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There are $3$ dice you roll one at a time, $X$ is the number of distinct numbers, as in,

$X=1$, you have $(1, 1, 1)$ since there is $1$ distinct # $X=2$, $(1, 2, 1)$ or $(2, 1 ,1)$ etc... $X=3$ all different as in $(1, 2, 3)$

Find the probability mass function of $X$.

I tried doing $(X=2) = 1- (X=1) - (X=3)$, but I cannot figure out $X=1$, there are $6^3$ different variations total I think, but how do I figure out the numerators?

Joao
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Johnson
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1 Answers1

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You're correct about the total number of possibilities.

For the sizes of the individual cases, it might help to break things up into patterns. For example, if $x=1$, every such triple is of the form $(a,a,a)$.

For $x=2$, every triple will look like exactly one of $(a,a,b)$, $(a,b,a)$, or $(b,a,a)$ where $a\neq b$. Suppose we choose $a$ first, then $b$. How many ways are there to choose $a$? How many then to choose $b$ after $a$ has been chosen?

For $x=3$, every triple looks like $(a,b,c)$ where $a \neq b$, $a \neq c$, and $b \neq c$. Choose $a$, then $b$, then $c$. You've already determined how many ways there are to pick $a$ and $b$, so how many ways remain to choose $c$?

As a check, the sizes of the three sets should add to $6^3 = 216$.

user172541
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