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Mathhammer for 2d6 pick highest.

7K views 5 replies 5 participants last post by  Mad Cat 
#1 ·
Hey everyone, quick question, how would you calculate the average result (not prob distribution) on 'roll 2d6 and pick the highest number', i.e average result on an ordnance pen tests, difficult terrain test etc.
 
#2 ·
Unfortunately, there is no "standard formula" that I know of, as 3D6 has a different formula from 2D6 or 4D6, etc. in short, what you are doing is calculating the odds of the first die rolling a number, and the odds of subsequent dice rolling less than that, keeping in mind that you cannot roll higher than 6.

This type of formula, simplified for 2Dn is:
(n+1)/n*[(2n+1)/3 - 1/2]

When n=6, the average roll is ~4.47

Welcome to Warhammer, the game that makes "worthless" math... nope, still worthless.
 
#3 ·
I was trying to quantify the effects of the new Farsight Enclave seismic fibrillator node.

So for the average infantry, this knocks off 1.5" of movement on the first turn, a bit more once you account for the possibility of also being in effect in subsequent turns.

Yeah... basically useless...

Thanks for the answer!
 
#4 · (Edited)
Well, let me run you through how I do it when I need to calculate this kind of thing (thank the Omnissiah for it being almost universally roll 2, pick the highest)
Open up your trusty spreadsheet program and type out on the side 1,2,3,4,5,6 on lines 2-7 in column A, then along B-G type in (once again) 1-6.
In B2-B7 fill it with the highest (granting 1, 2, 3, 4, 5, and 6), then in C2-C7 do the same (granting 2, 2, 3, 4, 5, 6) and so on. Then just count and take it as a fraction of 36. Since we're going highest with this I'll just give you the fractions (N A/36, where N is the number you want the roll to be and A is how many times that or something higher shows up)

1 36/36
2 35/36
3 32/36
4 27/36
5 20/36
6 11/36

Edit: Oh, darn, the average... Okay well time to multiply and sum...

4.4722.... Simply put, all I did was multiply by the number of appearances of each number, sum, and then divide by 36.... so yeah ~_~ And, yes, his way is better, but I never got into solid probability based math and modeling equations, I'm more into solving them, finding limits, diff eqs, etc. Kind of stupid not taking Stats courses now that I think about it...
 
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