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Battlefront WWII

Basic Probability Calculations

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:

Success 0.4 | Failure 0.6 | |
---|---|---|

Success 0.6 | S/S 0.24 | S/F 0.36 |

Failure 0.4 | F/S 0.16 | F/F 0.24 |

Success .76 | Failure .24 |

Success 0.76 | Failure 0.24 | |
---|---|---|

Success 0.2 | S/S 0.152 | S/F 0.048 |

Failure 0.8 | F/S 0.608 | F/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.

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'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 11:57AM

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