Let X be the first of two fair die rolls. Let M be the maximum of the
two rolls.
(a) Find the conditional probability mass function of M given X = x.
I made a table where the probability that any given value – k = 1, 2, 3, 4, 5, 6 – is the max of the two rolls.
$P(1) = 1/36, P(2) = 3/36, P(3) = 5/36, P(4) = 7/36, P(5) = 9/36, P(6) = 11/36$
I know that $P(M = m| X = x) = \frac{P(M = m, X = x)}{P(X = x)}$ and $P(X = k) = \frac{1}{6}$
How do I get $P(M = m, X = x)$?
Best Answer
$Pr(M=m\mid X=x) = \begin{cases}0&\text{if }1\leq m<x\leq 6\\\dfrac{1}{6}&\text{if }1\leq x<m\leq 6\\\dfrac{x}{6}&\text{if }m=x\\0&\text{otherwise}\end{cases}$
This can be seen since it is impossible for the maximum to be smaller than the first roll, that for the maximum to be bigger than the first roll the second die must match the maximum, and that for the maximum to be equal to the first die roll then the second die roll must be less than or equal to the first.
$Pr(M=m\cap X=x)$ can be calculated from the above using $Pr(M=m\cap X=x)=Pr(M=m\mid X=x)Pr(X=x)$