In my last column on the subject of probability theory, I promised to come back to the topic and look at some more examples. For what it is worth, here they are.

Old World of Darkness

Fix a number d between 1 and 10. Suppose that a number of 10-sided dice are rolled. The more talented a character is, the greater the number of dice rolled. A result of d or above on a die counts as one success. However, a die roll of 1 cancels a success. A botch occurs when there are more dice showing 1 than there are successes. Naturally, 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 {-1,0,1} and associated probabilities P(X=-1)=1/10, P(X=0)=(d-2)/10, and P(X=1)=(11-d)/10.

We can directly compute the expectation and standard deviation:

E(X)=(10-d)/10 and sd(X)= sqrt( 20 +10d -d^2) )/10

Thus, if we have random variables X1,...,Xk representing the number successes on various dice, then the expected number of total successes and standard deviation in the number are E(X1+...+Xk)=(10-d)k/10 and sd(X)= sqrt( 20 +10d -d^2)/10 respectively.

A botch occurs when X1+...+Xk<0. Actually, the probability of a botch is rather difficult to calculate exactly. However, the most likely way for a botch to occur is when one of the k dice shows a 1, and none of the other dice show successes. There are of course (with three or more dice) other less likely possibilities. Anyway, if we count only the botches of the type described, we see that the probability of a botch is approximately

k(d-1)^(k-1)/10^k

If anybody has an exact formula, I would be interested to know.

Value Matching Systems

There are at least two systems that I know of (Godlike and Weapons of the Gods) which involve rolling a number of 10-sided dice and counting the greatest number of dice showing the same number. Often, a matched pair indicates some level of success.

Now, let P(k) be the probabality of at least a matched pair when we roll k dice. Clearly P(1)=0 (we cannot have a matched pair when rolling one die) and P(11)=1 (there are only 10 possible digits, so there will always be a pair when rolling 11 dice).

When rolling k+1 dice, we get a matched pair either when we get a matched pair on the first k dice, or we do not get a matched pair on the first k dice, and the last die matches one of the k digits showing on the first k dice. We therefore obtain the equation:

P(k+1) = P(k) + (1-P(k))k/10

We know that P(1)=0. We can thus calculate P(2), then P(3), and so on. We obtain the following table.

Number of d10 Rolled	Probability of a pair
1	0
2	0.1
3	0.28
4	0.496
5	0.698
6	0.849
7	0.940
8	0.982
9	0.996
10	more than 0.999
11	1.0

Wrapping Up

One conclusion, which I hope was always implicit, is that there are many examples of systems where it is difficult to obtain analytic formulae for probabilities. However, a decent approximation is often possible.

In the next column, I want to look at a way of examining some basic issues which can be important before designing a campaign.