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View Full Version : Dice Math - Static Bonuses vs. Replacing Low Rolls



Sky
2016-11-30, 07:32 PM
Hello, Playground.

Was doing some thinking inspired by one of Pathfinder's Rogue Talents that lets you replace 1s on sneak attack dice with 2s. I did the math comparing the talent with a flat +1 for rolling 2d6 and found that replacing 1s with 2s only raises the average damage from 7 to 7 and 1/9th, while adding a flat +1 ups the average to 8.

However, it seems to me that replacing 1s with 2s would be more useful as you are rolling more dice, since you will be replacing 1s more often. My question is at what point does replacing 1s with 2s become better than a flat +1 bonus to damage, if ever?

Shackel
2016-11-30, 08:16 PM
At least in terms of d6s, around 6d6 is what you would need before replacing ones would, on average, mean more than a +1. In fact, that rings true for all dice: you would need a number of dice equal to the number of sides for it to equal out, since you're only adding 1/X, where X is the number of sides, to the average. 4d4, 6d6, 10d10, etc.

What it relies on, though, is less the average and more the "what-ifs", like what if on that 10d10 you roll 3 1s. Well, now it's a +3 to damage. At the same time, there's also only a 65% chance you'll roll a 1 at all. On 10 ten-sided dice.

My math may be wrong.

Knaight
2016-11-30, 08:30 PM
Hello, Playground.

Was doing some thinking inspired by one of Pathfinder's Rogue Talents that lets you replace 1s on sneak attack dice with 2s. I did the math comparing the talent with a flat +1 for rolling 2d6 and found that replacing 1s with 2s only raises the average damage from 7 to 7 and 1/9th, while adding a flat +1 ups the average to 8.

I'm not sure where you're getting this - the average should go from 7 to 7 1/3. Consider the averaging formulas for an n sided die:

(1+2+3+...n)*1/n would be the easiest way of writing it. That's just the sum of each possible result times the probability of the result. This formula simplifies to (n+1)/2, which is what actually gets used.

Now consider the case where the 1 is replaced with a 2, then do some simple math.
(2+2+3+...n)*1/n

Replace one of those 2s with 1+1
(1+1+2+3+...n)*1/n
Separate out one of the 1s.
1/n+[(1+2+3+..n)*1/n]

Clearly that gets you 1/n higher than you would otherwise get. At n dice it's equivalent to a +1. At 2n dice it's equivalent to a +2. So on and so forth.

Now for a more general case - say you replace that 1 with something other than a 2. Call this variable k. That gives you (k-1)/n higher than average. The obvious case here would be where instead of a 2 you use a reroll, where k thus equals (n+1)/2, assuming that a one rolled for said reroll stays a 1 and isn't rerolled itself. The infinite reroll case is a bit more complicated, and is handled like so:

Consider a die with the following sides [reroll, 2, 3,...n]. A reroll may be rolled up to infinite times, and the probability of a reroll is between 0 and 1 (obviously). An infinite geometric sum can be performed to see how many rerolls there are, with the formula p/(1-p). In said formula p would be (1/n), thus giving the formula (1/n)/(1-(1/n)). In addition to the rerolls there is also one first roll, so the total number of rolls is:
1+(1/n)/(1-(1/n))

A roll which rolls a reroll gives a zero on the roll itself, instead adding to the number of rerolls. So now we can consider the individual dice as showing [0,2,3,...n]. This averages out to (n+1)/2-(1/n). This is then multiplied by the number of rolls for the final formula:
[1+(1/n)/(1-(1/n))]*[(n+1)/2-(1/n)].