Probabilities: worked example

I will show you a worked example of a probability calculation, which will give the answer to a question which isn't answered on the page which led to this one: "What is the chance of getting a kill result (three hits) on 5d6?" which to non-Crossfire players is, "What is the chance of rolling a five or a six three times or more, when rolling 5d6?"

The magic formula used is this:

In this formula, n is the number of opportunities to score a hit (number of dice rolled), x is the number of hits/successful rolls we are asking about, p is the probability of a success on a single trial (1/3 in the case of a 5 or a 6 on 1d6), and f(x) is the probability of getting x successes in n trials. The curious expression n! means "n factorial". A factorial is the result of multiplying a number by the number below it, and the result by the number below that, and keeping doing this until one gets to multiplying by one. For example, 5!=5x4x3x2x1=120. Some useful factorials are: 0!=1, 1!=1, 2!=2, 3!=6, 4!=24, 5!=120, 6!=720, and 7!=5040.

So, in plain English, the formula is: Take the number of dice being rolled, find the factorial of it, and then divide the result by the result of n minus x factorial, times x factorial. Multiply the result of the first part of the equation by the second part, which is the chance of one success on one die, raised to the power of x. Multiply what you get as a result for multiplying part one by part two, by the result of p subtracted from one, raised to the power of n minus x.

Confused? Let's try plugging in some numbers:

So, for the question we asked above, p=1/3 (chance of a success on one die), n=5 (number of dice), and x=3 (number of successes needed for a kill).

We get: 5!/3!(5-3)!, times 1/3 to the power of three (1/3 cubed), times (1-1/3) to the power of (5-3).

This simplifies to: 120/6(2), times 1/27, times 2/3 squared.

Which simplifies to 10 times 1/27 times 4/9.

Which gives us the answer 40/243.

But we're not there yet...

You see, 40/243 is the chance that three dice will turn up as a five or a six, when five dice are rolled. But if four dice or all five dice were to come up 5s and 6s, then these too would be "kill" results. The question was "What is the chances of rolling a five or a six three times or more, when rolling 5d6?" To find this, we have to add to 40/243 the chances of 4 dice rolling hits, and of five dice rolling hits. Therefore we need to use the formula again for the values of x=4 and x=5.

The probability of getting three or more successes is prob(x>=3)=f(3)+f(4)+f(5).

If we use the magic formula at the top for x=4 (the chance that four dice will score hits) we get 120/4!(1!) times 1/3 to the power of 4, times 2/3 to the power of 1. This simplifies to 5 times 1/81 times 2/3 = 10/243.

Now for x=5: 120/120(0!) times 1/3 to the power of 5, times 2/3 to the power of 0. To work this one out, you have to know that 0!=1, and any number to the power of 0 is 1. This gives us 1 times 1/243, times 2/3 = 2/729.

So f(3)+f(4)+f(5)= 40/243 + 10/243 + 2/729 = 152/729.

If you want to express this as a percentage, then divide 100 by 729 and multiply by 152 = 20.85048%.

You might want to know the answer to "What is the chance of rolling at least one success with five dice?" In Crossfire terms, this would be the chance of a Pin result or better on 5d6. You could use the above method, and calculate the values for f(1), f(2), f(3), f(4), and f(5), and then add them all together. However, we know that all the possibilities added together must add up to 100%. Another way of putting this is: f(0)+f(1)+f(2)+f(3)+f(4)+f(5)=1. "f(0)" is the chance that after rolling five dice, not a single one has rolled a success. To find out the chance of a Pin or better, therefore, we can use the formula 1-f(0).

f(0)=5!/0!(5-0)! times 1/3 to the power of 0, times 2/3 to the power of 5-0. I'll tell you now that when x=0, the first part of the equation always comes out as 1, and the second part of the equation also comes out as 1. So we have 1 times 1 times 2/3 to the power of 5. This gives us 32/243.

32/243 now has to be subtracted from 1, giving us the answer: 211/243, or 86.831268%. You really are very likely to score a Pin result with 5d6.

Now you can work out all manner of your own probability calculations. To work out the chances of rolling sixes, rather than fives and sixes, just make p=1/6 instead of 1/3. To calculate 3d6+1P, use n=4 and then add one to the minimum value of x. Doubtless, though, you will soon want to know how to work out the answer to questions where the value of p is not constant. For example, "What is the chance of a fire group's causing a suppression, when the fire group is a machine gun rolling 4d6 and a rifle stand rolling 3d6?" To tackle this question, you must first calculate the chances of 3d6 and 4d6 causing a suppression or worse. I'll tell you that the answers are 7/27 and 33/81. These, then, are your two values of p. How to combine them? You have to work out all the combinations: one could suppress, the other could suppress, neither could suppress, both could suppress. In this case, there are four combinations. The chance that each combination occurs is one value of p multiplied by the other.

So the chance that only the MG suppresses is 33/81 (chance of two or more success in 4 trials) times 20/27 (chance that 3d6 does NOT score a suppress or worse) = 660/2187.

The chance that only the rifles suppress is 7/27 times 48/81 = 336/2187.

The chance that both suppress is 33/81 times 7/27 = 231/2187.

The chance that neither suppresses is 48/81 times 20/7 (we get these figures by subtracting the chance of a success from 1) = 960/2187.

To check that we haven't made a mistake, we can add all the numbers together: 660/2187 + 336/2187 + 231/2187 + 960/2187 = 2187/2187 = 1 (huzza!)

We are not interested, though, in the chance that neither causes a suppression, so we add up only the first three combinations: 66/2187 + 336/2187 + 231/2187 = 1227/2187 or as a percentage 56.104206%.

Now don't say that I never tell you anything.

I am enormously indebted to Andy Gordon, a professional statistician, for his providing me with the magic formulae, and instructions on how to use them.

So, do you reckon that you understand probability now? If you do, try the Monty Hall Dilemma in the Games section.



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