I am looking at creating a few new categories of weapons just to add a bit more reason to spend money on mundane equipment in the early game. Poor, and Masterwork.

My initial idea is that poor weapons will use the disadvantage mechanic on damage rolls, that is rolling twice and taking the lower, and masterwork weapons will use the advantage mechanic.

They won't be magic, and I will probably make poor and masterwork versions of magical weapons as well, but I don't want a masterwork weapon to overshadow a magical +1 weapon.

When rolling a D20 the accepted average seems to be that advantage equates to a +4/+5 bonus, so what would that average bonus be on damage rolls? Given the different damage dice it is probably best to stick with a 1d8 weapon in this case, but I would assume the advantage is less for small die, and more with larger die.

A good answer might include an anydice table or some kind of maths, but I find all that really hard to get my head around, so an answer along the lines of 'advantage when rolling a d8 equates to roughly x' is the best for me, with explanations in words rather than maths if you need to add anything else.

A really good answer will include anything else that I might need to consider but haven't added to the question.

Note: I have seen this question, which is similar, but if the answers there happen to cover my question I don't understand them.

  • 1
    \$\begingroup\$ Given @IlmariKaronen 's answer, the effect of this means that someone could purchase the damage equivalent of a +1 weapon for d6 and +2 for d12, so your Masterwork weapon should be priced accordingly. For something less powerful, perhaps you get advantage on the damage roll only should your natural hit roll (before adjustments) be an even number? \$\endgroup\$
    – Kirt
    Commented Jun 28, 2021 at 15:12
  • \$\begingroup\$ I will point out that the real difference between a poor weapon and a very good weapon is not the damage it does but the likelihood of it breaking during use. For a dull blade, you could just drop the die down from d8 to d6. \$\endgroup\$
    – Tysto
    Commented Sep 11, 2022 at 19:33

2 Answers 2


Here's a quick AnyDice program to show the effect of applying the D&D 5e (dis)advantage mechanic (i.e. roll twice, pick lowest/highest) to dice of various size. Based on the summary results, here are the average rolls in each case:

Die Normal Advantage Disadvantage Difference
d4 2.5  3.125 1.875 ±0.625
d6 3.5 4.4722 2.5278 ±0.9722
d8 4.5 5.8125 3.1875 ±1.3125
d10 5.5 7.15 3.85 ±1.65
d12 6.5 8.4861 4.5139 ±1.9861
d20 10.5 13.825 7.175 ±3.325

From this table we can see that — just considering the average damage per hit, and ignoring the distribution of results — (dis)advantage is equal to a bonus of about ±0.6 on a d4, ±1.0 on a d6, ±1.3 on a d8, ±1.7 on a d10, ±2.0 on a d12 and ±3.3 on a d20.

(This is lower than the often quoted "advantage ≈ +4 or +5" rule of thumb for d20 since it's based on the average roll rather than the chance of meeting a target DC that's typically somewhere near the middle of the 1–20 range. Due to the way the advantage mechanic works, it has the biggest effect compared to a flat bonus for mid-range target numbers.)

Addendum based on comments: For damage rolls with multiple dice, like a 2d6 greatsword, the magnitude of the effect depends on how you choose to generalize (dis)advantage to rolls using multiple dice. Off the top of my head, I can think of at least the following options for rolling NdX with (dis)advantage:

  1. Roll N pairs of dX, choose the highest/lowest die of each pair and add them up.
  2. Roll two sets of NdX, choose the set with the highest/lowest sum.
  3. Roll (N+1)dX, choose the N highest/lowest out of N+1.
  4. Roll 2dX and take highest/lowest, then add N−1 normal dX rolls.

This AnyDice program shows the average result of each of these methods of rolling NdX with advantage. For example, for 2d6 the averages look like this:

Roll Average Difference
normal 2d6 7.00
2 rolls of d6-with-advantage 8.94 +1.94
highest of two 2d6 rolls 8.37 +1.37
highest 2 of 3d6 8.46 +1.46
d6 with advantage + d6 7.97 +0.97

For method #4 the average difference between a normal roll and one with advantage is of course the same as for a single dX roll, while for method #1 it's N times that. Methods #2 and #3 tend to be somewhere in between. In fact, for N = 2 method #3 seems to give results exactly halfway between methods #1 and #4, although this does not continue to hold for higher values of N.

  • \$\begingroup\$ Where’s the 2d6 for the greatsword? \$\endgroup\$
    – Dale M
    Commented Jun 28, 2021 at 12:43
  • 5
    \$\begingroup\$ @DaleM: That depends on how you want to implement it. If you roll each d6 separately with (dis)advantage, it's twice as much as for 1d6 (i.e. ±1.94). If you roll just one d6 with (dis)advantage, it's obviously the same as for 1d6 (±0.97). And if you roll 3d6 and take two highest/lowest, it's about halfway in between (±1.46). \$\endgroup\$ Commented Jun 28, 2021 at 13:03
  • \$\begingroup\$ @IlmariKaronen The precedent of 'Great Weapon Fighting' would be separate 'advantage' on each individual die of the composite roll (although not true advantage since the rolls are sequential and you have to keep the second roll even if lower). The natural contrast to that is, I think, rolling two separate 2d6 totals and taking the higher/lower of the two - and I'm not sure that was one of the options you compared? \$\endgroup\$
    – Kirt
    Commented Jun 28, 2021 at 15:07
  • 4
    \$\begingroup\$ @Kirt: Good point, I missed that option. Apparently it's slightly weaker than "highest/lowest 2 of 3d6", with a difference of ±1.37 compared to normal 2d6. \$\endgroup\$ Commented Jun 28, 2021 at 17:10
  • 1
    \$\begingroup\$ I wonder whether it would be worth mentioning possible rarity in relation to mastery, too, e.g. a weapon smith who can produce weapons with mastery where the effective bonus is >+3 ought in principle to be a "very rare" find indeed. Weapon smiths producing items with master where the weapons are closer to +2 "rare" and closer to +1 "uncommon". Though, there are no d20 weapons I know so maybe rarity ought to be looked at differently here. Hmmm.... \$\endgroup\$
    – Senmurv
    Commented Jun 28, 2021 at 21:12

I am aware that you did not ask for a math-based answer, and Ilmari already did a wonderful job, but I decided to provide one anyway, since in my honest opinion the math approach is one of the best strategies to generalize and it allows one to use the results in other contexts.

Using the Order Statistics is the way.

When one has to deal with advantage/disadvantage mechanics, one has to consider the beloved order statistic. The formulae are given below, while at the bottom of this answer I provide the mathematical details.

Given two rolls of a die with \$d\$ faces, \$X_{(2)}\$ denotes the outcome of rolling with advantage and \$X_{(1)}\$ the outcome of rolling with disadvantage. The expected values are, respectively, $$ {\rm E}[X_{(2)}] = \frac23d+\frac12-\frac{1}{6d}, \quad {\rm E}[X_{(1)}] = \frac13d+\frac12+\frac{1}{6d} $$ which provide the same results as Ilmari's answer. Below is a plot which depicts the behaviour of these functions for dice of an arbitrary number of sides; the case for \$d=20\$ is highlighted. Note that \${\rm E}[X_{(2)}]\$ and \${\rm E}[X_{(1)}]\$ are not linear functions of the number of faces, but, on the other hand, for large \$d\$ the last term (\$1/(6d)\$) is close to 0.

Plot of the expected values.

Denoting with \$X\$ the result of a straight roll of a die with \$d\$ faces, the expected values of the differences between rolling with adv/disadv can be calculated as $$ {\rm E}[X_{(2)}-X]=\frac{d^2-1}{6d}, \quad {\rm E}[X_{(1)}-X]=\frac{1-d^2}{6d} $$ These differences are strictly greater than 1 (in absolute value) for \$d>6.16\$, as shown in Ilmari's table. Roughly speaking, when \$d=6k\$ (i.e. when the number of faces is a multiple of 6), then the difference between the average straight roll behaves as \$\pm k\$, depending on whether you are considering adv or disadv. When the die has a number of faces which is not a multiple of 6, then depending on how you approximate results (flooring, ceiling or rounding) you may consider an higher or lower difference. For example, for \$d=20\$, the difference is 3.325, which you may consider as \$\pm\$3 (floor or round operation) or \$\pm\$4 (ceiling operation).

An interesting fact is that the function describing the expected value of a straight roll is the average of the two functions describing the expected value of rolling with adv/disadv: $$ \begin{eqnarray} {\rm E}[X] &=& \frac12d +\frac12\\ &=& \frac12\left( \frac23d+\frac12-\frac{1}{6d} + \frac13d+\frac12+\frac{1}{6d}\right)\\ &=&\frac12\left({\rm E}[X_{(2)}] + {\rm E}[X_{(1)}]\right)& \end{eqnarray} $$

Mathematical details

Consider a fair die with \$d\$ faces: since the outcome of such a die follows a discrete uniform distribution, the chance of getting the result \$x\$ is given by the probability density function (pdf) \$f(x)=\displaystyle\frac{1}{d}\$ and the cumulative distribution function (cdf) of \$x\$ is \$F(x)=\displaystyle\frac{x}{d}\$. Recall that the cdf \$F(x)\$ provides the probability of getting a result of less than or equal to \$x\$.


Denote with \$X_{(2)}\$ the maximum outcome of two rolls of a die: using the formula for discrete distribution the probability of being \$x\$ (the highest outcome) is given by\$^1\$ $$ \begin{eqnarray} P(X_{(2)}=x)&=& F(X)^2-(F(x)-f(x))^2\\ &=& 2f(x)F(X)-f(x)^2 \end{eqnarray} $$ Substituting the formulae for the cdf and the pdf and after some algebra one has that $$ P(X_{(2)}=x) = \frac{2x-1}{d^2} $$ It is now possible to compute the expected value: $$ \begin{eqnarray} {\rm E}[X_{(2)}] &=& \sum_{x=1}^dxP(X_{(2)}=x)\\ &=&\frac{1}{d^2}\left( 2\sum_{x=1}^dx^2 -\sum_{x=1}^dx\right)\\ &=& \frac23d +\frac12 -\frac{1}{6d} \end{eqnarray} $$ where we used the fact that $$\displaystyle\sum_{x=1}^dx^2=\frac{d^3}{3}+\frac{d^2}{2}+\frac{d}{6}, \quad \displaystyle\sum_{x=1}^dx=\frac{d^2+d}{2} $$ This formula gives the same results as stated in Ilmari's answer. Now, compute the expected value of the difference between rolling with advantage and normal rolling, denoted with \$X\$: $$ \begin{eqnarray} {\rm E}[X_{(2)}-X] &=& {\rm E}[X_{(2)}] -{\rm E}[X]\\ &=& \frac23d +\frac12 -\frac{1}{6d}- \frac{d}{2}-\frac12\\ &=& \frac{d}{6} -\frac{1}{6d}\\ &=& \frac{d^2-1}{6d} \end{eqnarray} $$


Using the same formula as the previous case, one can compute the expected value\$^1\$ of \$X_{(1)}\$, which is the outcome of rolling with disadvantage: $$ P(X_{(1)}=x) = f(x)^2+2f(x)-2f(x)F(x) $$ Then $$ P(X_{(1)}=x) = \frac{1-2x}{d^2} +\frac{2}{d} $$ which leads to $$ \begin{eqnarray} {\rm E}[X_{(1)}] &=& \displaystyle\sum_{x=1}^dxP(X_{(1)}=x)\\ &=& -\frac{2}{d^2}\displaystyle\sum_{x=1}^dx^2 +\left(\frac{1}{d^2}+\frac{2}{d}\right)\sum_{x=1}^dx\\ &=& \frac13d +\frac12 +\frac{1}{6d} \end{eqnarray} $$ The expected value of the difference between a straight roll and a roll with disadvantage is $$ \begin{eqnarray} {\rm E}[X_{(1)}-X] &=& {\rm E}[X_{(1)}] -{\rm E}[X]\\ &=& \frac13d +\frac12 +\frac{1}{6d}- \frac{d}{2}-\frac12\\ &=& -\frac{d}{6} +\frac{1}{6d}\\ &=& \frac{1-d^2}{6d} \end{eqnarray} $$

\$^1\$ The computation is left as an exercise for the reader.

  • \$\begingroup\$ Interesting answer, i am curious how you might approach the other part of the OPs problem of providing advantage - where the masterwork weapon never exceeds +1 (outshining a magic weapon). Maybe it’s a altogether new answer? \$\endgroup\$ Commented Sep 7, 2021 at 23:59
  • \$\begingroup\$ Thanks @AmethystWizard. I covered that aspect in my answer, didn't I? In the paragraph starting with "These differences are strictly greater than 1" I considered how much large is the increasing wrt straight rolls. Maybe it is not so clear, may a table help? \$\endgroup\$
    – Eddymage
    Commented Sep 9, 2021 at 6:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .