Handling critical hits easily when using average damage

Bloodcinder 06/12/2018. 5 answers, 1.717 views
dnd-5e damage critical-hit

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).

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.

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.

5 Answers


Rubiksmoose 06/13/2018.

[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\$


Another way to phrase this is:

[Number of dice] + [Maximum possible damage]

As shown mathematically in EightAndAHalfTails' answer and described in Glen_B's answer, there is another way to describe the above. It is helpful to note this since this way may be easier to understand/remember and this technique is only really helpful if it is something the DM can remember.

So, I showed above that the max damage can be expressed as:

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

Or if we replace the words with variables:

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

However, if you expand that multiplication you get:

\$xy + x + z\$

which we can 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:

$$[Number of Dice] + [MaxDamage]$$

Examples

  • 5 (1d6+2):

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

  • 23 (3d12+4): $$3+(36+4)=3+40=43$$


FenrirG 06/12/2018.

1st Double the average.
2nd Subtract the damage modifier.

"it includes the damage bonus twice." So subtract once.

5 (1d6+2)

  1. ×2=10
  2. -2=8

Alternative to be more accurate.

This is only necessary for rolling odd number of dice. The average for odd number of dice will always have 0.5 which is rounded down for regular damage, so for 1d6, the average on a critical hit would be 7 (= 3.5 × 2), but taken as 6 if you round down the average of a regular hit first. Here is a way to include the 0.5s with 3 mental arithmetic operations:

  1. Double the average damage
  2. Subtract the modifier
  3. Add 1

Crit with a 3d12+4 damage (average 23)

  1. ×2 = 46
  2. -4 = 42
  3. +1 = 43 (equal to average of 6d12+4)

Southpaw Hare 06/12/2018.

The average of double NdX is N×(X+1). For 1dX, this is simply X+1.

For instance, the average damage of a critical hit with 1d6 is 7.

If all you're doing is doubling dice on a critical hit, the formula is easy to remember: simply take the number of sides on the die and add one.

The average roll on a die of x sides is (x+1)/2. Since you're multiplying the number of dice by 2, this negates the division by 2, thus leaving you with x+1.

As an exhaustive list of this formula applied to all damage dice typically used:

  • d2 = 3
  • d3 = 4
  • d4 = 5
  • d6 = 7
  • d8 = 9
  • d10 = 11
  • d12 = 13

When calculating your critical hits, just substitute in this number for your doubled dice first, and then add your bonuses/penalties only once.


EightAndAHalfTails 06/12/2018.

I can think of two ways, depending on how accurate you want to be and much mental arithmetic you're willing to do.

Method 1 (3 operations)

The critical hit damage for an attack dealing (\$x\$d\$y\$ + \$z\$) damage is given by:

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

or

\$xy + x + z\$

Prove this by considering the expected value of a d\$y\$: \$((y+1)/2)\$.

Thus, the expected value of \$x\$d\$y\$ is \$x((y+1)/2)\$.

This is the term we double on a critical hit:

\$2x((y+1)/2) = x(y+1) = xy+x\$

Then we add on the bonus damage \$z\$ a single time, giving the quoted formulae.

Method 2 (2 operations, sometimes less accurate)

The critical hit damage for an attack dealing \$m\$ (\$x\$d\$y\$ + \$z\$) damage is approximately:

\$2m - z\$

This method doubles the listed average damage and then subtracts the bonus damage that was erroneously included twice.

The inaccuracy in this method stems from the fact that the listed average damage is always an integer. However, the actual statistical expected value is not always an integer. In particular, the expected value is non-integer when \$xy + x\$ is odd. Since \$y\$ is always even (all D&D dice have an even number of sides\$^1\$), this is the same as saying \$x\$ is odd.

In the case where \$x\$ is odd, the listed average damage is rounded down, and hence undershoots the actual expected value by 0.5. This means that our doubled value will undershoot the actual expected value by 1.

If you consider this to be important, then you could go for...

Method 2a (1 parity check + 2 or 3 operations, exact)

The critical hit damage for an attack dealing (\$x\$d\$y\$ + \$z\$) damage is given by:

\$2m - z\$ if \$x\$ even

or

\$2m - z + 1\$ if \$x\$ odd

  1. Some attacks may use d3s or other odd-sided dice. Luckily the expected value of odd-sided dice is always even, so this method introduces no error in those cases.

Glen_b 06/13/2018.

Critical damage doesn't happen enough times in a combat that you should feel the need to be accurate*, but what about the rule "an average crit is the number of dice more than the maximum damage"?

Ignoring +'s for the moment: The sum of any number of D&D dice is symmetrical about its mean. The mean is the average of the minimum and the maximum. The minimum is always the number of dice you rolled. Hence the average of "double the die roll" is "min + max". Now we need to incorporate the +'s. If we just add that on it leaves the maximum in the right place, but if we include it in the minimum we will have added the +damage twice. So we want something like "the minimum possible on the dice + the maximum roll" which is much more memorable as "the number of dice + the maximum roll".

This works for any number of dice plus a modifier.

In the case of 1d6+2 it's one die, so "one more than maximum" is 9, which is in fact the same as the average of double damage + modifier.

Or consider an attack that does 2d4+1. That's two dice and the maximum is 9, so an average crit should do 9+2.

If you somehow had different dice and needed them all to be doubled (like d4 + d8), this rule still works.

If you have different dice but for some reason only some of them get to be doubled, put the average damage of the ones that don't double with the +s and you only count the number of dice that get to double in the above rule.

So for example imagine you had some situation where you had 2d8 + d6 + d4 + 3 where the 2d8 and the d6 got doubled by the crit but (for whatever potential reason) the d4 did not - the non-doubling d4 would count as +2. Then you'd get max damage is 16+6+(2+3) = 27, and the number of dice that double is 3, so an average crit is 30 (the true average is 30.5 which rounds down to 30 in the usual way)

* "Average" damage will tend to to be too low (it's 1 point too little damage per two hits on anything with an odd number of dice), so bumping up crits a little to compensate for usually being too low isn't necessarily a problem. If you want to say that a crit does double the average damage, you're still going to be doing too little damage in the long run -- averaged over all attacks -- any time there are an odd number of dice (of the normal types) and just slightly too much when there are an even number of dice; overall you should not have anything to worry about because it should still average less than the true average if you had rolled.

[Or if you want to make your life super easy, just take the maximum damage. Yes it's too low, but so is "average damage"]


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