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Find the porbaility mass function of $Z~\overset{\text{def}}=~Y-X$.

Would it be the $$\sum_{y}^{} p(y)= 1$$ to find Y, in which we add all the probabilities of the pips being larger? Same goes for the X value.?

The problem i'm stuck on is how can I show which one is smaller and which one is bigger? Since if I get the smallest pips (1,1), then (1,2) would be considered bigger.

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    There are only $36$ possible rolls, just write them all down. Then you'll have the answer and you can see if you can find better ways of getting to it. – 2017-02-23
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    $(1,1)(1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)$ – 2017-02-23
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    Right. So, what is $Z$ in each case? – 2017-02-23
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    Perhaps the issue is that you don't understand the definition of $Z$. What it is: you take your roll, $(a,b)$ and then you look at the result. Say $a≤b$. Then $Z=b-a$. If, on the other hand, $a>b$ then $Z=a-b$. Thus if you get $(2,3)$ then $Z=3-2=1$. – 2017-02-23
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    So I would have to do the Z case for every single roll? – 2017-02-23
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    Yes. Well, there are faster ways to do it, but writing it out for each roll should take less than a minute. Then, like I say, you'll know what the answer is and you can try to come up with more efficient ways to compute it. – 2017-02-23
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    Just to make sure I will get this: $0+1+2+3+4+5+1+0+1+2+3+4+2+1+0+1+2+3+3+2+1+0+1+2+4+3+2+1+0+1+5+4+3+2+1+0=70?$ – 2017-02-23
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    Well, you don't mean to add them up. The idea is to compute the probability of each case. So...what is the probability that $Z=0$? $Z=1$? and so on. – 2017-02-23
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    $P(Z=0)= \frac{1}{6}, P(Z=1)=\frac{5}{18}$ and so on? – 2017-02-23
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    Yes, that looks right. To be clear: while I'm sure there are some ways to do it more efficiently, the enumerative method is really not so terrible. I suppose if these were hundred sided dice we'd need an alternate method. – 2017-02-23
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    True, so my answer would just be finding the probability of $Z(0,1,2,3,4,5)?$ – 2017-02-23
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    Yes. You just need to list those probabilities, which at this point you can simply read off. – 2017-02-23
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    Thank you for the guidance. – 2017-02-23

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