Conditional Probability Mass Function with fair die

Question

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 $P(M = m, X = x)$?

Answer 1

$Pr(M=m\mid X=x) = \begin{cases}0&\text{if }1\leq m<x\leq 6\\\dfrac{1}{6}&\text{if }1\leq x<m\leq 6\\\dfrac{x}{6}&\text{if }m=x\\0&\text{otherwise}\end{cases}$

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 $Pr(M=m\cap X=x)=Pr(M=m\mid X=x)Pr(X=x)$

Answer 2

I don't know what you think the notation $P(M \mid X)$ means, since $M$ is a random variable rather than an event in this question.

Let's write $P(M = m \mid X = x)$ instead. You already have an answer showing how to compute this probability directly, which I think is the easiest way to solve the problem. But your idea for solving it works too.

Writing the events explicitly in the probability formulas (that is, writing $M=m,$ which names an event, rather than $M,$ which names only a variable), your formula for conditional probability becomes $$P(M=m \mid X=x) = \frac{P(M=m,X=x)}{P(X=x)}.$$

You correctly state that $P(X=x)$ for each $x \in \{1,2,3,4,5,6\}.$

If your table is the usual $6\times6$ matrix in which each column represents one of the six possible outcomes on one die and each row represents an outcome of the other die, then you should be able to read off the values of $P(M=m,X=x)$ for each $m$ and $x$ just by counting cells in the table and dividing by $36.$

For example, there are four cells in which $M=4$ and $X=4,$ so $P(M=4,X=4) = \frac4{36} = \frac19.$

There is just one cell in which $M=5$ and $X=4,$ so $P(M=5,X=4) = \frac1{36}.$

And there is no cell in which $M=3$ and $X=4,$ so $P(M=3,X=4) = 0.$

You should be able to generalize this relatively easily to all cases where $m = x,$ all cases where $m > x,$ and all cases where $m < x.$ I would still prefer the direct solution given in the other answer, but this way will come up with the same results.

Answer 3

Let $Y$ be the result of the second die.

How do I get $P(M=m,X=x)$?

It is the probability that the maximum equals $m$ when given that the first die equals $x$.

$$\mathsf P(\max\{X,Y\}{=}m\mid X{=}x) = \begin{cases}\mathsf P(Y{=}m) &:& m\gt x\\[1ex]\mathsf P(Y{\leq}m)&:& m=x\\[1ex]0&:&\text{otherwise}\end{cases}$$