Question

Given a throw of three dice, let X,Y,Z be the number of dots showing on each. What is P(X<Y<Z)?

Answer 1

By symmetry this is

1/3! P(X,Y,Z are all different)


P(X,Y,Z are all different) =

1- P(X,Y,Z are not all different)

X,Y,Z are not all different if and only if one or more of the three conditions are satisfied:

c1: X = Y

c2: X = Z

c3: Y = Z

Terefore:

P(X,Y,Z are not all different)=

P(c1 OR c2 Or c3)

The prinicple of Inclusion and Exclusion says that:

P(c1 OR c2 Or c3) =

P(c1) + P(c2) + P(c3) -

[P(c1 And c2) + P(c1 And c3) +
P(c2 And c3) ] +

P(c1 And c2 And c3)


We have P(c1) = P(c2) = P(c3) = 1/6

and

P(c1 And c2) = P(c1 And c3) =
P(c2 And c3) = P(c1 And c2 And c3) = 1/36

Therefore:

P(c1 OR c2 Or c3) = 3/6 - 2/36 = 16/36 = 4/9


P(X,Y,Z are all different) = 1-4/9 = 5/9


P(X<Y<Z) = 1/6 * 5/9 = 5/54

Answer 2

Another way to compute this is by counting the number of ways you can make the three numbers diffrent. If you choose X first, then you have 6 choices for X. If Y is next then there are 5 choices for Y lft and 4 for Z. So, there are 6*5*4 possibilities. The probablity of any particular outcome is 1/6^3, so the probability is:

6*5*4/6^3 = 20/36 = 5/9

The probability that X, Y and Z are in the correct order is 1/6 times this probability which is:

5/9 * 1/6 = 5/54