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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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4 Answers 4

up vote 22 down vote accepted

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 = 4.1̅6

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

Die   Avg.    Δ
---   ----   ----
d4    3.00   0.50
d6    4.1̅6   0.6̅6
d8    5.25   0.75
d10   6.30   0.80
d12   7.3̅3   0.8̅3

Dice are independent. 2D6 will have an average value of 2 * 4.1̅6, or 8.3̅3.

Common weapon average damage (Great Weapon Fighting):

  • Greatsword (2d6): 8.3̅3 (Δ1.3̅3)

  • Greataxe (1d12): 7.3̅3 (Δ0.8̅3)

  • Longsword (1d10): 6.30 (Δ0.80)

  • Smite (level 1, 2d8): 10.50 (+ weapon damage) (Δ1.5)


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

2d6 distribution 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 dX , if you will). Great Weapon Fighting will increase the average roll on your die by 1-(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 (Y/2)*[1-(Y/X)]. 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 dX is [(X^2 + X)/2] / X. This is the same as (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 dX. 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 dX.

  2. Adding the average value on the original dX two times.

  3. Dividing this number by X.

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

[-1 -2 + (X + 1) / 2 + (X + 1) / 2] / X = 
[   -3 + (X + 1)                  ] / X = 
         (X - 2)                    / X =
          1 - (2 / X)

General case (short proof)

Let us suppose that a dX 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*(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 dX and the average for the modified dX [the one where we re-roll if the first roll turns up any of the Y lowest values]) is equal to:

    [ -(Y^2 + Y) / 2 + Y * (X^2 + X) / 2  ] / X =
Y * [ -(Y   + 1) / 2 +     (X   + 1) / 2) ] / X =
       (Y / 2)           * (X   - Y)        / X =
       (Y / 2)           * [1   -        (Y / X)]
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does rpg SE support MathTEX? – Skyler Sep 25 at 8:40
@Skyler No, but this answer has been added to the list of answers that would benefit from MathJax in the feature-request to enable it. – SevenSidedDie Nov 13 at 18:57

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