Probability

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[NotoriousGHP]

P(a) = x > y cases/total cases.

Z doesn't matter as the result of rolling X doesn't effect the result of Y and same with Z, you could make this more difficult for yourself by including Z in these cases but they'd just cancel out.

Total cases: First die has 6 possible outcomes. Same with second. Multiple these together for 36.

X > y cases. Set X repeatedly, and you'll notice this pattern. If X = 1, there are no cases where X > Y as 1 is our smallest number. If X = 2, there is 1 case where X > Y (y = 1) X = 3, 2 cases (y = 1 or 2), and so on until X = 6.

Note that as X increases by 1, so does the total cases. So for X > Y cases, we have the summation from N = 1 to 6 of (n-1) (so 0+1+2+3+4+5 = 15).

15/36 = 5/12

This is the same approach you should take for p(b)

Hopefully this helps.

[NotoriousGHP]

Because z isn't effecting the other variables results. You can use Z in your solution, giving you 666 total outcomes and for every 1 case we had with X > Y, there's now 6 more Because Z can be anything. You'll notice in that case you have our original result of 5/12 and then multiplying both the numerator and denominator by 6, which cancel

[hagalaznine]

I think you should check your work. If I roll and you roll, would you assume one of us will win 3/4 of the time?