In the One D&D Character Origins UA, it mentions that rolling a nat 20 on an attack roll grants inspiration. This can then be used to give the PC advantage on their next attack, giving them a higher chance of rolling a nat 20 again. While this only a 5 percentage point increase in crit chance, it does affect hit chance quite a bit, this could affect damage quite a bit.

What kind of effect are we talking about here? Is it going to be a big deal? In what situations will it be more different or least? Should I be worried about characters snowballing?

I did some quick math using a level 1 character with 1d8 weapon +4 damage +6 hit against a CR1 enemy and found the following;

  • They look like they are dealing about 1.7% more damage on average
  • Having low or high damage die or damage bonus doesn't make a difference
  • Having a high dice target makes the damage % higher - a 5 higher target meant roughly 1.3 more percentage points, a 10 higher target meant that much more again.
  • Elven accuracy at level 1 does seem to increase the damage by about 1.7 percentage points
  • Features that improve your crit chance see a slight increase in damage, but not much maybe .1 percentage point.

Overall it seems like the you are going to find it hard to find even a 5% increase in damage thanks to this new rule. Is that analysis correct or is inspiration on nat 20 to hit a bigger deal than I think?


5 Answers 5


I think you are asking for an objective measure of the effect of the rule. That needs to be answered with mathematical evidence, and little can be learned from particular character examples. Here I focus on the maths, and leave the comments for other people.


The math involved in an exact answer is complicated, but the "feedback loop" (i.e. the recurrence) converges very rapidly to values very close what you get without inspiration just 0.25% above for rolling a 20 and below for rolling a 1. So, there will be a measurable but small consequence from this rule.

After many rolls, the probabilities of rolling n is approximately

$$ p(n) = (4.72 + 0.026 \times n ) \,\% $$

For example, if you start without advantage, the firts roll all numbers are equally probable (5%), the second roll the probability of rolling a 1 or a 20 are 4.76% and 5.23% respectively. After infinite rolls, where you use the inspiration immediately goes to 4.75 and 5.25% respectively. Very similar to the second roll.

Calculating the expected damage, can be done considering this probability. Just multiply the probability for each outcome with the corresponding damage, and add all the results.


The expected damage, or average damage, is the sum of all the possible damages weighted by their probability.

$$ \sum_{n=1}^{20} p(n) h(n) $$

where p(n) is the probability of rolling n, and h(n) is the damage rolling n, so zero if n<AC, and double if n=20.

For instance if the damage is (1D8 + 2), and a 10 hits, the average damage is 3.57500 without considering double damage on 20, 3.90000 with double damage on 20, and 3.99587 with inspiration rule and double damage on 20.

More simple, assuming a m and above hits, then the sum would be

$$ (\text{Damage}) \times \left( p(20) + \sum_{n = m}^{20} p(n) \right) $$

The factor on the right depend on m the roll that hits, depending on AC and modifiers, and is given in this table.

$$\begin{array}{cccc} \text{Roll that hits} & \text{Factor with Inspiration} & \text{Factor without Inspiration} & \text{Additional damage} \\\hline 2 & 1.005 & 1. & 0.48\% \\ 3 & 0.957 & 0.95 & 0.72\% \\ 4 & 0.909 & 0.9 & 0.97\% \\ 5 & 0.86 & 0.85 & 1.22\% \\ 6 & 0.812 & 0.8 & 1.47\% \\ 7 & 0.763 & 0.75 & 1.72\% \\ 8 & 0.714 & 0.7 & 1.96\% \\ 9 & 0.664 & 0.65 & 2.21\% \\ 10 & 0.615 & 0.6 & 2.46\% \\ 11 & 0.565 & 0.55 & 2.7\% \\ 12 & 0.515 & 0.5 & 2.95\% \\ 13 & 0.464 & 0.45 & 3.19\% \\ 14 & 0.414 & 0.4 & 3.44\% \\ 15 & 0.363 & 0.35 & 3.68\% \\ 16 & 0.312 & 0.3 & 3.92\% \\ 17 & 0.26 & 0.25 & 4.15\% \\ 18 & 0.209 & 0.2 & 4.38\% \\ 19 & 0.157 & 0.15 & 4.58\% \\ 20 & 0.105 & 0.1 & 4.75\% \\ \end{array}$$

Math analysis

Let's consider the D20 Discrete Uniform Distribution, any outcome is equally probable, so the probability is 5% for any number from 1 to 20.

$$ P_{D20}(n) = \frac{1}{20} $$.

Plot 1: Discrete Uniform Distribution for a D20
Discrete Uniform Distribution Plot, all with value 0.05 in the range 1 to 20
This is the simplest case of rolling a D20, each outcome is equally probable.

The "Advantage D20", that is taken by the maximum of two independent D20 rolls is given by

$$ P_{A D20}(n) = \frac{(2 \times n - 1)}{400} $$

This is no longer a fixed number, but a function of the number n. So for example the probability of getting a 1 with advantage is now 0.25% and rolling a 20 is 9.75%, compared with the fixed 5% without advantage.

Plot 2: Discrete Distribution for a D20 with advantage
Advantage discrete probability distribution. The values grow linearly from 1/400 to 39/400
By selecting the largest outcome from two dice the distribution now grows linearly from 0.25% for 1 to 9.75% for 20

Now consider you started without advantage and roll some unknown number, for the second roll (k=2), consider that there is an 5% (1/20) that the previous roll gave advantage and 95% (19/20) that it didn't. Now the "Inspiration D20" P(n,2) needs to consider this two cases, with and without advantage, weighted by the probability of each case.

$$ P_{I}(n,2) = 0.95 \times \frac{1}{20} + 0.05 \times \frac{(2 \, n - 1)}{400} = 0.047375 + 0.00025 \times n $$

Plot 3: Inspiration Discrete Distribution
Inspiration discrete probability distribution.
Almost flat distribution, in mainly the Discrete Uniform Distribution and a bit of the Discrete Distribution for a D20 with advantage.

Now we can generalize for the roll number k as a function of the outcome from the previous roll k-1,

$$ P_{I}(n,k) = \sum_{i=1}^{19} \left(P_{I}( i ,k-1) \times \frac{1}{20} \right)+ P_{I}( 20 ,k-1) \times \frac{(2 \, n - 1)}{400} $$

The recurrence here becomes complicated to do by hand, using Wolfram Mathematica

sol = First@ RSolve[
         p[1,n] == 1/20,
         p[k,n] == Sum[p[k-1,i],{i,1,19}]/20+p[k-1,20](2 n -1)/400
      , {n,1,20}
   , Evaluate@Table[p[k,n],{n,1,20}]
   , k

I'm omitting the long mathematical expression here, as it doesn't add much.

The solution depends on the roll number and on the initial condition, but stabilizes very fast to the limit when k is very big, which can be calculated taking the limit when k goes to infinity to obtain.

$$ p(n) = \left(\frac{600}{127} + \frac{10 n}{381}\right) \,\% $$

Now the table of values is as follows

$$ \begin{array}{ccc} \text{Roll outcome} & \text{Probability fraction} & \text{Probability percentage} \\ \hline 1 & \frac{181}{3810} & 4.751 \% \\ 2 & \frac{182}{3810} & 4.777 \% \\ 3 & \frac{183}{3810} & 4.803 \% \\ 4 & \frac{184}{3810} & 4.829 \% \\ 5 & \frac{185}{3810} & 4.856 \% \\ 6 & \frac{186}{3810} & 4.882 \% \\ 7 & \frac{187}{3810} & 4.908 \% \\ 8 & \frac{188}{3810} & 4.934 \% \\ 9 & \frac{189}{3810} & 4.961 \% \\ 10 & \frac{190}{3810} & 4.987 \% \\ 11 & \frac{191}{3810} & 5.013 \% \\ 12 & \frac{192}{3810} & 5.039 \% \\ 13 & \frac{193}{3810} & 5.066 \% \\ 14 & \frac{194}{3810} & 5.092 \% \\ 15 & \frac{195}{3810} & 5.118 \% \\ 16 & \frac{196}{3810} & 5.144 \% \\ 17 & \frac{197}{3810} & 5.171 \% \\ 18 & \frac{198}{3810} & 5.197 \% \\ 19 & \frac{199}{3810} & 5.223 \% \\ 20 & \frac{200}{3810} & 5.249 \% \\ \end{array} $$

which is not very different from the regular 5% of getting any number.

Now, going back to the fact that the probability depends on the roll number, we know it tends to the values shown before, but how is the probability the roll just after a roll we know started with advantage, or without. This is given in this plots.

If you start without inspiration

Plot 4: Probability trend from first to 4rth roll starting without advantage
Many traces start all at 0.05 and spread to a stable value.
Notice that all the probabilities start the same (5%) and spread quickly.

If you start with advantage.

Plot 5: Probability trend from first to 4rth roll starting with advantage
Many traces start all widely spreader and converge to a narrow stable spread.
Notice that the probabilities start speeded like the Discrete Distribution for a D20 with advantage but converge quickly to the smaller spread of the Inspiration discrete Distribution.

We see here that regardless if we start with or without advantage by the second roll the probabilities are almost indistinguishable from the limit (settled) case.

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    \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Someone_Evil
    Commented Aug 30, 2022 at 20:11

This mechanic has some effect on DPR, which grows linearly on the target roll.

The difference in DPR for a target roll \$t\$ is given by $$ \Delta_{\text{DPR}}(t) = \sum_{i=t}^{20} \frac{20}{421-2i}- \frac{21-t}{20} + (C-0.05) $$ where \$C=0.0524934\$ is the crit chance.

\$t\$ \$\Delta_{\text{DPR}}(t)\$ \$\Delta_{\text{DPR}}(t)\%\$
2 0.0056 0.5593
3 0.0076 0.8033
4 0.0094 1.0488
5 0.0110 1.2956
6 0.0124 1.5439
7 0.0135 1.7935
8 0.0143 2.0444
9 0.0149 2.2967
10 0.0153 2.5501
11 0.0154 2.8046
12 0.0153 3.0600
13 0.0149 3.3160
14 0.0143 3.5723
15 0.0134 3.8282
16 0.0122 4.0826
17 0.0108 4.3335
18 0.0092 4.5771
19 0.0072 4.8041
20 0.0050 4.9869

The above table is computed when the expected damage is 1 and the modifier is 0, in order to check the "pure" influence of this mechanic.

Mathematical derivation

The derivation is organized in 3 parts: first of all, the explicit formula for the chance to crit is given under the new mechanics. Secondly, the reasoning is generalized to any target roll. Finally, the final formula is provided, which leads to the table at the top of the answer with \$d=1, m=0\$.

Chance to crit

Let's analyze how the probability of getting a critical hit changes when the new mechanic about inspiration is applied.

Denote with \$P(k)\$ the probability of getting a 20 on \$k\$ d20 rolls. Obviously, $$ P(1) = \frac{1}{20}. $$ Then, the probability of getting a 20 on the second roll is given by $$ \begin{split} P(2) =& \frac{1}{20}P(\text{1st roll did not crit}) + \frac{39}{400}P(\text{1st roll did crit})\\ =& \frac{1}{20}(1-P(1)) + \frac{39}{400}P(1)\\ =& \frac{1}{20} + \frac{1}{20}\frac{19}{20}P(1). \end{split} $$ The reason why I splitted \$\frac{1}{20}\frac{19}{20}\$ will become clear later on. We can compute then $$ \begin{split} P(3) =& \frac{1}{20}P(\text{2nd roll did not crit}) + \frac{39}{400}P(\text{2nd roll did crit})\\ =& \frac{1}{20} \left(1-P(2)\right)+ \frac{39}{400}P(2)\\ =& \frac{1}{20} + \frac{1}{20}\frac{19}{20}P(2)\\ =& \frac{1}{20} + \frac{1}{20}\frac{19}{20}\left(\frac{1}{20} + \frac{1}{20}\frac{19}{20}P(1)\right)\\ =& \frac{1}{20}+ \frac{19}{20}\left(\frac{1}{20}\right)^2+\left(\frac{19}{20}\right)^2\left(\frac{1}{20}\right)^3 \end{split} $$

One may compute then \$P(4),P(5), \dots\$: this will lead to find the formula for \$k\$ rolls, after some algebraic manipulation $$ P(k) = \frac{1}{20}\sum_{i=0}^k\left(\frac{19}{20}\frac{1}{20}\right)^i=\frac{1}{20}\sum_{i=0}^k\left(\frac{19}{400}\right)^i $$ This can be proven also by induction. For a very large of roll, we can compute then the limit for \$k\to\infty\$: $$ C = \lim_{k\to\infty}P(k) = \frac{1}{20}\sum_{i=0}^\infty\left(\frac{19}{400}\right)^i = 0.0524934. $$

Then, the probability to crit slightly increases: this theoretical computation is in accordance with rheman's answer.


Let's now generalize the above computation for a number \$t\in [1,20]\$. Denote with \$P_a(k,t)\$ the probability of rolling \$t\$ with advantage in \$k\$ rolls under the new mechanic. First of all, let's recall that the probability to roll a number t with advantage on a single roll is $$ P_a(t) = \frac{2t-1}{400}. $$ Fix \$k=2\$: the probability to roll \$t\$ with advantage the 2nd roll or to get a new inspiration is given by $$ \begin{split} P_a(2,t) &= \frac{1}{20}\left(1-P(1)\right) + \frac{2t-1}{400}P(1)\\ &= \frac{1}{20} + \frac{2t-21}{400}P(1), \end{split} $$ being \$P(1)=1/20\$. Following the same reasoning above, for a enough large number of rolls one obtains that the probability \$P_a(t)\$ of getting \$t\$ or advantage is given by $$ P_a(t) = \frac{20}{421-2t}. $$ For \$t=20\$ we obtain again the result for crit chances. The results are summed up in the following table.

\$t\$ \$P_a(t)\%\$
1 4.7733
2 4.7962
3 4.8193
4 4.8426
5 4.8662
6 4.8900
7 4.9140
8 4.9383
9 4.9628
10 4.9875
11 5.0125
12 5.0378
13 5.0633
14 5.0891
15 5.1151
16 5.1414
17 5.1680
18 5.1948
19 5.2219
20 5.2493

The results are slightly different from rherman's since they are using a linear approximation.

The probability to get a roll less or equal to \$t\$ is $$ P_\leq(t) = \sum_{i=1}^t \frac{20}{421-2i}, $$ for which there is not explicit&simple formulation (see here and here for a deeper insight).

The probability hence to get a roll at least equal to \$t\$ is then $$ P_\geq(t) = 1-\sum_{i=1}^{t-1} \frac{20}{421-2i} = \sum_{i=t}^{20} \frac{20}{421-2i} $$

Final formula

We can gather all the computation here, and using the formula provided by Someone_Evil we obtain the DPR for a particular target number \$t\$: $$ \text{DPR}_a(t) = P_\geq(t)(d+m) + C\,d = \sum_{i=t}^{20} \frac{20}{421-2i}\,(d+m) + C\,d $$ where \$d\$ and \$m\$ are the expected damage and the modifier, respectively.

Compare it with the DPR formula for a target \$t\$: $$ \text{DPR}(t) = \left(\frac{21-t}{20}\right)(d+m) + \frac{d}{20}, $$ the difference reads as $$ \sum_{i=t}^{20} \frac{20}{421-2i}- \frac{21-t}{20} + (C-0.05) $$

which leads to the table at the top of the answer.


The effect is negligible.

I've done horrible things to an excel spreadsheet. This is really easy to brute force using excel formulas. In cell A1, type:


This gives a uniform distribution on the integers from 1 to 20. Next, in cell A2, we type:


If the cell above was a 20, it rolls with advantage, else it rolls a single d20. This is where the naughtiness comes in. I then just copied A2 down to A1048576, then copied the column across seven more columns for a total of 8,388,608 trials, then averaged the space:

Average d20 roll Average with Inspiration on a 20
10.500 10.675

With +6 to hit against an AC 15 target, this takes the probability of a hit from 60% to 61.2%, or the expected damage of 1d8+4 from 5.1 per attack to 5.2.

The limitation to this answer is that it doesn’t provide any meaningful insight into the underlying math. But it gives a concrete number, so that’s cool.

  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$ Commented Aug 30, 2022 at 1:25
  • \$\begingroup\$ I think this answer overlooks the increase in damage from the increased likelihood of critical hits. The effect might be negligible in the "to hit" stats but gut feel is that the DPR increase will be more noticable. \$\endgroup\$
    – linksassin
    Commented Aug 30, 2022 at 2:22
  • 2
    \$\begingroup\$ @linksassin Fair, I’ll rerun the numbers tomorrow. \$\endgroup\$ Commented Aug 30, 2022 at 2:22

Should I be worried about characters snowballing?

No. In the best case scenario for a player character, they are using Elven Accuracy to increase their chances of rolling another crit (and gaining another inspiration to trigger elven accuracy) on top of having another source of advantage for multiple rolls per turn (say 2 attacks, and a familiar - inspiration for one attack, familiar for the other), as well as playing a Champion fighter to crit on a 19 or a 20. This gives them on average 2/20x6, or 12/20 chance of critting on one of those two attacks and resetting the inspiration, on a 5th level elven fighter with 2 feats built purely to do this and with no way of capitalizing on those crits in any meaningful way.

Meanwhile a 2nd level barbarian with reckless attack can get advantage any time they damn well please for no resource expenditure beyond hp - hp they are happy to lose at half the rate anyone else does while fulfilling their role as 'tank'. Likewise a darkness warlock with elven accuracy gets it all the time, a rogue gets it on the one attack/turn they care about, etc. This helps the champion fighter a little bit, but I haven't seen someone suggest a champion fighter in an optimization build yet, so like, eh.

What I would be more worried about is players rolling random skill checks to pick up inspiration used for more decisive situations. Not a problem if you ask for rolls, which is actually the DM's prerogative, rather than people saying 'I roll X', but if you have the latter at your table you may find people diverging off to Investigate fishmonger's stalls and otherwise dick around until they fish out a 20 so they can use inspiration to Deceive the guards at the gate they need to get past. Likewise, the first round of combat may result in a flurry of inspiration as players use up the inspiration they've earned in social and investigation checks.

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    \$\begingroup\$ Critting on a 19 doesn't grant inspiration. You have to roll an actual nat20 for that. Critting on a 19 or 18 makes advantage more valuable in terms of average damage output, though. Especially if they're a multi-class paladin and/or have some superiority dice (fighting style and/or feat) to pump into crits to get extra damage dice doubled. A crit-fishing Champion build sounds pretty gimmicky, this probably doesn't make it good :P \$\endgroup\$ Commented Aug 29, 2022 at 18:37

It won't affect damage much

You can get an exact, in-depth mathematical treatment from @rhermans answer. This one is more of a ballpark, but hopefully easier to understand, and trying to provide some absolute expected numbers. All averages, of course.

Normal crit chance of 5%

The value of advantage depends on the opponent's AC, at the typical about 65% chance to hit, is worth about 22.75% increased chance to hit. The average damage you can expect from your attack varies wildly, but is somewhere in the ballpark of level plus 7 damage on average.

That means, if you happen to score a critical, you can expect to get about 23% of level plus 7 extra damage on the next roll, about 2 damage on level one, or 22 damage on level 20.

To look at the absolute damage this is worth, you need to multiply this with your chance to crit of 5%, which means essentially no extra damage on level one, and about 1 point of damage on level 20.

Each follow-up attack has the same low chance of critting, so damage from the second follow-on attack that you could add to the total in "feedback-loop mode" would already be attenuated down to about 5% * 5%, or a quarter of a percent, which yields nothing to add in absolute terms, no matter what level you are on.

Optimized chances

As the inspriration rule currently is worded, it only grants inspriation on a natural 20, not on a critical. So the only way to push this is to use something like Elven Accuracy, which would double the chance to not quite 15%, from three rolls if you have advantage.

Even if we generously assume that you can get inpiration advantage on the first attack of the fight from a prior critical roll, that allows you to roll 3d20 in the hope to score a critical. That would give you 2 points net at best from the first follow-on attack in absolute terms. On the second follow-on, you'd be down to a 15% * 15%, less than have a point of net damage. So even in best case, about 2 points of extra damage total.

(Even if the feature would be reworded to allow any critical to inspire, you likely would get best about 5 points total, from an Elven Accuracy high level champion fighter to spread the crit range).

What to use inspiration for to maximize its effect

In many cases, the best combat use for inspiration is likely not to burn it for extra damage on an attack roll (although there surely can be some situation where exactly that is the best thing to do), but instead conserve it to maximize your chances that you'll save and not be banished, held, polymorphed, frightened or some other effect that removes you from combat, and negates one or several turns worth of total damage output.

  • \$\begingroup\$ It's not clear whether critting on a 19 would enhance the feedback effect. You'd benefit more from advantage when you have it, but a nat19 probably wouldn't grant inspiration even if you have class features that make it a crit. OTOH it's possible they'd phrase it as getting inspiration when a d20 roll is a critical success or something. (I assume they'd word it so you don't get inspiration from auto-critting, e.g. on a paralyzed opponent, although an Assassin rogue getting advantage after critting a surprised opponent would be narratively appropriate.) \$\endgroup\$ Commented Aug 29, 2022 at 18:15
  • \$\begingroup\$ A high-level Champion with Elven Accuracy would be the best case, of course, with higher chances to chain nat20s, and each attack with advantage having a pretty significant chance to do extra damage by critting even without generating inspiration. They couldn't also be a Half Orc (Savage Attacks) or high-level barbarian (Brutal Critical) for extra benefits from crits, since Elven Accuracy requires a different race. Oh right, and Elven Accuracy only works on Dex attacks (or int/wis/cha; a level of Hexblade allows Cha. But barbarian Brutal Critical specifically requires Str) \$\endgroup\$ Commented Aug 29, 2022 at 18:23
  • \$\begingroup\$ @PeterCordes You are right again. As it is currently worded, it grants inspiration only on a nat20, not on any crit. So extending your crit range would not help, only Elven Accuracy. I will amend my erring ways :)... \$\endgroup\$ Commented Aug 29, 2022 at 18:25
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    \$\begingroup\$ @PeterCordes, yes it is, from D&D Beyond, here \$\endgroup\$ Commented Aug 29, 2022 at 18:34
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    \$\begingroup\$ Phrasing note: you're using "sandbag" to mean "reserve", "stockpile", "hoard", or "save". That's not exactly how its metaphorical use in English normally works. I think the meaning you were going for is the sense of sandbagging your damage output, i.e. underperform to fool someone. But you're not doing it to fool someone (with your ability to drop the sandbags and go full speed later), you're doing it to save inspiration for something else (which is tactically wise). And you definitely wouldn't say you're "sandbagging your inspiration". \$\endgroup\$ Commented Aug 30, 2022 at 10:54

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