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How much is advantage really worth?


Continuing our previous discussions on the Advantage and Disadvantage mechanic in DnD 5e, we are going to examine a straightforward question — how much is advantage worth?

On the surface, the question is open ended — it requires defining “worth” in some way. A decent interpretation is “some measure of how much advantage increases your chance of success”. In our post on probability rules of thumb for the mechanic, we give general statements on the outcome of the dice rolls, such as “you are more likely to get a 20 than rolling below a 7”. It gives a feel for relative likelihoods, but still leaves us stuck thinking in terms of probabilities, which suffer from being challenging to interpret.

Another popular way to summarize advantage is to ground it in familiar DnD terms as a bonus. Bonuses are numbers added (or subtracted) from dice rolls depending on how good (or bad) a character is at a certain action, e.g. a barbarian will usually have a higher bonus for swinging a greataxe than a wizard and therefore a higher chance to hit.

Framing advantage as a bonus has led to the claim that “advantage gives you about a +5 bonus to your roll” — is this true? People quickly point out that such a rule is an oversimplification, but how wrong is it? Let’s dig into the math and explore!

The math

As per usual, we will work out the solution for nn sided dice and substitute in d20s at the end. We have two random variables that we want to look at:

XadvAdv(n)XDice(n)X_{adv} \sim \mathrm{Adv}(n) \qquad X \sim \mathrm{Dice}(n)

The left being rolling with advantage, which we explored here, and the right being a simple nn sided dice roll, explored here. We want to show when these two have the same probability of success by adding a bonus, bb, to the simple dice roll. “Success” in DnD meaning “rolling at or above a target value”. We can represent the target value as xx and write:

P(Xadvx)=P(X+bx)P(X_{adv} \geq x) = P(X + b \geq x)

In the above, we are most interested in the bonus, bb. That tells us how much advantage is “worth” by drawing a direct equivalence with adding a modifier to a normal roll.

We are going to use results from the previous posts linked above to avoid going through any derivations, specifically the cumulative probability expressions:

P(Xadvx)=(xn)2P(Xx)=xnP(X_{adv} \leq x) = \left(\frac{x}{n}\right)^2 \qquad P(X \leq x) = \frac{x}{n}

We will rearrange our expression that equates advantage and a bonus to make use of these results, like so:

P(Xadvx)=P(X+bx)P(Xadvx)=P(Xxb)1P(Xadvx1)=1P(Xxb1)P(Xadvx1)=P(Xxb1)\begin{aligned} P(X_{adv} \geq x) &= P(X + b \geq x) \\ P(X_{adv} \geq x) &= P(X \geq x - b) \\ 1 - P(X_{adv} \leq x - 1) &= 1 - P(X \leq x - b - 1) \\ P(X_{adv} \leq x - 1) &= P(X \leq x - b - 1) \\ \end{aligned}

Now we plug in to get:

(x1n)2=xb1n\left(\frac{x-1}{n}\right)^2 = \frac{x-b-1}{n}

Simple algebra allows us to solve for bb:

b=x1(x1)2nb = x - 1 - \frac{(x-1)^2}{n}

One requirement that I will impose is that bb must be a whole number. The outcome of a dice roll can only ever be a whole number, and fractional bonuses will never increase your probability of success. This means that the values of bb that we can find from this will only be approximate in most cases, but they will represent the bonus that gets as close as possible to the probability of success when rolling with advantage. We can summarize our result as:

badv=x1(x1)2n\boxed{ b_{adv} = \left\lfloor x - 1 - \frac{(x-1)^2}{n} \right\rceil }

Where the notation a\lfloor a \rceil means rounding to the nearest whole number.

The results

By inspecting badvb_{adv}, we can get some of the main properties — first, the bonus depends heavily on the target roll. As it is a quadratic expression, the graph is a downward facing parabola. By finding the maximum of badvb_{adv}, we determine where advantage gives the biggest bonus. We denote the maximum to be b^adv\hat{b}_{adv} and find:

b^adv=n4x^=n2+1\hat{b}_{adv} = \frac{n}{4} \qquad \hat{x} = \frac{n}{2} + 1

Substituting in n=20n=20, we see an important result — the maximum bonus of advantage is +5 and occurs when the target roll is an 11. So there is some truth to the statement that advantage is worth +5, it just happens to be the maximum possible bonus and occurs when you need to roll about average.

The best way to get a complete view of the bonus from advantage is to look at the graph.

We immediately see some important observations that can be made. First, if your target roll is anywhere between 8 and 14, then +5 is a good rule of thumb. For all of those targets, the best approximating bonus is +5. Outside of those target rolls in the center, the equivalent bonus drops off significantly. When trying to hit target rolls of 17 or higher, advantage is worth about +3 or less. At the extreme, advantage only gives you a bonus of about +1 to hit a target of 20.

It can be helpful to think about why advantage has this behavior of being equivalent to a strong +5 bonus when you need to hit around an 11 and dropping off when you need to hit higher. It comes down to what is actually happening when you apply the effects of advantage versus adding a bonus.

When you roll with advantage, you shift the mass of the distribution so that low rolls become rarer and high rolls become more likely, but the possible outcomes stay the same. However when you roll with a bonus, it has the effect of shifting the distribution uniformly — with a +5 bonus, it’s no longer possible to roll below a 6 and rolls up to a 25 can be achieved. This shifting causes bonuses to be a stronger force than advantage when trying to hit targets at the edge of possibility.

Okay…so what is it actually worth?

One thing that’s clear from the above is that the relative value of advantage depends heavily on the target roll. Without knowledge of typical target rolls, it can be hard to give a hard and fast rule. To really dig into this question, exploring the rules of a system to figure out a typical target roll for success would be necessary (perhaps a worthwhile exploration in a future post).

That being said, we can get somewhere with just a little bit of reasoning. If we pretend that we are completely ignorant of common target rolls, then we can approximate the overall badvb_{adv} with the median value over all target rolls. Doing so we find:

Med(badv)=4\mathrm{Med}(b_{adv}) = 4

An equivalent bonus of +4 is not much different than the commonly used +5, but it is a more conservative rule of thumb. A +5 represents the highest bonus effect of advantage, whereas +4 represents an average over the whole range of outcomes. The more extreme the target roll, the weaker the bonus gets. In practice, it seems likely the the majority of target rolls would fall somewhere near the middle of the distribution, and therefore the commonly accepted +5 rule of thumb effectively captures the typical behavior of rolling with advantage.

A short note on disadvantage, it is fairly easy to show that all the work above applies to disadvantage as well, you just need to make the bonuses negative — a good exercise for the intrepid reader!

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