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Battlefront WWII
Basic Probability Calculations


How likely is the outcome of a given event? The wargamer often wants to know this as he contemplates launching an assault. There are several approaches to this question:

Enumeration

Perhaps the easiest to understand, but the hardest for humans to use quickly, is the 'enumeration' approach. The close combat calculator is an example of this approach. There are 10 possible results for the attacker and the defender, yielding a matrix of 100 total outcomes. Using the computer we figure out every possible combination, count the cells that yield a particular result and then divide this total by 100 to give us the net probability. While a bit clumsy for human use, the enumeration approach is always possible. The close combat calculator is a special case of the interaction of two independent actions (the two die rolls), each of which has 10 equally likely outcomes (at least if the dice are "honest"). Any particular cell in the table requires a 10% (probability=0.1) likely roll by the attacker AND a 0.1 likely roll by the defender. Whenever you deal with an AND situation like this, the probability is the product of the outcomes that are required to produce it. A particular cell in the close combat table thus has a probability of 0.1 x 0.1=0.01 of occurring. Since there are 100 such cells, the total probability is 100 x 0.01 = 1.0. This makes sense, SOME outcome will occur (probability = 1.0). However, many of the individual cells in the matrix are equivalent. With a 0 modifier, if the attacker rolls a 1 and the defender rolls a 10, the combat result is the same as if the attacker rolled a 1 and the defender a 9. We must analyze the enumeration results to see what they mean. The close combat calculator does this and combines the 100 separate cells into the 5 possible outcomes.

Enumeration with unequal probabilities

Enumeration can be used even if the probabilities are not equal. For example, assume that we have three attacks, one of which has a 40% chance of success, one of which has a 60% chance of success, and one of which has a 20% chance of success. What is our overall chance of success? We build our first matrix with attack 1 across and attack 2 down the left:
 Success 0.4Failure 0.6
Success
0.6
S/S 0.24S/F 0.36
Failure
0.4
F/S 0.16F/F 0.24
When we analyze this matrix, we see that any success result counts as a success. The F/S and S/F and S/S results all are counted as successful. Thus the first matrix yields.
Success .76Failure .24
We now build a second matrix with the results of the first two attacks across the top and the third attack down the left:
 Success 0.76Failure 0.24
Success
0.2
S/S 0.152S/F 0.048
Failure
0.8
F/S 0.608F/F 0.192

Analyzing this second matrix yields a total chance of success of 0.152 + 0.048 + 0.608 = 0.808 and a chance of failure of 0.192 Once again, the total probability adds up to 1.0.

Simple mathematical calculation for AND probablities

Building matrices like those above can be painful but it is instructive. If you have a simple win/lose situation there is actually a much easier way to calculate the odds, but it requires a reversal in the way you look at the problem. In the above situation, you will win if the first attack wins OR the second attack wins OR the third attack wins.
OR probabilities are harder to calculate directly.
Instead, you can turn the problem around and say that you will lose if the first attack loses AND the second attack loses AND the third attack loses. As we have seen before, AND probabilities are performed by simple multiplication.
In the first attack, I win 0.4 of the time and will lose 1.0 - 0.4 = 0.6 of the time.
I will lose the second attack 0.4 of the time and the third attack 0.8 of the time.
I will lose all three of them 0.6 x 0.4 x 0.8 = 0.192 of the time, which is the same result that we found using the enumeration method.

Audie Murphy on Defence

Here is an example of the calculation method. Audie Murphy is defending against a German close assault conducted by 4 units with a with a 0 modifier. Audie (being the war hero that he is) will roll a 10 as a defender. This means that an assaulting force consisting of a single unit will be destroyed (be beaten by 4 or more) if it rolls a 6 or less (0.6 probability). What are the chances that an attacking force consisting of four units at zero modifer will lose a unit?
Audie's die roll of 10 is applied to all 4 attacks and the best result for the attacking force will apply to all.
For the attacking force to lose a unit, the first attack must roll a 6 or less AND the second attack must roll a 6 or less AND... (etc.).
The chance of this occurring is 0.6*0.6*0.6*0.6=0.1296, so the chance of losing a unit with 4 attackers is reduced to about 13%.
Similarly, what are the chances that our four units will take out Audie in an exchange so he dies in a blaze of glory?
In this case an exchange (even roll) occurs 0.1 of the time for each attacking die roll (a roll of 10).
The chance of this NOT occurring is 1-(0.1)=0.9 For Audie to survive, attack 1 must fail AND attack 2 must fail AND... (etc.) so we can use our multiplication of AND probabilities. The probability of Audie retiring to Hollywood are 0.9 x 0.9 x 0.9 x 0.9 = 0.6561. So Audie has only about a 66% chance of retiring and making his Westerns after the war.

I hope that people find the above instructive. Note that this is far from all-inclusive. Entire textbooks have been written about probability theory. However, if you can convert your estimate into an AND situation you can calculate the estimate the chance of success fairly rapidly.

This page was last updated on 10/28/2013 at 08:57AM

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