I was looking at the Great Weapon Fighting style and thinking about how much extra damage it actually gave. I figured out that the amount of extra damage you get for a die with n sides is: (n-2) / 2.
That's the TL;DR; Here's how I derived that formula which I thought might be interesting.
First, there is a neat mathematical trick we'll be using so we'll start with that. The story goes that the genius of Carl Friedrich Gauss was discovered as a boy in elementary school when he, along with the rest of the class, was assigned the task of adding up all the numbers from 1 through 100 as busy work. The clever boy realized that he could pair 1+100, 2 + 99, 3 + 98, etc all the way to 50 + 51 making 50 pairs of 101 for a total of 5050. The formula for adding up all the numbers from 1 through n is: n(n+1) / 2
I'm sure everyone here is used to calculating the average roll for an n sided die by using (n+1)/2. That's a sort of reduced, simplified form of how you calculate the average result. What you really do is multiple each possible result by the chance of it happening.
That looks like: 1/n + 2/n + 3/n + ... + n/n
or: (1 + 2 + 3 + ... + n) / n
We know from above what the sum of 1 through n is, so we can substitute that in: (n(n+1) / 2) / n
The n's on the top and the bottom cancel each other out and we get the familiar: (n+1) / 2
Now lets look at what happens when we can reroll a 1 or 2 (but only once). Lets go back to the original formula we had for average: (1 + 2 + 3 + ... + n) / n
We're going to replace the 1 and 2 at the start of the formula because they get rerolled. And what is the expected value to put there? The average roll for an n sided dice or n+1/2. So, we can replace the 1+2 with 2 average rolls and end up with: ((n+1) + 3 + ... + n) / n
That can tell us the average value, but we were looking for how more damage you got versus without the style. So we can subtract the normal average roll from the average with reroll and that looks like:
(((n+1) + 3 + ... + n) / n) - ((1 + 2 + 3 + ... + n) / n)
They have the same denominator and most everything just cancels each other out
( ((n+1)+ 3 + ... + n) - (1 + 2 + 3 + ... + n) ) / n
( (n+1) - (1 + 2) ) / n
Voila! ( n-2 ) / n
With that you see the average bonus you get for a given die size:
d12 = 10/12 = .83
d10 = 8/10 = .8
d8 = 6/8 = .75
d6 = 4/6 = .66
As you can see, the style should maybe be called "Two Dee Six Fighting" because using a Greatsword or Maul is definitely the best way to try to extract value from it. A character with the style will average exactly one more point of damage per attack by using a 2d6 weapon instead of a d12 (which is kind of interesting).
You might be curious about what would happen if you allowed continuous rerolling of 1's and 2's as a house rule. A good trick in that case, so as to not waste time actually rolling dice over and over, is to use a die one size smaller and add 2 to the result. So, if you were using a Halberd you could roll d8+2 producing a range of values from 3-10. This method also makes it easy to see that continuous rerolls of 1's and 2's will boost average damage by 1 per die.
That's the TL;DR; Here's how I derived that formula which I thought might be interesting.
First, there is a neat mathematical trick we'll be using so we'll start with that. The story goes that the genius of Carl Friedrich Gauss was discovered as a boy in elementary school when he, along with the rest of the class, was assigned the task of adding up all the numbers from 1 through 100 as busy work. The clever boy realized that he could pair 1+100, 2 + 99, 3 + 98, etc all the way to 50 + 51 making 50 pairs of 101 for a total of 5050. The formula for adding up all the numbers from 1 through n is: n(n+1) / 2
I'm sure everyone here is used to calculating the average roll for an n sided die by using (n+1)/2. That's a sort of reduced, simplified form of how you calculate the average result. What you really do is multiple each possible result by the chance of it happening.
That looks like: 1/n + 2/n + 3/n + ... + n/n
or: (1 + 2 + 3 + ... + n) / n
We know from above what the sum of 1 through n is, so we can substitute that in: (n(n+1) / 2) / n
The n's on the top and the bottom cancel each other out and we get the familiar: (n+1) / 2
Now lets look at what happens when we can reroll a 1 or 2 (but only once). Lets go back to the original formula we had for average: (1 + 2 + 3 + ... + n) / n
We're going to replace the 1 and 2 at the start of the formula because they get rerolled. And what is the expected value to put there? The average roll for an n sided dice or n+1/2. So, we can replace the 1+2 with 2 average rolls and end up with: ((n+1) + 3 + ... + n) / n
That can tell us the average value, but we were looking for how more damage you got versus without the style. So we can subtract the normal average roll from the average with reroll and that looks like:
(((n+1) + 3 + ... + n) / n) - ((1 + 2 + 3 + ... + n) / n)
They have the same denominator and most everything just cancels each other out
( ((n+1)
( (n+1) - (1 + 2) ) / n
Voila! ( n-2 ) / n
With that you see the average bonus you get for a given die size:
d12 = 10/12 = .83
d10 = 8/10 = .8
d8 = 6/8 = .75
d6 = 4/6 = .66
As you can see, the style should maybe be called "Two Dee Six Fighting" because using a Greatsword or Maul is definitely the best way to try to extract value from it. A character with the style will average exactly one more point of damage per attack by using a 2d6 weapon instead of a d12 (which is kind of interesting).
You might be curious about what would happen if you allowed continuous rerolling of 1's and 2's as a house rule. A good trick in that case, so as to not waste time actually rolling dice over and over, is to use a die one size smaller and add 2 to the result. So, if you were using a Halberd you could roll d8+2 producing a range of values from 3-10. This method also makes it easy to see that continuous rerolls of 1's and 2's will boost average damage by 1 per die.
