Victory in the Pacific

Battles of Numbers
Markov Chain Modeling
in Victory in the Pacific

by Mircea Pauca

This article aims to fill a practical gap much needed by players of “Victory in the Pacific”: prediction of many-to-many-units combat. It was inspired by William Burch’s “Hard Numbers” article series which gives exhaustive answers for some cases of one-on-one combat. He correctly recognizes that the complete space state for larger battles becomes intractably large, having to count all possible combinations of partial damage on each unit.

This article uses Markov chains (see below) as theoretical instruments and, as a simplifying assumption, only counts the number of units able for combat after each round (not sunk, disabled, or crippled [i.e., carriers without airpower]). The current version of these spreadsheets can track up to 6 units for one side and 4 (of better quality) for the other.

Starting from probabilities to take the enemy out of combat in small engagements (1:1, 2:1, …) these spreadsheets calculate the probabilities of each possible final state from given battle conditions, summarizing them as win/loss probabilities, average number of combat-able units surviving the battle, and loss ratios. Results can be compared with the deterministic Lanchester Equivalent Value (see below for a summary).

Case 1: US carriers 1-3-7(4+) vs. Japanese Land-Based-Air (LBA) 3-4-*
Case 2: Allied light carriers 0-2-7(2+) or 0-2-7(2) vs. Japanese LBA 3-4-*
Case 3: Japanese “typical” carriers 1-2-* (3+) vs. Allied LBA 2-4-*
Case 4: Allied LBA 2-4-* vs. Japanese LBA 3-4-*
Case 5: Allied cruisers 1-1-7 vs. Japanese cruisers 1+2-7
Case 6: One Allied cruiser vs. one Japanese cruiser 1+2-7 (in detail, no approximation). 

Download ZIP file with Excel tables detailing all of the above calculations.

 

Known limitations to this approach:

 

What is a Markov Chain?

It’s a practical and fairly simple mathematical concept. Suppose a system with “n” (finite number of) possible states. The system can pass from a given state to others with various probabilities – we suppose these probabilities don’t depend on any previous states and that the probabilities remain constant in time. In other words, the system has no memory except the current state. In these examples, there are n=35 states corresponding to all of the combinations of 7 attacker states {0,1,2,3,4,5,6} with 5 defender states {0,1,2,3,4}. Some states are absorbing states meaning the system, once it gets into one of them, remains there forever.

The heart of the model is the Transition Matrix A, which contains in each cell on row “i” and column “j”, the probability to pass to the next state “j”, given that the current state is “i”. If the probabilities of being in each state “i” at time step “t” are written as a column vector of probabilities P(t), then the distribution of probabilities for each state at the next step can be calculated in just one matrix multiplication:

            P(t+1) = AT P(t)           where T is transposition (rows<=>columns)

and is the same as expressing it element by element:

P(j;t+1) = SUM(i) [ A(i,j) * P(i;t) ]

In the examples, the transition matrix A is calculated by multiplying the probabilities of independent events – for each combination of losses for given enemy fire. For instance:

A (6,4 -> 5,2) = Prob (1 hit from 4 defenders) * Prob (2 hits from 6 attackers).

Probabilities of each damage level for each side keeps account of the exact numbers of firers and targets in each sub-battle, and their effects combined by discrete convolution. For instance, 5:3 is actually two 2:1 fires combined with one 1:1, and 5:2 is a 3:1 combined with a 2:1. The inferior side always fires one or more 1:1.

Starting at time-step 0 with a certain situation (say 6:4 with probability 1, all others with 0) the spreadsheet calculates the probabilities of each state after 1,2,… 15 time steps.

The simulation horizon 15 was chosen empirically so that the probability to remain in a continuing state (non-absorbing) be very small, less than 0.000005.

The absorbing states include, obviously, all cases of total victory (one side with 0 survivors) and the 0:0 draw by mutual annihilation. It also occurs if one side recognizes its inferiority and withdraws. This is an important pyramid of sub-decisions for each side, which needs to be known before making decisions with larger forces. For instance, will you break combat with 1CV against 1LBA ? With 2:2 ? 3:2 ? and so on. The defender can also withdraw if badly outnumbered (usually 4:1 or worse).

A cautious attacker, giving up if odds are not decisively favorable, lowers the probability of “success” (so more narrowly defined), but reduces own losses more than the enemy’s – and improves the loss ratio. The best decision rule in each case depends on the relative importance of the territorial “prize” compared to more losses (see below).

 

Lanchester’s Equivalent Value

 Lanchester’s classic model of battles (1916) describes the continuous evolution in time of the strengths of two battling forces X(t) and Y(t). The firing effectiveness of each side is given by coefficients “kx” and “ky”. For instance, “kx” means how many Y units are taken out of combat by one X unit in one unit of time.

dX/dt = - ky * Y(t)

dY/dt = - kx * X(t)

Being a system of differential equations, this assumes infinitely fine-grained strengths and time and no randomness. It also assumes homogeneous forces on each side and linear effectiveness of fire – each new firer can find a target without interfering with other friendly units’ effect, and if an enemy is out of combat all friendly units know this and switch to other targets instantly. But it can still form a basis of good reasoning for our game.

Lanchester’s Equivalent Value is the initial ratio of force X(0) / Y(0) which makes both sides attrit each other exactly in the same proportion, so both sides will be annihilated –asymptotically in a long time.

LEV = sqrt (ky/kx)

In this comparison of quantity and quality, quantity is quadratically more important! If one Y unit is 4 times as deadly as X, then X:Y=2:1 will be “equivalent” in their death.

If one side is stronger relative to the opponent than the equivalent ratio says, they will improve even more their situation, slowly at first, then increasingly faster and easier, in a spiral of positive feedback. The trajectory in the state space (X,Y) is a nice parabola. The final strength of the winning side (assuming X wins) is:

X(final) = sqrt (X(0)2 – ky/kx*Y(0)2) = sqrt ( X(0)2 – (Y(0)*LEV)2 )

This famous conclusion called Lanchester’s Square Law shows very clearly the power of concentration. If 5 units battle 3 enemies of similar quality (LEV=1), 4 of the first side will survive victorious with no enemies around!

Another useful measure is the “speed of combat” which is the geometric mean of both sides’ effectiveness. If the forces are equivalent, they both decay exponentially with this “speed”. It is important because if different types of units from each side are fighting, the “fastest” combinations dominate until one of the “fast” opponents disappears first.

V = sqrt (kx*ky)

It’s also important in choosing between targets of different types – a rule would be to aim the “fastest” enemy first – the one on which a given amount of fire saves the most damage from enemy return fire. Another principle would be “comparative advantage” from international trade economics. Aim on the target type (from several opposing types) that is relatively easier to hit, compared to what other friendly types can do.

“Two armies that fight each other are one large army that commits suicide” (Henri Barbusse, The Fire)

This is a good description of attritional combat near equivalence (as WWI was), but this is NOT the region that any commander will want to put its forces into. With crushing superiority, combat becomes an one-sided “execution”. With practically possible forces, combat is actually a linear combination of “suicide” and “execution” as shown below by the solution of the Lanchester system. The first part shows the average force of both sides decaying, the second part one side’s “edge” of superiority increasing until the enemy collapses. 

X(t) = ½ * (X(0) + Y(0)*LEV) * exp(-V*t) + ½ * (X(0) – Y(0)*LEV) * exp (V*t)

Sources:

Rob Alexander, “Models, Gaming, and Simulation”
http://classweb.gmu.edu/ralexan3/SYST683/archive.htm

James Taylor, “Advanced Aggregate Combat Modeling”
http://www.npsnet.org/~jtaylor/MV4656.html

 

Practical considerations in using Lanchester theory

With the mechanism of “Victory in the Pacific” none of the assumptions hold!

For these reasons, Lanchester equivalent values are given for comparison only. Usual probabilities to win with initial force ratios close to LEV are around 40% - not 50%; the difference comes from the decision to stop the battle and call it a “loss” if becoming weaker at any moment due to random fluctuations. So I recommend to add a force surplus of 50-100% over the LEV in each battle to cover some bad luck and still have good chances to win, shorten the battles and reduce your side’s casualties.

The Markov chain model I propose solves all these difficulties, and gives You, the Commander, the probabilities of every possible end state, and a summary. How much risk you accept is your decision !

Another practical dilemma appearing often in VitP is what to do if the enemy deploys with equally force in two zones, and you can get, say, 1.5 : 1 (over the LEV) in both or 3:1 in one of them. The deciding factor should be the relative importance of territorial and positional “prizes” of victory compared to keeping your units alive. With two weaker attacks, you can gain 2, 1 or 0 prizes with typically heavy losses in both actions for both sides. With one strong attack, you can get one prize with much better chances, and much less loss (perhaps 1/4) because only half the enemy fires at start, then will retreat if possible. Overall enemy losses could be lower too as only part of them are exposed.

The defender, knowing this, will modify his deployment taking into account the “prizes.” Stronger defenses for the more important prize -  so as to make both equally unattractive to the assailant. Or if this leads to unacceptable loss in either battle, defend only one prize, and if risking a loss too, defend a prize unattractive enough so that the attacker will avoid it and take cheaply the easier ones.

Thank you for your time and consideration! Comments, criticism, suggested improvements are welcome!

Mircea Pauca (mircea.pauca@gmail.com), Bucuresti, Romania


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