Sooo, running tables on odds, I came up with a formula that seems to add up to determine the.....I think it'd be the mean, I'm terrible at math...but the roll with the highest odds at any rate:
M = Y(X/2) + (Y/2)
X = Dice Range; d4, d8, etc.
Y = Multiplier; 2d, 5d, etc.
M = Mean or median or whatever.
So if I want the average on, say, our good friend Mr 20th Lvl Disintegrate, 40(3) + 20 = 140?
Does that work out right? Is there a better way to express it?
RPG Superstar 2010 Top 16
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Yes.
The average roll on an n-sided die is (1+2+3+...+n)/n. That is, ((n^2 + n)/2)/n, which simplifies to (n+1)/2.
In your symbology, you're rolling Y dX. The average roll of a dX is (X+1)/2. If you roll two dX, the average would be twice that. Three dice, three times the average of one die.
So the average roll for Y dX
= Y * (X+1)/2
= Y * (X/2 + 1/2) by the distributive law of division over addition
= Y * X/2 + Y * 1/2 by the distributive law of multiplication over addition
= Y (X/2) + Y/2
Thanks a lot, Chris!
So...is 1 added to X-sides of a dice because the range doesn't include zero?
I was also wondering whether or not the average would be rounded down for in-game utility. 10+ to AC instead of a defense roll would suggest that, but I'm not really sure.
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Thanks a lot, Chris!
So...is 1 added to X-sides of a dice because the range doesn't include zero?
I was also wondering whether or not the average would be rounded down for in-game utility. 10+ to AC instead of a defense roll would suggest that, but I'm not really sure.
Rolls in D&D are rounded down.
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RPG Superstar 2010 Top 16
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No problem, Laddie. This is what I do for a living.
The "n+1" in the formula comes from a common mathematical identity: If you add up the numbers 1 + 2 + 3 + ... + n, the sum will equal half the product of n times n+1. There are a lot of proofs of this formula; the simplest use the principle of induction.
If 6-sided dice read 0,1,2,3,4,5, then the sum would equal half the product of n times n-1.
Fractions round down in 3rd Edition D&D by fiat, but you should round the final result, not any intermediate values. So a game utility calculating the effects of a 9d6 fireball ought to calculate 9 * (6+1)/2 = 9 * 3.5 = 31.5, rounding off to 31. (And not 9 * 3.5, rounding the intermediate value of 3.5 down to 9 * 3 and then getting 27.)
No problem, Laddie. This is what I do for a living.
The "n+1" in the formula comes from a common mathematical identity: If you add up the numbers 1 + 2 + 3 + ... + n, the sum will equal half the product of n times n+1. There are a lot of proofs of this formula; the simplest use the principle of induction.
If 6-sided dice read 0,1,2,3,4,5, then the sum would equal half the product of n times n-1.
Fractions round down in 3rd Edition D&D by fiat, but you should round the final result, not any intermediate values. So a game utility calculating the effects of a 9d6 fireball ought to calculate 9 * (6+1)/2 = 9 * 3.5 = 31.5, rounding off to 31. (And not 9 * 3.5, rounding the intermediate value of 3.5 down to 9 * 3 and then getting 27.)
Thanks again!
The reason I asked was because I'm tinkering with a point-buy system. In that context, I would probably want to price a combination at the fraction, but then the upper levels of chance on a range would rate as preferable to a flat bonus in most gamers' heads...