There has been some talk in the VC forum recently about lizardman ld and how it relates to normal tests. I said i would give a conclusive report and here it is. Make sure the tables are read carefully. I drafted this in word first and not all of it copied over as i'd have liked but i made a few changes and it should read ok now. Be warned that it does take a certain knowledge of mathematics to read:)

Probabilities with tossing two and three fair dice

Questions:

1) Say I roll two dice, and I want the sum of the two numbers to be less than or equal to five. The probability of this isn't too difficult to work out.

2) Now say I roll three dice and I pick the lowest two scores, what is the probability of these two adding up to less than or equal to five? And what is this probability equivalent to on two dice? i.e. Is getting less than or equal to seven on two dice equal in probability to getting less than or equal to five on three dice when you pick the lowest two?

3) How do these relationships change as I change the desired number?

The probabilities are not difficult to work out for independent fair dice. Let X be the score on a single die. The distribution of the score is given by the probability function

P(X = x) = 1/6, for x = 1, 2, 3, 4, 5, 6.

If we have two dice, the joint probability function is

P(X1 = x1, X2 = x2) = 1/62, for x1 = 1, 2, 3, 4, 5, 6 and x2 = 1, 2, 3, 4, 5, 6.

Then the probability function of the sum Y = X1 + X2 is easily found to be as shown in the table below:

Table 1: Probability function of the sum of the scores on two dice

y:2 3 4 5 6 7 8 9 10 11 12

P(Y = y)*6^2:1 2 3 4 5 6 5 4 3 2 1 respectively!

Thus, the answer to question 1 is given by P(Y <= 5) =(1 + 2 + 3 + 4)/36 = 10/36.

Now consider three dice.

The joint probability function for the scores on three dice is given by

P(X1 = x1, X2 = x2, X3 = x3) = 1/63,

for x1 = 1, 2, 3, 4, 5, 6, x2 = 1, 2, 3, 4, 5, 6, and x2 = 1, 2, 3, 4, 5, 6.

Now suppose that we order the three scores (X1, X2 and X3) such that the ordered values are represented by X(1), X(2) and X(3) where equal values are allowed, i.e. X(1) <= X(2) <= X(3).

Now, the joint probability function of the ordered values needs careful handling because of the possible equalities. The joint probability function is given by the following expressions where equalities and strict inequalities must be carefully observed.

P(X(1) = x1, X(2) = x2, X(3) = x3) = 6(1/6)3, for x1 < x2 < x3,

= 3(1/6)3, for x1 = x2 < x3,

= 3(1/6)3, for x1 < x2 = x3,

= (1/6)3, for x1 = x2 = x3,

The probability function for the sum of the lowest two, Y =X(1) + X(2), is now given in the following table (you need to carefully add up all the possible probabilities from the joint density function â€“ an interesting exercise in itself):

Table 2: Probability function of the sum of the two lowest scores on three dice

y:2 3 4 5 6 7 8 9 10 11 12

P(Y = y)*6^3:16 27 34 36 34 27 19 12 7 3 1 respectively

Thus, P(X(1) + X(2) <= 5) = (16 + 27 + 34 + 36)/63 = 113/216.

The second part of question 2 was â€˜Is getting less than or equal to seven on two dice equal in probability to getting less than or equal to five on three dice when you pick the lowest two?â€™

We see from Table 1 that for two dice

P(X1+ X2 <= 7) = (1 + 2 + 3 + 4 + 5+6)/62 = 21/36 = 126/216.

So these two probabilities are not the same.

You can answer question three for other values by reference to Tables 1 and 2.

So there we have it!

Ciao

Stonehambey