# [RPG] Handling critical hits easily when using average damage

critical-hitdamagedicelessdnd-5e

I use average damage for my monsters almost exclusively. If a goblin's stat block says its weapon attack deals 5 (1d6+2) damage, it deals 5 damage.

Critical hits come up often enough to need a way to deal with them easily. One technique is to double the average. The example goblin would deal 10 damage. This technique is easy (it only requires 1 mental arithmetic operation) but not accurate (it includes the damage bonus twice).

I run most sessions with just 2 d20's at hand, and bringing extra dice just for when critical hits occur would be a waste of space and time.

Is there a diceless technique for averaging critical hit damage that is both easy and accurate? Let's define an "easy" technique to require no more than 3 mental arithmetic operations and no homework ahead of time (in general, it needs to be something that is memorable and isn't too mentally demanding or it defeats the purpose).

The Dungeon Master's Guide includes a technique that counts as easy and accurate (p. 248), but it involves rolling some of the dice, so it doesn't meet my diceless criteria.

Please help me be as lazy as possible without feeling pangs of mathematical guilt.

There are two mathematically equivalent ways to think about/remember the technique I am presenting here. Since this question seeks to find a method that makes a DM's work easier and it only works if the DM can remember and understand it I will outline both variations.

# [Number of dice] × [Number of sides+1] + Modifier

The expected damage for any single die is (x+1)/2 where x is the number of sides the die has.

Since we are dealing with critical damage here, that number is always doubled so for a single die it simplifies to (x+1)+modifier damage.

For more dice you simply multiply that number before adding the modifier. Thus, the general rule:

$$[Number Of Dice] \times [NumberOfSides +1] + modifier$$

### Examples

• 5 (1d6+2):

\$1\times7+2=9\$

• 23 (3d12+4):

\$3\times13+4=43\$

# [Number of dice] + [Maximum possible damage]1

The previous technique took the form:

\$x \times (y+1) + z\$

Where \$x\$ is the number of dice you are rolling, \$y\$ is the number of sides on the die and \$z\$ is the damage modifier.

If you expand that multiplication you get:

\$xy + x + z\$

which we can then rearrange to be:

\$x + xy + z\$

Notice that \$xy + z\$ is actually equivalent to the maximum damage for the attack. For example, the maximum of \$3d6+4\$ is \$3 \times 6+4\$.

Thus, we can simply express the answer as:

$$[NumberOfDice] + [MaxDamage]$$

### Examples

• 5 (1d6+2):

\$1+(6+2)=1+8=9\$

• 23 (3d12+4):

\$3+(36+4)=3+40=43\$

1 - This is an elaboration and clarification on the idea that EightAndAHalfTails' answer and Glen_B's answer presented.