War is a state of armed conflict between countries, societies, and informal groups, such as insurgents and militias. It is generally characterized by extreme aggression, destruction, and mortality, using regular or irregular military forces so that a lot of money and lives are spent in fighting the war. Warfare refers to the common activities and characteristics of types of war or of wars in general. It is still unavoidable in the contemporary era [1]. The defeated side usually suffers heavy losses, since war usually results in significant deterioration of infrastructure and the ecosystem, a decrease in social spending, famine, large-scale emigration from the war zone, and often the mistreatment of prisoners of war or civilians. Thus,it is essential for a country to estimate its winning probability based on the military power of itself and its opponent. A country can decide whether to join the war or can know how to prepare for the war and take action such as deploying the fighting forces during the war, based on the winning probability estimation.

There are many factors that can influence the final result of a campaign, such as strategy, weather, topography, information, war chest, and logistics supply [1-5]. With the advancement of technology, the amount of armed forces that a country owns plays an important role. Say, in the case of ocean war, owning some aircraft carriers would play a dominant role in winning against its opponent. First, this paper considers only single type of armed forces. In particular, two parties each with a certain number of armed forces are considered. Though it is normal that the party with a greater amount of armed forces wins, this is not always the case. It is not unusual in history that some stronger troops are defeated by weaker ones. Therefore, this paper assumes that the result of warfare is stochastic for each unit of time. A dynamic modeling process is used to model the evolution of a campaign, which ends if all the armed forces of any party is extinguished. The final result can be the winning of any parties or a draw, where both parties have their armed forces extinguished.

Quite a few previous studies discussed the winning probability in a war or a contest [6-10]. However, they are all about a single battle, not a long-lasting campaign, which is a dynamic process that includes many battles. In a long-lasting campaign, the side that faces the relatively superior situation at the beginning of the war may also have a chance to take a turn for the worse. Some previous models used for campaign analysis have noticed the long-lasting time [11-12]. However, they all emphasize the command and control aspects of the campaign, with representation of sensor and communication systems, information flows, command decision making, combat, and perception of the battlefield. Differently, this paper focuses on a quantitative evaluation of the winning probability for each party in a long-lasting campaign. In particular, the process of a campaign is modeled based on a Markov process [12-15], where the result of the campaign at a specific time point is stochastic and depends on the result of the campaign at the last time unit.

The remainder of this paper is organized as follows: in Section 2, the basic assumptions are introduced, and then we model the process of a long-lasting campaign and present the methods to calculate the probabilities of the final states. In Section 3, the proposed method is used for some special cases to show how our model works. Section 4 concludes this study and proposes some future possible researches.

Here, we consider that there are two sides, A and B, in a campaign, as shown in Figure 1. For simplicity, we also use A and B to denote the numbers of armed forces that party A and party B own. The armed forces are assumed to be uniform, and both sides do not have any reserved forces or assists.

It is assumed that the two parties fight every day, and during each day each party may lose some units of armed forces. For mathematical tractability, it is assumed that the number of armed forces annihilated each day is an integer. The campaign continues until the armed forces of at least one party have been extinguished.

The numbers of armed forces owned by both parties in the beginningare denoted as M and N, respectively. Herein, we construct a Markov chain that hasthe following states:

In these state, those with $A=0$ and $B\ne 0$ are the states that means “A wins”. Similarly, those with $A\ne 0$ and $B\text{=}0$ are the states that means “B wins”. On the other hand, (0, 0) indicates that the campaign ends with draw. Finally, the rest of the states indicatethat the ending is still uncertain and the campaign will continue. The transition relationships of the four kinds of states are shown in Figure 2.

As shown in Figure 2, States A, B, and D are the ending states, which are the absorbing states. Once the campaign situation falls into these states, it will never have the chance to change.

Furthermore, the state probabilities can be summarized by a vector of $\left( M+1 \right)\times \left( N+1 \right)$ as

Where$L=\left( M+1 \right)\times \left( N+1 \right)$,${{Q}_{1}}={{p}_{M,N}}$,${{Q}_{2}}={{p}_{M,N-1}}$, ⋯, and ${{Q}_{l}}={{p}_{0,0}}.$

In addition, the state transition matrix is used to describe the state change that happens in each time unit as:

Where each ${{p}_{\left( i,j \right),\left( ii,jj \right)}},$$0\le ii\le i\le M,\text{ }0\le jj\le j\le M$ is the probability that (A,B) equals (ii,jj) given that it equals (i,j) at the last time unit. Inthe transition probability matrix, the sum of every row is equal to 1 by definition.

In this matrix, the absorbing probabilities are set for those states that any side has no residual armed forces. The mathematical expression for this case is shown in Equation (4).

This indicates that the state willbe maintained once the armed forces of any side are extinguished, that is, the state shown in (4). It is noted that ${{p}_{\left( \text{0},0 \right),\left( \text{0},0 \right)}}=1$ holds from (4), which corresponds to a draw ending.

Then, we can have the state probability at the K^th day after the campaign begins, $Q\left( K \right)$, as shown below:

Where $Q\left( 0 \right)$ indicates the initial state.

As discussed above, the ending condition is that the fighting resources of any side are totally eliminated. Otherwise, the campaign continues until the ending condition is obtained.

From (5), we can obtain the probability of A winning in K days, which is the sum of all the probabilities of the states that A wins during the first K days, as shown below:

Where 1(TRUE)=1 and 1(FALSE)=0.

Similarly, we can have the probability of B winning in K days, and it can the mathematically expressed by

The probability of a draw is the sum of all the states that the fighting resources of both sides are total eliminated. The probability of this case can be expressed by

The last situation is the state corresponding to an uncertain outcome, where both sides still have some armed forces left. The probability of this situation can be expressed by

The sum of the four probabilities for any K should be 1, that is

Note that PA(K), PB(K), PD(K), and PU(K) represent the probability of A winning, B winning, a draw, and an uncertainty that happens during the time from day 1 until day K. If one only wants to calculate the probability of these events that happen on a particular day K, then the transition matrix needs to be changed as below:

Where

$\left\{ \begin{matrix} & {{{\tilde{p}}}_{\left( \text{0,}Y \right),\bullet }}=\text{0}\ ,\quad \quad \quad \quad \text{ }Y=0,\cdots ,N \\ & {{{\tilde{p}}}_{\left( X,0 \right),\bullet }}=\text{0}\ ,\ \quad \quad \quad \ \ \ X=0,\cdots ,M \\ & {{{\tilde{p}}}_{\left( X,Y \right),\bullet }}={{p}_{\left( X,Y \right),\bullet }}\ ,\quad \ \ \ X=1,\cdots ,M,\text{ }Y=1,\cdots ,N \\ \end{matrix} \right.$

It can be seen that the “1” in ${{p}_{\left( \text{0,}X \right),\left( 0,X \right)}}$, $X=0,\cdots ,N$, and ${{p}_{\left( Y\text{,0} \right),\left( Y,0 \right)}}$, $Y=0,\cdots,M$ of the original matrix has been changed to “0” in order not to take into account these ending events that happen before day K. The rest of the values are the same as in the T.

Then, we can have the state probability transiting from non-ending states on theK^th day after the campaign begins, $\tilde{Q}\left( K \right)$, as shown below:

Where $\tilde{Q}\left( 0 \right)\text{=}Q\left( 0 \right)$ is the initial state distribution.

Then, the probability of A winning and B winning just on theK^th day after the campaign begins can be calculated by

and

Similarly, the probability of a draw andan uncertainty can be calculated by

and

Similarly, the sum of the four probabilities also should be 1:

Then, we can obtainthe probability distribution of the ending time, ET, as shown below:

Accordingly, we can calculate the expected duration of this campaign, ED, as shown below:

As a summary, we use a flowchart to show the process of how our model works, as shown in Figure 3. Here, the ending of the campaign is set as the probability of an uncertainty being less than 0.0001.

In this section, we set two special cases as examples to show the working process of the proposed model. The basic background is that there are two sides in a long-lasting ground campaign, A and B. In the beginning, each side has several armed forces.

We first consider a fair start, whereboth sides are assumed to have two armed forces. Then, we can have the state space of this case, as shown below:

In these states, (2, 0) and (1, 0) are the states that A wins. Similarly, (0, 2) and (0, 1) are the states that B wins. On the other hand, (0, 0) indicates that the campaign is tied and the rest of the states indicatethat the campaign will continue.

Then, the state probability can be expressed by a vector of nine elements as

Where ${{Q}_{1}}={{P}_{2,2}}$, ${{Q}_{2}}={{P}_{2,1}}$, ⋯, and ${{Q}_{9}}={{P}_{0,0}}.$

Next, according to Equation (21), the initial state vector is $Q\left( 0 \right)=\text{ }\left[ 1,0,\cdots ,0 \right]$, which indicates that the number of armed forces of the two sides are (2,2).

For convenience, we set a probability matrix for this special case. Note that the probability of annihilating armed forces is set to be higher for the side with more armed forces. This is consistent with intuition. The probability matrix we used here is shown as following. Last but not least, the probabilities of transition to other states from ending states including A wins, B wins, and a draware set to zero.

Then, according to (11), we can also determine the transition probability matrix for one day, $widetilde{T}$, which does not accumulate the probability of events that happened before the day of concern as:

Finally, using the model proposed in Section 2, we can obtain various results for Case 1, as shown in Section 4.

Second, we consider an unfair start. In this case, one side has two armed forces while the other one only has one armed force. Then, we can obtain the state space of this case, as shown below.

In these states, (2,0) and (1,0) are the states that A wins. However, the situation is critical for B, where only (0,1) is the state that B wins. Others are similar to Case 1: (0,0) indicates that the campaign is tied; the rest of states indicatethat the campaign will continue.

Then, the state probability can be expressed by a vector of six elements as

Where ${{Q}_{1}}=\text{ }{{P}_{2,1}}$, ${{Q}_{2}}=\text{ }{{P}_{1,1}}$, ⋯, and ${{Q}_{6}}=\text{ }{{P}_{0,0}}$.

Next, according to Equation (25), the initial state vector is $Q\left( 0 \right)=\left[ 1,0,\cdots ,0 \right]$, which indicates that the armed forces of two sides are (2,1).

For comparison, we build the transition probability matrix for Case 2 using some information from the transition probability matrix of Case 1. In particular, when the transition process is the same as Case 1, the probability in the matrix also uses the same value as Case 1. The transition probability matrix for Case 2 is shown below:

Similarly with Case 1, according to Equation 11, we can also determine the transition probability matrix for one day, $\tilde{T}$, as shown below.

Finally, using the model proposed in Section 2, we can obtain various results for Case 2, as shown in Section 4.

In this section, we first display the results we calculated using the transition matrix, $T.$ Tables 1 and 2 show the probabilities of different states during the first K days for Case 1 and Case 2 respectively.

From Tables 1 and 2, we can see that the winning probabilities of A and B as well as the probability of a draw are increasing with time, while the probability ofan uncertainty is decreasing with time.

The ending states will converge to a constant after a period. The case of (2,2) converges to (0.436,0.436,0.128) after about 20 days, while the case of (2,1) converges to (0.72,0.16,0.12) after about 17 days. It is noted that the campaign of (2,1) ends faster than that of (2,2). In fact, there is a large chance that the (2,2) may need a few days to transit into (2,1) or (1,2) first before it can finally transit to (2,0), (0,2), (1,0), (0,1), or (0,0). Actually,astime elapses, the campaign has an increasingly higher probability to have already ended before the day of concern.

Next, we display the results we calculated using transition matrix $\tilde{T}.$ Tables 3 and 4 show the probability of different results that happen on the K^th day after the campaign begins and the probability of an uncertaintyuntil day K, for Case 1 and 2 respectively.

From Tables 3 and 4, we can see that the winning probabilities of A and B as well as the probability of a draw first increase to peak value on day 2 and then decrease with time, while the probability of an uncertainty only decreases with time from the beginning.

Unlike Tables 1 and 2, all the four probabilities infinitely tend to 0. In other words, the probabilities will converge to (0, 0, 0, 0) after a period. The case for (2, 2) converges to (0, 0, 0, 0) after about 20 days, while (2, 1) uses about 17 days.

Finally, according to Equation 19 and the results from Tables 3 and 4, we also calculate the ending probability of each day, as shown in Figure 4. Then, the expected duration of the two cases can also be obtained, which are 3.4 days for Case 1 and 2.6 days for Case 2.

This paper proposes a dynamic model to estimate the winning probability for a long-lasting campaign. In particular, two parties each with a number of armed forces in the beginning are considered. The Markov process is used to describe the changing of their armed forces with time. The probability for each party to win and the probability of a draw ending are analysed. Numerical examples are presented to illustrate the applications. The results show that the winning chance for each side will finally converge to a constant value.

This work can be extended in multiple ways in the future. First, this paper considers single type of armed forces, which can be extended to combine different types of armed forces. Second, the logistic consumption of the armed forces can be incorporated. Third, the optimal allocation of resources into building different types of armed forces and providing logistic services can be investigated.