Some weeks ago, I went over some numbers that Skiltao mentioned on his blog (Part 1 & Part 2). I didn't do any rigorous mathematical analysis on them—what I did was more akin to recognizing the geometric qualities of dice probabilities—but I did manage to illuminate the qualities of the dice rolls and why certain relationships existed between them. Whenever I was done, Skiltao asked about alternative dice relationships, for example, if different-sized dice would have similar relationships or how dice which have repeated numbers would be affected.
The first question is straightforward. The odds of a tie between, say 1D10 and 3D4 being expressed as a certain value on a 1D10 + 3D4 roll is easy to test.
1D10 ranges from 1 to 10, while a 3D4 ranges from 3 to 12. The odds of a tie are the compounded probabilities of a set of values on the table below:
1D10 1D4 #1 1D4 #2 1D4 #3
[ 3] (1) (1) (1)
[ 4] (2) (1) (1)
[ 5] (2,3) (2,1) (1)
[ 6] (2,3,4) (2,1) (1,2)
[ 7] (3,4) (3,2,1) (1,2)
[ 8] (3,4) (3,2) (1,2)
[ 9] (3,4) (3,4) (3,2,1)
[10] (4) (3,4) (2,3)
I'm sorry that's not as clear as the 2D6 table from part one, but you get into matrices after a while.
None the less, the 1D10 is as fully reversible as 1D6. You can flip the range of probabilities, [3-10], to get an equal set of probabilities, [8-1]. The summation for [8-1] and (3-10), yields 11 by matching [3, 8), [4, 7),...[8, 3).
So yes, as long as the lower set can be flipped, the relationships between the probability of a tie between two sets of dice will be equal to the odds of a particular summation of their results.
Now, whenever we compare the odds of 1D10 beating 3D4, we have a different relationship.
Again we're doing pairs of odds:
1D10 3D4 1D10 3D4 1D10 3D4 1D10 3D4 ... 1D10 3D4
(4) [3] (5) [3] (6) [3] (7) [3] ... (10) [3]
(5) [4] (6) [4] (7) [4] ... (10) [4]
(6) [5] (7) [5] ... (10) [5]
(7) [6] ... (10) [6]
(4) [3] (5) [3] (6) [3] (7) [3] ... (10) [3]
(5) [4] (6) [4] (7) [4] ... (10) [4]
(6) [5] (7) [5] ... (10) [5]
(7) [6] ... (10) [6]
... (10) [7]
... (10) [8]
... (10) [9]
Again-again, we can turn that into a set of probabilities of getting a results equal to less than, say 10, on the summation between 1D10 and 3D4.
1D10 3D4
[1] (3,4,5,6,7,8,9)
[2] (3,4,5,6,7,8)
[3] (3,4,5,6,7)
[4] (3,4,5,6)
[5] (3,4,5)
[6] (3,4)
[7] (3)
Because, for a 1D10 [1-7] = [3-10], we can see that the odds of getting a ten or less on 1D10 + 3D4 = the odds of 1D10 beating 3D4.
Now, for irregularly numbered dice, 2,2,3,4,5,5--known as XD<NNNNNN>, where 1D6 = 1D<123456>--it's a different matter.
The odds of 1D<112344> tying 2D<112344> is equal to:
1D<1...4> 2D<1...4>
[2] (2)
[3] (3)
[4] (4)
Now, the odds are swankier, but still symmetrical. [1-3] is still equal to [2-4]. So the odds of 1D<112344> equaling 2D<112344> is equal to getting a sum of 5 on 3D<112344>. Special note: I don't have to crunch the actual odds. This is all geometry of probabilities, which is something I'm making up as I go and is probably wrong. If you were hoping for actual numbers and formulas and equations you are shit out of luck.
Right, so part two subsection two: are the odds of 1D<112344> beating 2D<112344> equal to the odds of rolling less than or equal to N on 3D<112344>? Probably.
1DN* 2DN* 1DN* 2DN*
[3] (2) [4] (3)
[4] (3)
[4] (3)
Again, we can turn that into a set of probabilities of getting results equal to less than, say 4, on the summation between the two.
1DN* 2DN*
[1] (2,3)
[2] (2)
Brace yourselves: it's the same fucking thing, so yeah.
But I'm not sure if those answers are in keeping with the spirit in which the original questions were asked. I took a moment and reconciled the 1D<112344> and 1D6 intersection, and because the geometry of the probabilities are both symmetrical, I'm pretty sure the relationships would hold true.
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*Seriously, fuck that.
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