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Advent

Probability Theory 2

by Paul Mitchener
Sep 20,2005

 

Methods of Campaign and Adventure Design 12

By Paul Mitchener

Probability Theory 2

Thanks to everybody who made comments on the last column. In this column, I want to look at calculations in some more complicated examples. The examples this time basically all involve rolling several dice and adding the results, sometimes in a slightly disguised form.

The meaning of standard deviation

Before getting to the first of my main points, I want to explain a little more what the standard deviation of a random variable means. In the last column, I defined the standard deviation, and said that it was a measure of a random variable's degree of randomness.

Jay Verkuilen mentioned the 67%/95% rule in the feedback forums. What this rule says is that for a `nice' random variable, approximately 67% of all possible outcomes are no further from the expectation than the standard deviation, and approximately 95% of all possible outcomes are no further from the expectation than twice the standard deviation.

I will explain exactly what I mean by `nice' in a future column where I talk about Gaussian distributions and the central limit theorem. I believe that any random variables we will meet here are `nice' enough that this rule more or less applies.

Independence

In the last column, I gave the formal definition of a discrete random variable, and indicated how a die can be seen as a random variable. When we have several dice, we have several random variables.

Let us suppose that we have two random variables X and Y, with ranges of values {a,b,...} and {a',b',...} respectively. Then as well as the probabilities P(X=a), P(X=b),... and P(Y=a'), P(Y=b'), ... , there are also associated probabilities of the form P(X=a and Y=a'). The sum over all possible combinations of these joint probabilities must be equal to one.

The analysis of the joint behaviour of the random variables X and Y becomes far simpler when they are independent, that is to say

P(X=a and Y=a') = P(X=a).P(Y=a')

for all possible values a and a'.

Independence means that the values of the random variables X and Y do not affect each other. Random variables representing dice are always independent from each-other; the result of an individual die roll should not affect the result of another roll.

When our random variables X and Y are not independent, we can further define conditional probabilities:

P(X=a | Y=a') = P(X=a and Y=a')/P(Y=a')

The conditional probability P(X=a | Y=a') is the probability that X is equal to a if we already know that the result of Y is a'. Conditional probabilities are related to ordinary probabilities by the partition theorem:

P(X=a) = P(X=a|Y=a')P(Y=a') + P(X=a|Y=b')P(Y=b') + P(X=a|Y=c')P(Y=c') + ...

Sums

Let X and Y be two random variables. Then we can form the sum X+Y; this sum is a single random variable, with various associated probabilities which can be calculated. Further calculations give us the following result.

Theorem: Let X and Y be independent integer-valued random variables. Then the expectation and variance of the sum X+Y are given by the formulae E(X+Y) = E(X) + E(Y) and var(X+Y) = var(X) + var(Y) respectively.

As usual, the standard deviation can be calculated as the square root of the variance. Going slightly further, if we have a number of independent random variables X1,X2,...Xk, then we have the formulae E(X1+...+Xk) = E(X1)+...+E(Xk) and var(X1+...+Xk) = var(X1)+...+var(Xk)

If the random variables all have the same values and probabilities as some random variable X, then our formulae can be further simplified E(X1+...+Xk) = k.E(X) var(X1+...+Xk) = k.var(X)

Applying our formulae from last weeek, when each random variable represents an N-sided die, we see that

E(X1+...+Xk) = k(N+1)/2 and var(X1+...+Xk) = k(N^2-1)/12

Examples

As an illustration, here are some calculations (to one decimal place) using some fairly common dice types, so that expectations and (especially) standard deviations can be compared for dice methods with similar ranges. I have included 4d3 since it is the method used in Fudge, albeit disguised.

Using the comments made at the start of this column on the approximate meaning of the standard deviation, these calculations should give some feeling for how various dice rolling methods tend to behave.
Dice Type Expectation Standard Deviation
d20 10.5 5.8
2d10 11 4.1
3d6 10.5 3.0
2d6 7 2.4
4d3 8 1.6

Dice Pools

I now want to look at an example of a dice pool system, where a number of 10 sided dice are rolled. The more talented a character is, the greater the number of dice rolled. A result of 7 to 9 on a die counts as one success, and a result of 10 counts as two successes. Successes on the various dice are added; the more successes there are, the better.

The number of successes arising from a single die roll can be seen as a random variable X, with range of values {0,1,2} and associated probabilities P(X=0)=6/10, P(X=1)=3/10, and P(X=2)=1/10.

We can directly compute the expectation and standard deviation:

E(X)=1/2 and sd(X)= sqrt(9/20)=0.67

Thus, as above, if we have random variables X1,...,Xk representing the number successes on various dice, then

E(X1+...+Xk)=k/2 and sd(X)= sqrt(9k/20)

Probabilities of Events

Sometimes, as well as means and expectations, we want to look at individual probabilities. For example, with the above dice pool system, we might want to know what the probability of no successes is.

As already mentioned, the random variables X1,...,Xk representing the successes on various dice are independent. Thus

P(X1+...+Xk =0) = P(X1=0 and X2=0 and... Xk=0) = P(X1=0).P(X2 =0)...P(Xk=0)

Thus the probability of no successes is P(X=0)^k = (6/10)^k

In fact, there is a further complication of the dice pool system we are examining, namely that a result of no successes when one or more dice roll a 1 is called a botch. A botch represents a nasty or amusing failure in the game. It is useful to know how probable a botch is, since having this probability available gives the GM some indication of how bad the effects should be.

The event of a botch is obviously not independent of the event that we have no successes; a botch can only occur when we have no successes. However, if we have no successes, the probability that a particular die does not show a 1 is 5/6. Thus the probability that we do not have a botch when there are no successes is (5/6)^k; the individual dice are still independent.

It follows from the partition theorem that the overall probability of a botch is (6/10)^k (1-(5/6)^k). The following table tells us various probabilites to two decimal places.
Number of d10 Rolled Probability of no successes Probability of botch
1 0.6 0.1
2 0.36 0.11
3 0.22 0.09
4 0.13 0.07
5 0.07 0.05
6 0.05 0.03
7 0.03 0.02
8 0.02 0.01
9 0.01 0.01
10 0.01 0.01

There are two observations to make here. The first, and most obvious, is that as more dice are rolled, the probability of no successes and the probability of a botch both decrease (with a minor glitch for botches when moving from 1d10 to 2d10). The second observation is that if a character achieves no successes, the more skilled he is, the more likely the failure is to result in a botch.

Wrapping Up

In the next column I want to talk about continuous random variables, the Gaussian distribution and the central limit theorem. I also promise to think about some of the systems which are less tractable to analysis, such as various roll and keep mechanisms.

If requested, I can give further details of calculations and other clarifications in the forum. TQo0~^DҒt< ek&Ǿ$\۵ZFȃuwݝIŃU QYir2HR2.u3MFoعq]4#A`pP5(b& )b)ⰾp7(i<[-2gL#5[f g?*rVGf8*)s'+20ϟ̑F}KB<7wSL\gbvm9WiRބYŜvd y0'p2I_Fc2>#o A )VL[Qk?3`)<У[(*W.JH ?tXCt谙 X:@ \0w ~LqĤE-rFkYœj4q 5AQ6[AxG [>w|?( fХθY䝛$c=_qNĦoǸ>O_|&/_Mi7"宥CЧk0dӷLh;TmuCGU-!Ul{ h<\bQX.~"O2*yPcz!ŠGg

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