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Help me with math. How do I count the probability that xDy rolls

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File: ten sided die.jpg (24KB, 373x336px) Image search: [Google]
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Help me with math. How do I count the probability that xDy rolls higher than xDy+z
For example: 2D10 vs. 2D10+5
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File: useless die.jpg (76KB, 570x453px) Image search: [Google]
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>>157838
Numerically, I'm afraid.

You work out the probability series for 2D10 (2:0.01, 3:0.02, 4:0.03 ... 11:0.1 ... 19: 0.02, 20: 0.01), and for 2D10+5 (7:0.01, 8:0.02,.......), then for each entry in the first series you multiply its probability by the sum of the probablities of the values in the second series that are lower.

Then you sum this third series, and that's your probablity that the first dice will roll higher than the second.

Concrete example, taken from Space Marine.

Some tactical marines (CAF 1) close-assault some gretchin (CAF 0). The marines get 2d6+1, versus the gretchin's 2D6. How likely are the marines to win?

Value, probably on 2D6, probablity on 2D6+1
2,1/36,-
3,2/36,1/36
4,3/36,2/36
5,4/36,3/36
6,5/36,4/36
7,6/36,5/36
8,5/36,6/36
9,4/36,5/36
10,3/36,4/36
11,2/36,3/36
12,1/36,2/36
13,-,1/36

Chance of winning with a:
2,-
3,-
4,3/36*1/36=3/1296
5,4/36*(1+2)/36=12/1296
6,5/36*(1+2+3)/36=30/1296
7,6/36*(1+2+3+4)/36=60/1296
8,5/36*(1+2+3+4+5)/36=165/1296
9,4/36*(1+2+3+4+5+6)/36=84/1296
10,3/36*(1+2+3+4+5+6+5)/36=186/1296
11,2/36*(1+2+3+4+5+6+5+4)/36=60/1296
12,1/36*(1+2+3+4+5+6+5+4+3)/36=33/1296

Total probability = (3+12+30+60+165+84+186+60+33)/1296=633/1296=0.48=48%
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>>157981
Derp, I calculated how likely the marines are to lose, but you get the idea.
>>
>>157981
Thanks, man.
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>>157838
You'll probably have to actually write a table of all the possible rolls, because it's a non-uniform distribution. However, something to keep in mind with opposed rolls is that "is A higher than B" is equivalent to "is A-B greater than 1". So the end result is equivalent to "is 2d10-(2d10+5) greater than 1. Note that in this case you CANNOT simply have the 2d10s cancel out, even if it wasn't for the '+5', because die rolls are independent and don't represent a single variable with a fixed value.

>>157981 provides a good solution to solve this problem by hand, but if you find yourself doing this sort of thing a lot, you might want to check out anydice.com, which lets you set up complex die rolls and do greater than/less than comparisons and see the distributions.
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