Thursday, September 27, 2012

Weird thing about the chance one roll has to beat another:

A while ago, EastwoodDC posted a spreadsheet for calculating the chance of rolling any given result on a number of whatever sized dice. I started looking at the chances one roll had to beat another and I noticed something weird.
When rolling 1d6 against 2d6, the chance that they'll tie is equal to the chance of rolling exactly 7 on 3d6, and the chance that 1d6 will beat 2d6 is equal to the chance of rolling 6 or less on 3d6.
When rolling 2d6 against 3d6, the chance that they'll tie is equal to the chance of rolling exactly 14 on 5d6, and the chance that 2d6 will beat 3d6 is equal to the chance of rolling 13 or less on 5d6.

When rolling 3d6 against 4d6, the chance that they'll tie is equal to the chance of rolling exactly 21 on 7d6, and the chance that 3d6 will beat 4d6 is equal to the chance of rolling 20 or less on 7d6.
So I'm pretty sure that, anytime you roll Nd6 against (N+1)d6, the chance they'll tie is equal to the chance of rolling N*7 on (N+N+1)d6 and the chance that Nd6 will win is equal to the chance of rolling (N*7)-1 or less on (N+N+1)d6.

Messed around a little more, and it looks like 1d6 ties 3d6 as often as 4d6 will roll exactly 7; 1d6 beats 3d6 as often as 4d6 rolls 6 or less; 2d6 ties 4d6 as often as 6d6 will roll exactly 14; and 2d6 beats 4d6 as often as 6d6 rolls 13 or less. So I expect that, when rolling Nd6 against (N+2)d6, they'll tie as often as (N+N+2)d6 rolls exactly N*7, and that Nd6 will beat (N+2) as often as (N+N+2) rolls (N*7)-1 or less.

Wait. Those look the same. Can I just follow this pattern for any number of d6?
Nd6 ties (N+M)d6 as often as (N+N+M)d6 rolls exactly (7*N).
Nd6 beats (N+M)d6 as often as (N+N+M)d6 rolls (7*N)-1 or less.
And then, since the "7" there is a result of the average outcome of a single d6, can I extend it to other sizes of dice?
NdD ties (N+M)dD as often as (N+N+M)dD rolls exactly ((D+1)*N).
NdD beats (N+M)dD as often as (N+N+M)dD rolls ((D+1)*N)-1 or less.
Seriously just guessing at this point (could probably figure it out by hand with more spreadsheets, or by pure math from Pascal's Triangle) but I think this is going to work.


EDIT: VanVelding has tackled the math for this here and here.
EDIT 2: and here.

4 comments :

  1. I turned this over a bit and made a blog about it. How's Wednesday look for you?

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  2. Wednesday's good, I should be free of tests and homework by then.

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  3. You probably have it right. I'd have to do the math too.

    One of these days I will write this up with a function to reference values from Pascal's triangle, then things like this will be easier to write down. You are pretty close to doing just that already.

    Dan

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  4. Heh, yeah, I'm still getting used to spreadsheets.

    Still though, I can't shake the feeling that we should be able to convert the values into a short equation without looking directly at Pascal's triangle. Riemann sums, or simple integrals, or maybe weight an average of the Normal distribution against something else? I dunno.

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