I saw Skiltao talking about odds on his blog
and asked myself why that would happen. It's been a while since I did
much with Battletech and had to wrestle with numbers like this (because
it's always Battletech). I'd suggest reading and understanding his original post before continuing because this thing runs long without me recapping anything. Besides, Skiltao's a pretty good guy and you should read his(?) blog.
This is part two. Part one is here, but they're mostly unrelated. The second part of Skiltao's post was how the odds of 1D6 beating 2D6 were equal to the odds of rolling a 6 or less on 3D6. In order for that to happen, you've got to roll a three or greater on the 1D6 and anywhere from two to five on the 2D6. I'm expressing these probabilities as (3-6) and [2-5], respectively.
Really though, you need pairs of these odds:
1D6 2D6 1D6 2D6 1D6 2D6 1D6 2D6
(3) [2] (4) [2] (5) [2] (6) [2]
(4) [3] (5) [3] (6) [3]
(5) [4] (6) [4]
(6) [5]
So the odds of getting a 1D6 result greater than a 2D6 result is the sum of the odds of these pairs.
To get a 6 or less on 3D6, just like last week's analysis, you have to look at the set of 1D6 and 2D6 that make up the 3D6:
1D6 2D6
1 [2,3,4,5]
2 [2,3,4]
3 [2,3]
4 [2]
Now you can take that upside down 2D6 pyramid, rotate it clockwise ninety degrees and note how cleanly it lines up with the 2D6 results on the first table.
The only remaining difference is the odds of the 1-4 run versus the 3-6 run, but as we proved last week, the probabilities are symmetrical. Ergo, (1-4) = (3-6).
According to Skiltao's equations, the odds that 239 D10s will beat 349 D10s should equal the odds of getting 2628 or less on 588 D10s.
Testing that with this method:
Odds of a 239D10 beating 349D10 are (350-2390) and [349-2389], given that they're properly paired.
But, because the odds are symmetrical, you can say (350-2390) = (239-2279).
So, when you slide (239-2279) up to [349-2389] and add 239+2389, you get 2628. And just to be sure, you can add 2279+349 and get 2628 again.
Done.
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5 comments:
Do you people have a facebook fan page? I searched for one on facebook or twitter but could not discover one, I'd really like to become a fan!
Sadly, I don't have either. Readers aren't plentiful enough to justify a facebook page and I'm not clever enough to justify a twitter.
So "D" in my equations = "sides on die +1" because it's the highest possible result added to the lowest possible result, right? So would a d10 numbered from 0-9 have the same set of die combinations as dice numbered from 1-10 (perhaps, looking at your charts, rotated 90-120 degrees)?
Also, any thoughts on how you'd change the equation to calculate rolls involving mixed die types, like 1d10+3d4? Or irregularly numbered dice, like a d6 numbered 223455?
...Not that I'm trying to take advantage of your recent mathiness or anything.
It's based on similar sets of probabilities, which are determined by the highest and lowest results. Using 0-9 die would have the same number of results as 1-10 die, but the you'd subtract 3 from the sum of three dice.
As for dice of different sizes and different numbering, it's not really coming to me off the top of my head. I'll work on it.
Cool. Was looking at WoD a while ago, and 0-9 has become relevant to my interests.
Don't sweat too much over the mixed sizes thing. My interest there is of the idlest, most theoretical kind.
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