sqrt{2}$ and either

$\frac{B}{R}\leq 1/\gamma_{i}=\frac{u_{i}}{2(2-\sqrt{2})}$

or

$\frac{u_{i}}{2(\sqrt{2}-1)}=1/\beta_{i}\leq\frac{B}{R}\leq
1/\alpha_{i}=1-u_{i}/2$

  * iii)

$u_{i}\geq 2-\sqrt{2}$ and $\tfrac{B}{R}\leq 1/2$

Therefore, in these cases the upper and lower bound coincide. This implies
that the value of the game equals

$\sum_{i=1}^{n}\phi_{i}\left(\frac{B}{R}\right),$

since this is the upper bound we found. In particular, when all $u_{i}$’s are
$0$, the value is

$\frac{1}{2}\frac{B}{R}.$

We get back to this in Section 6. In the other cases the bounds do not
coincide, so we only obtain a range of possible game values. Figure 6 shows
the lower and upper bounds for a fixed choice of $w_{i},u_{i}$ as a function
of $B/R$.

(a) $u_{1}=\frac{3}{10},u_{2}=\frac{4}{10},u_{3}=\frac{5}{10}$

(b) $u_{1}=\frac{3}{10},u_{2}=\frac{4}{10},u_{3}=\frac{5}{10}$

(c) $u_{1}=0.31,u_{2}=0.33,u_{3}=0.35$

(d) $u_{1}=0.31,u_{2}=0.33,u_{3}=0.35$

Figure 6: The value of the GL-MS with 3 fields, for different values of
$u_{i}$. All three fields have value 1/3. The right figures are zoomed-in
versions of the left figures.

######  Remark 3 (Comparison with known results).

When all $u_{i}$’s are set to $0$, we obtain the following unique second-stage
strategy: $B_{i}=w_{i}B$ and $R_{i}=w_{i}R$. Then on each field a single field
General Lotto game is played with the same resource ratio $B_{i}/R_{i}=B/R$.
The total expected payoff still equals $\tfrac{B}{2R}$ whenever $B\leq R$ and
$1-\frac{R}{2B}$ otherwise. This is (like in Remark 1) consistent with
Corollary 1 of Kovenock and Roberson [12]. The difference between our case and
theirs is that they treat the game as a one-shot game. Apparently changing the
game to a multistage variant did not affect the value.

##  5 Measuring information versus strength

In the preceding sections, we have determined the optimal strategies and value
for games with fixed amounts of resources and information, effectively
addressing our first main research question. For this section we transition to
our second main research question: The influence of information versus
strength and the weapons mix problem. The weapons mix problem consists of
choosing the optimal combination of information (scouting drones) and
resources which provide strength (troops). In this section we present three
different ways to compare the influence of information against the influence
of resources for $GL\text{-}S$, ending with the weapons mix problem. Two of
these ways are similar to methods that were independently used in [15].

###  5.1 Influence ratio

One way to measure the influence of information versus strength is to compare
the increase in game value under the addition of information against the
increase of the game value under the addition of resources. To this end we fix
$R>0$ and compute for each pair of $B,u$ the ratio of derivatives of the game
value with respect to $B$ and $u$, respectively. This yields the following:

$\displaystyle V_{u}$ $\displaystyle:=$ $\displaystyle\frac{\partial
V}{\partial u}=\begin{cases}0&\frac{B}{R}<u\\\ \frac{1}{2}&u<\frac{B}{R}<1\\\
\frac{(1-u)(2B/R-1-u)}{2\left(\tfrac{B}{R}-u\right)^{2}}&1<\frac{B}{R}\end{cases}$
$\displaystyle V_{B}$ $\displaystyle:=$ $\displaystyle\frac{\partial
V}{\partial B}=\begin{cases}\frac{1}{R}&\frac{B}{R}<u\\\
\frac{1}{2R}&u<\frac{B}{R}<1\\\
\frac{(1-u)^{2}}{2R(B/R-u)^{2}}&1<\frac{B}{R}\end{cases}$

This allows us to define the influence ratio IR:

$\displaystyle IR$ $\displaystyle:=$
$\displaystyle\frac{1}{R}\frac{V_{u}}{V_{B}}=\begin{cases}0&\frac{B}{R}<u\\\
1&u<\frac{B}{R}<1\\\
1+\frac{2}{1-u}\left(\tfrac{B}{R}-1\right)&1<\frac{B}{R}\end{cases}$

The graph of the influence ratio is shown for $R=1$ in Figure 7. The diagonal
line $u=B$ between blue and purple marks the ‘Exploitation Line’, i.e. under
this line additional information is valuable because it can be used, but above
this line there are simply not enough resources. Using this, the graph can be
divided into 3 parts. Directly above the exploitation line we have the upper
triangular region, where the influence ratio is 0, which means adding
information does not increase the value of the game. Directly below the
exploitation line is the lower triangular region, where the influence ratio is
exactly 1, which means everywhere in the lower triangle, increasing either $u$
or $B$ by the same amount, leads to the same increase in value. And on the
right side we have the right rectangle, where the influence ratio is larger
than 1, which means the value increases more for an increase in $u$ than for
the same increase in $B$. Note the interesting behaviour around the point
$(1,1)$. The reason is that for $B<1$, going from information $u=1-\epsilon$
to $u=1$ is not valuable at all, because there are not enough resources to use
the information. However, for $B>1$ the increase of information from
$1-\epsilon$ to $1$ is extremely valuable. The reason for this is that when
$u=1-\epsilon$ adding resources will never yield a value of $1$, where adding
just the final $\epsilon$ bit of information does yield game value $1$.

Figure 7: Heatmap of the Influence Ratio, as a function of resources $B$ and
information $u$. A higher influence ratio indicates that information is more
valuable.

###  5.2 Resources needed to attain a given game value

Another way to quantify the influence of information is by studying how many
resources are needed to obtain a certain game value given a level of
information. The resource ratio $B/R$ that is needed to attain game value $v$
given information $u$ equals

$\frac{B}{R}=\begin{cases}u+\frac{(1-u)^{2}}{2(1-v)}&u\leq 2v-1\\\
2v-u&2v-1\leq u\leq v\\\ v&u\geq v\end{cases}$

Note that for any given fixed value $v$, the equation above exactly represents
the contour line of the value function for this particular value $v$. In
Figure 8, we display several of these contour lines. This visualization also
provides insight into how to maximize the game value with minimal additional
information or resource allocation. Specifically, this can be achieved by
moving perpendicular to the contour lines. Notably, in the top-left region, we
observe once again that adding information does not enhance the game value;
instead, only the addition of resources can lead to an increase.

Figure 8: Contour plot of the value of $GL\text{-}S$ as a function of $B/R$
and $u$. The arrows indicate the direction in which the value increases most
steeply.

###  5.3 Optimising information vs strength

In this section, we delve into the weapons mix problem, which revolves around
a fixed budget $D$ to be divided between resources and information. We assume
$R=1$ and set the cost of resources $B$ to $1$ per unit and the cost of
information at $c$ per unit. This yields the restriction that $B+cu\leq D$.
Using the solution found to $GL\text{-}S$ in Section 3, we can determine the
$B$ and $u$ that maximise the game value under this constraint, thereby
identifying the optimal allocation of the budget between resources and
information.

Varying the total budget $D$ produces the graphs in Figure 9. The line
corresponding to $c=100$ was included to illustrate the case in which
information is so expensive that it is not used at all, resembling the ‘no-
information case’.

The top-left graph of Figure 9 shows the maximum attainable value for a budget
$D$. It’s evident that a higher cost $c$ of information always leads to a
lower game value.

Moving to the top-right graph, we see that for $c<1$ it is optimal for any
budget $D$ to invest in both resources and information, with both steadily
increasing for higher budgets until a game value of $1$ is reached. For $c>1$
it is better to only invest in resources for when the budget is small.
However, if we increase the budget, there is a certain point after which it
also becomes worthwhile to invest in information as well. Interestingly, from
that point on Blue will actually start decreasing the amount spent on
resources, as his investment in information increases faster than the increase
in budget. The decrease in resources bought can be seen in the bottom-left
graph.

Finally, note from the top-left graph that a budget $D\geq 1+c$ ensures the
the game value is $1$. This outcome is understandable since $c$ can be spent
on buying information, thereby acquiring perfect information $u=1$. With
perfect information, Blue only needs to spend $R=1$ on resources to always
match Red and thus always win. Since a budget of $1+c$ guarantees Blue always
wins, the surplus budget need not be used. The unused budget is shown in the
bottom-right graph.

Figure 9: Let $D$ be the fixed budget to be spent between information and
resources. Here we plot the characteristics as function of the total budget
$D$, for several different costs of information $c$.

##  6 Interpretation

In the previous sections we have tackled the two main research questions
outlined at the start of this paper: the weapons mix problem and efficient
strategies when using scouts. Now, our focus shifts towards interpreting these
questions and their respective solutions within a military context,
particularly concerning the scenario of troops providing firepower and
scouting drones providing information. In this section we outline general
guidelines for optimal strategies, the most effective combination of troops
and scouting drones, and whether to prioritise additional troops or additional
scouting drones. Underlying these guidelines are the insights obtained in
various parts of the paper. We use the strategies and values calculated for
the $GL\text{-}S$ model and employ the three distinct ways to compare
information and strength within these models.

###  6.1 Efficient strategies with scouts

We start with guidelines for efficient strategies when the number of resources
and the amount of information are predetermined. The situation we analyse is
as follows: Red and Blue will face each other in a number of battles. Both
have a given amount of resources and additionally Blue has a given number of
scouts, which roughly translates to a probability to gain information about
Red’s allocation of resources.

Blue and Red have different challenges in formulating their strategies. For
Blue, the strategy can be split into two sub-strategies: what to do if Blue
gains information, and what to do if Blue doesn’t gain information. How does
Blue best convert his scouting advantage into the battle? Red aims to provide
a strategy that is as scout-resistant as possible, recognising the possibility
of her allocation to be scouted to Blue. Is this achieved by playing more
stable and always sending the same number of resources, which causes Blue to
learn nothing extra when he gains information? Or is this done by playing more
risky and pooling a large part of resources together for a large attack,
hoping that this one is not revealed?

The answer to all of these questions depends on the probability of Blue
gaining information and the resource ratio $B/R$. There are three distinct
cases, each with its own set of strategies for Blue and Red. We discuss these
cases separately and provide insights based on Theorem 2, which describes an
optimal solution to $GL\text{-}S$.

#### Red heavily outnumbers Blue, who has many scouts $(B/R<u)$

In this case Blue’s optimal strategy involves only allocating resources when
information about Red’s allocation is obtained, conserving resources
otherwise. Blue does not have enough resources to match Red every time the
field is revealed, but should try to match as often as possible. Red’s best
response is to always allocate the exact same number of resources, thus
neutralising the advantage Blue has when he gets information about Red’s
allocation.

#### Red outnumbers Blue, who has few scouts $(u<B/R<1)$

Here Red uses the same strategy as she would in a standard General Lotto Game,
always allocating resources but randomising the exact amount slightly. Blue
can match Red’s allocation whenever it is revealed and still have resources
left in his resource budget. Therefore, Blue should also allocate resources
occasionally when he has no information about Red’s allocation. If Blue
decides to allocate resources, he should follow Red’s strategy for determining
the exact number.

#### Blue outnumbers Red $(1<B/R)$

Now Blue can match Red’s allocation whenever it is revealed and additionally
allocate resources every time it is not revealed. The exact amount should be
randomised slightly. For Red, it is best to allocate resources even less often
than in a standard General Lotto game. This means that Red should often not
allocate any resources, and occasionally allocate a large number, while still
randomising the exact amount. The level of risk taken by Red increases with
both the ratio with which she is outnumbered and how often Blue gains
information on her resource allocation. If Red decides to allocate resources,
she should follow Blue’s strategy for determining the exact number.

###  6.2 Weapons mix problem

Now that we have provided guidelines for efficient strategies given fixed
amounts of resources and information, we discuss guidelines for when the
amounts of resources and information are not fixed, and can instead be chosen
under a fixed budget. First we first discuss the influence of information and
resources, respectively, and how they relate to military practice and
experience. Then we consider the budget constraints, thereby addressing the
weapons mix problem.

#### The value of information

To analyse the value of information, recall Figure 2. Two modes of behaviour
exist, depending on whether Blue has more $(B/R>1)$ or less $(B/R<1)$
resources than Red. In the first case, an increase in information always leads
to to an increase of value. However, in the second case, there is a cutoff
point: initially, extra information is beneficial, but beyond the cutoff
point, there is no additional gain. The main takeaway from these
considerations is that information only provides value as long as there are
enough resources in strength available to take advantage of this information.

We see something similar in military practice, for instance in the war
following the Russian invasion of Ukraine. Ukraine has a great intelligence
position due to help from many allies. However, at some point more
intelligence does not yield any better results. Knowing exactly what the
opponent does is not helpful when there are no resources to counter them.

#### The value of resources

For a deeper analysis of the value of resources, we turn to Figure 1. Again,
there is a clear break point. When $B/R$ is smaller than $u$, all of the Blue
forces are used to counter Red forces when they are observed, hence they are
always employed effectively. However, when $B/R$ is larger than $u$, some of
the Blue resources are used in situations where Red is not observed. These
Blue resources are used less effectively. This results in a smaller increase
in game value after the break point.

Similarly, in military practice, deploying troops is much more effective when
the strength of the opponent in the area is known. As soon as all available
information on the opponent has been exploited, any new troops are used in
more uncertain situations and are therefore less effective. We find this
reflected in the Russian-Ukrainian war, where Russia has many more troops and
weaponry to deploy but has an inferior intelligence position. This makes the
deployment of Russian troops less effective.

#### Budget constraints

Finally we arrive at our second research question: how to optimally distribute
a military budget over resources and information. We studied this question
most directly in Section 5.3, of which the main results are shown in Figure 9.
From this figure, we can extract three regimes, which we show in Table 2.

  * (I)

In this case information is cheap compared to resources. This corresponds to
$c<1$ in Figure 9. In this case it is directly worthwhile to buy information,
so as budget increases resources and information are bought equally.

  * (II)

Here information is expensive (so $c>1$) and there is a low budget. In this
case it is not worthwhile to invest in information and it is best to spend all
of the budget on resources.

  * (III)

As in case (II) information is expensive (so $c>1$), but there is also a high
budget. Now it is again worthwhile to buy information. In fact, with a higher
budget some resources are exchanged for more information (as one can tell from
the decreasing lines in the bottom-left graph of Figure 9).

The main takeaway here is that if information is cheap, you should always buy
it, along with resources. If information is expensive, you should focus on
resources first and only buy information if your budget is high enough.

|  Information
---|---
Cheap | Expensive
Budget | Low | (I) Combine | (II) Only resources
High | (III) Focus on information
Table 2: Preferred strategy in cases of high and low budget and with cheap and
expensive information. Derived from Figure 9.

##  7 Discussion

In this paper, we introduced the General Lotto Game with Scouts, a resource
allocation game enhanced with the possibility of gaining information on the
opposing player though scouting. We provided a complete solution for the game
concerning a single field. Subsequently, we extended our analysis to a
multistage version, establishing upper and lower bounds for its value and
determining conditions on when they coincide. Additionally, we introduced
several metrics to quantify the trade-off between information and resources.
Finally, we interpreted these findings and drew qualitative conclusions,
summarised as follows:

  * •

Information is only useful when there are enough resources to exploit it.

  * •

Resources are employed more efficiently in the presence of information.

  * •

When information is cheap, one should always buy it alongside resources. When
it is expensive, one should first focus on buying resources and only invest in
information if the budget is high enough.

  * •

When Blue receives information, it is always best to call as often as
possible.

  * •

Red should play very stable if she heavily outnumbers Blue, and very risky if
Blue heavily outnumbers her.

In this section we discuss the model and results. We conclude with some
suggestions for future work.

###  7.1 Discussion of the model and results

#### Lotto constraint before or after observing Red’s resource allocation

The characterising feature of a General Lotto model is the constraint that the
total allocated budget should equal a fixed constant in expectation. Since our
studied situation involves potential information revelation, one faces a
decision when this constraint should be applied: one can require the expected
total resources to equal a constant before or after potentially revealing
Red’s resource allocation.

The second option might seem more natural, but has certain curiosities in the
General Lotto framework. For instance: whenever Red allocates less resources
than Blue’s total budget, she always loses. Therefore Red will likely
alternate between allocating nothing and allocating significantly more than
Blue’s budget. Additionally, Blue has only one sensible response when Red’s
allocation is revealed, namely matching Red with the highest possible
probability.

In this paper, we chose for the former definition: requiring the budget
constraint before the allocation is revealed. This allows Blue to decide how
much of the budget he wants to allocate in the information and no-information
case, which in turn allows us to study where he can use his budget most
effectively. The only drawback is that since Blue’s allocation depends on
Red’s, his allocation distributions depends on Red’s allocation distribution.

#### Constant price of information

In Section 5.3, we examined a fixed budget $D$ to be divided between resources
and information. We assigned information a fixed cost of $c$ per unit.
However, this straightforward approach implies that the cost of increasing
information from 0 to 0.1 is identical to the cost of increasing from 0.9 to
1. In practical scenarios, obtaining basic information about a situation is
often much cheaper than acquiring the final details.

This choice for a fixed cost model significantly impacts the perceived value
of information in comparison to resources. In particular in the top-right part
of Figure 7, the value of information is much larger than the value of
resources. This large difference stems from the fact that even a marginal
increase in information can elevate the game value to 1, a feat unachievable
solely through an increase in resources.

To better reflect real-world conditions, one could consider assigning
information a non-linear cost. For instance, introducing a cost function such
as $F(u)=1/(1-u)-1$ would result in the cost of acquiring information
increasing with each unit purchased. Consequently, perfect information becomes
impossibly expensive to attain, thereby imposing a practical limit on
information acquisition.

###  7.2 Future work

We conclude this work with potential avenues for extending our research beyond
its current scope. While we have already mentioned adjusting the Lotto
constraint and the introduction of a non-linear price of information, there
are several additional options for future research.

First, one could consider more complicated scenarios for the single field
case. For instance, one could investigate scenarios where the information
obtained by Blue is partial, perhaps by revealing each of Red’s units
independently with certain probabilities. Another intriguing possibility is to
introduce false signals, complicating the accuracy of the information gained
by Blue and adding an element of uncertainty.

Second, it could be interesting to study extensions to the multistage version
of the game. For example, one could explore allowing Blue to optimise the
detection probabilities $u_{1},..,u_{n}$ under a fixed budget constraint.
Additionally, each field could have a ‘visibility’ parameter, which influences
how hard it is for Blue to obtain information, resulting in varying
information costs per field. Another option is to give Blue a fixed amount of
scouts, which he must then distribute strategically among the fields. I.e. if
blue has only two scouts, only the two fields he sends these to could be
revealed. These extensions might give a ‘search game dynamic’ where a trade-
off will be sought between allocating resources and search capacity to more
and less visible fields.

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