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Two fair dice are rolled. Find the joint probability mass function of $X$ and $Y$ when $X$ is the smallest and $Y$ is the largest value obtained on the dice.

Reasoning this out (but leaving out the possibly irrelevant details), my work shows

$$\begin{align} &\sum_{x\ \in\ X\ \ } \sum_{y\ \in \ Y\ :\ x \le \ y} \left( \left(\frac{1}{6-x+1} \right) \left(\frac{2(6-x+1)-1}{36} \right) \right) \\ & = \sum_{y\ \in\ Y\ \ } \sum_{x\ \in \ X\ :\ x \le \ y} \left( \left(\frac{1}{6-x+1} \right) \left(\frac{2(6-x+1)-1}{36} \right) \right) \\ & = 1 \end{align}$$

which seems to imply we have found the PMF as asked in the problem, but I only know the last line follows because Mathematica said so. How can we show that this double summation is indeed equal to $1$ by hand?

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    $\begingroup$ Add up the fractions? $\frac{11}{216}+\left(\frac{11}{216}+\frac{9}{180}\right)+\cdots$ $\endgroup$
    – Steve Kass
    Commented Nov 21, 2013 at 22:12
  • $\begingroup$ @Steve Kass, nice, but for more general problems (where the possibilities aren't limited to the sides of 6-sided dice), how would you show the same? $\endgroup$
    – NaN
    Commented Nov 22, 2013 at 6:27

1 Answer 1

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Like Steve said, add up the fractions since there is only a really finite space $|6 \times 6$ to consider.

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