# How much damage does Great Weapon Fighting add on average?

The Great Weapon Fighting Fighting style states the following:

When you roll a 1 or 2 on a damage die for an attack you make with a melee weapon that you are wielding with two hands, you can reroll the die and must use the new roll. The weapon must have a two-handed or versatile property for you to gain this benefit.

How much does this ability increase the average damage of its wielder?

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I've forgotten the formal proof for this, but hopefully this is correct:

Consider a D6 (for the sake of concrete language).

When you roll a 1, you reroll the die and keep the result. This produces an average value of 3.5, and happens 1/6 of the time.

When you roll a 2, you reroll the die and keep the result (even if it's lower). This produces an average value of 3.5, and happens 1/6 of the time.

When you roll a 3, you keep the result. This produces an average value of 3, and happens 1/6 of the time.

And so on.

This gives the following formula for the average of the D6: $(3.5 + 3.5 + 3 + 4 + 5 + 6) / 6 = 4.1\bar{6}$.

Working similar formulas for the other dice, we get this table:

\begin{array}{lccc} \hline \text{Die} & \text{(standard) Avg.} & \text{GWF Avg.} & \Delta \\ \hline \text{d4} & 2.5 & 3.00 & 0.50 \\ \text{d6} & 3.5 & 4.1\bar{6} & 0.6\bar{6} \\ \text{d8} & 4.5 & 5.25 & 0.75 \\ \text{d10} & 5.5 & 6.30 & 0.80 \\ \text{d12} & 6.5 & 7.3\bar{3} & 0.8\bar{3} \\ \hline \end{array}

Dice are independent. 2D6 will have an average value of $2 \cdot 4.1\bar{6} = 8.3\bar{3}$.

Common weapon average damage (Great Weapon Fighting):

\begin{array}{lcc} \hline \text{Weapon} & \text{Avg. GWF dmg} & \text{improvement w/ GWF}\\ \hline \text{Greatsword (2d6)} & 8.3\bar{3} & 1.3\bar{3} \\ \text{Greataxe (1d12)} & 7.3\bar{3} & 0.8\bar{3} \\ \text{Longsword (1d10)} & 6.30 & 0.80 \\ \text{Smite (level 1, 2d8)} & 10.50 & 1.50 \\ \qquad \text{(+ weapon damage)} \\ \hline \end{array}

Observations:

• The ability works out to about a +1 to damage.

• It scales to almost a +3 when smiting. The more dice you add (high level smite, for example), the better the ability.

• The bonus is "swingy." It can range from a -2 to a +10 on 2D6, for example.

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is delta the increase in average damage from GWF? If so, can you mention that explicitly – Premier Bromanov Jun 5 at 19:45
@PremierBromanov Delta means "change in value". – AceCalhoon Jun 5 at 20:25
not everyone is as learned in math, thats usually why they ask these questions – Premier Bromanov Jun 5 at 22:05

AceCalhoon's answer has the numbers, but I think it is useful to illustrate it with probability graphs. Here is how the probability distributions of two common damage dice, d12 and 2d6, change.

D12: The results 1 and 2 simply become very unlikely, boosting the probability of the rest.

2D6: Here the effect doesn't look linear. 2-5 all become much less likely. The peak is skewed from 6-8 to roughly 7-10.

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Using the same math as AceCalhoon, here is your relative damage increases in percentage.

d10: +14.5% damage

d8: +16.6% damage

d6: +19% damage

d4: +20% damage

However this only applies to damage dice, not flat damage bonuses. How many damage dice or flat bonuses you have depends greatly on your build (Great Weapon Master's +10 damage has no synergy, but Paladin's Smite and Crusader's Mantle do).

In general, if you use a greatsword, you can expect around a 10% to 15% overall damage boost from this fighting style.

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I'm a math guy and I know most people aren't so I'll spare the gritty details unless someone actually wants to see a proof.

Suppose you have a die of size $X$ (a d$X$ , if you will). Great Weapon Fighting will increase the average roll on your die by $1-\frac{2}X)$. So the bigger the die, the more your average damage increases, although this increase can never be larger than 1.

In general, suppose you have a die of size $X$ and you get to re-roll it once whenever it lands on any of the $Y$ lowest numbers (for Great Weapon Fighting $Y$ would be 2, for the Halfling's Lucky racial feature $Y$ would be 1, etc.). Then the increase in the average roll on your die is equal to $\frac{Y}{2} \cdot \left[1-\frac{Y}{X}\right]$. Note that this formula only makes sense if $Y$ is less than $X$.

Below is a short proof which is not technically correct but is much easier to follow than a complete proof. Again, if anyone would like a more detailed proof just let me know.

# Short proof

If you want to find the average roll on a die, you add up the values on each of its faces and divide by the total number of faces. It's a known mathematical formula that the sum of numbers from 1 to $X$ is equal to $(X^2 + X)/2$. So the average roll on a d$X$ is $\left[(X^2 + X)/2\right]/X = (X + 1)/2$.

With Great Weapon Fighting you are allowed to re-roll all 1s and 2s on a damage die. This is equivalent to replacing the "1" and "2" on the die with the value of its average roll (for a d6, for example, using GWF would be equivalent to rolling a 6 sided die where the faces were labeled "3.5", "3.5", "3", "4", "5", "6").

Suppose we had such a die (the modified one described in the previous paragraph), and we wanted to find out the difference between its average value and the average value of the original d$X$. How would we proceed? We would do this by:

1. Subtracting 1 and 2 from the sum of the numbers on the faces of the original d$X$.

2. Adding the average value on the original d$X$ two times.

3. Dividing this number by $X$.

Putting this together, the number we are looking for (the difference between the original d$X$ average and our modified GWF die average) is equal to

\begin{eqnarray} &\left.\left[-1 -2 + \frac{X + 1}{2} + \frac{X + 1}{2}\right]\right/X\\ =&\frac{-3 + (X + 1)}{X}\\ =&\frac{X - 2}{X}\\ =&1 - \frac{2}{X} \end{eqnarray}

# General case (short proof)

Let us suppose that a d$X$ is to be rolled and, if the die comes up showing any of the $Y$ lowest values, then the die is re-rerolled exactly one time. What is the expected value for such a die? As above, we proceed as follows:

1. Subtract the values "1" through $Y$ from the sum of the numbers on the faces of the original die. This value (the value of the numbers we are subtracting) is equal to $(Y^2 + Y)/2$.

2. Add the average value of the original die roll $Y$ times. This is equivalent to adding $Y \cdot (X + 1) / 2$ to the sum of the numbers on the original die.

3. Divide this number by $X$.

Putting this together, the number we are looking for (the difference between the average for the original d$X$ and the average for the modified d$X$ [the one where we re-roll if the first roll turns up any of the $Y$ lowest values]) is equal to:

\begin{eqnarray} &\left.\left[-\frac{Y^2+Y}{2}+Y \frac{X+1}{2}\right]\right/X\\ =&\frac{Y}{2} \cdot \frac{-(Y+1)+(X+1)}{X}\\ =&\frac{Y}{2} \cdot \frac{X-Y}{X}\\ =&\frac{Y}{2} \cdot \left(1-\frac{Y}{X}\right) \end{eqnarray}

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