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Two fair dice are rolled. Let X and Y denote the largest and smallest values obtained, respectively. Compute the conditional mass function of Y given X = i, i = 1, 2, 3, 4, 5, 6. Are X and Y independent? Why?

I am stuck on this question. Well I know that the conditional probability is $$ P(X|Y) = \frac{P\{ X=x, Y=y\}}{P\{Y=y\}} $$ but I'm not really sure what to do with in this question. I think we let j the the values of Y so we have j < i, and I dont know how to continue.

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  • $\begingroup$ The pmf of Y is uniformly distributed between the points available. These are clearly not independent, consider when $X=1$ $\endgroup$ – Connor James Feb 29 '16 at 2:47
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Suppose for instance that $X=2$. Then the possible outcomes of the two rolls of the dice are $(1,2),(2,1)$, and $(2,2)$, all with equal probability.

This tells us that the possible values of $Y$ (given that $X=2$) are 1 and 2, and that $\mathbb{P}(Y=1|X=2)=\frac{2}{3}$, while $\mathbb{P}(Y=2|X=2)=\frac{1}{3}$. The same method works for the other values of $X$.

To elaborate on my answer, if $X=i$ then there are $2i-1$ outcomes, namely the $2i-2$ pairs $(i,j)$ and $(j,i)$ for $j<i$ and the outcome $(i,i)$. Each of these outcomes has equal probability. Therefore $\mathbb{P}(Y=j|X=i)=\frac{2}{2i-1}$ if $1\leq j<i$, and $\mathbb{P}(Y=i|X=i)=\frac{1}{2i-1}$.

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  • $\begingroup$ could you provide a few more pointers? So I noticed that for $X = i$ there are $i$*$n - 1$ possible outcomes. And $Y$ has $i$ possible values. Should I try to think how to relate this to $P\{X=i,Y=j\} $? $\endgroup$ – karambit Feb 29 '16 at 3:05

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