Boolean Advantage & Disadvantage

The advantage and disadvantage rule was introduced in Dungeons & Dragons, Fifth Edition. It is explained as follows: when you have advantage, roll two dice and take the higher of the two; when you have disadvantage, roll two dice and take the lower of the two. The rule abstracts certain factors of character ability tests that prior (in Third Edition) would have been handled by adding or subtracting minute value from the roll. Instead the dungeon master need only decide if a character is advantaged or disadvantaged to such a degree that the roll warrants the appropriate procedure.

In this post, I want to offer a new view on how this mechanic impacts chances of success by looking at it in terms of boolean (binary) logic and probability. This has implications for how we combine rolls in general, whether we view overall success as lacking failure (disadvantage) or overall failure as lacking success (advantage).

Still no working computer. :c

The Rules Part

An ability test or check is a binary pass-or-fail mechanic. The player or dungeon master rolls the twenty-sided die greater than or equal to some number, representative of the difficulty of the attempted task, in order to succeed at that task. The roll is modified directly only by character ability bonuses in the six usual categories. Advantage and disadvantage, as I said, abstract all other factors by allowing or imposing a second roll (in favor or disfavor of the acting character, respectively).

However, the phrasing of the rule seems to obscure its outcome, at least in the context of character ability tests (which is the only context for this rule, as far as Fifth Edition is concerned). Rather than being a matter of getting a higher or lower number than usual (the oft-cited figure is that advantage or disadvantage modify the result on average by ~5 pips), it is a matter of applying new logic to the ability test procedure. This is due to the binary or boolean nature of ability tests.

Having advantage, on paper, means that you select the higher of two dice when attempting an ability test. Let us consider the potential outcomes. The first is that both dice give a high enough value to succeed at the attempted task. The second is that only one die gives such a high value; however, since we want the higher die, this means that the character succeeds at the task. The character only fails if neither die succeeds. Thus we can rephrase advantage as follows: roll two dice for at least one success. Alternatively, roll one die as usual, but reroll if you fail.

Having disadvantage is the opposite of the above, on paper and in practice. You are supposed to accept the lowest of two dice. What this means, however, is that if one die fails, then you have already failed. This is because the other die can either be higher than the first (and is thus negated), or it is lower than or equal to the first (so it has the same outcome). The only good outcome of disadvantage is that you succeed at both rolls, since if only one roll fails then that roll is necessarily lower than the other. Therefore we can rephrase the rule as follows: you must pass two rolls in order to succeed at the task.

The Math Part

We can use this boolean understanding to come up with chances of success given advantage or disadvantage. Suppose two rolls X and Y. Advantage means that X or Y (or both) must pass in order for the task to succeed; this is called logical disjunction, or “OR”. Disadvantage means that both X and Y must pass in order for the task to succeed; this is called logical conjunction, or “AND”. Keep in mind that a regular ability test has just one roll, R. However, the probability of X and Y are each equal to the probability of R. This is represented as:

p(X) := p(R)
p(Y) := p(R)

Disadvantage is the easiest to calculate. What is the probability of success (S) if a character has disadvantage? It is the probability that the roll R succeeds twice rather than just once. This means we multiply p(R) times itself, since dice A and B are just “clones” of R:

p(S) := p(X) * p(Y)

Advantage is slightly more difficult. We have to calculate the chance that at least one die succeeds, which is different from the chance that both dice succeed (which is what disadvantage is). We calculate this by first figuring out the chance that both dice fail, and then subtracting this from 100% for the chance that both dice don’t fail. Got a headache yet?

p(S) := 1 - (1 - p(X)) * (1 - p(Y))

Let’s assume our original chance of success is an even 50%. With disadvantage, that chance dwindles to 25% (which is 50% times 50%).

With advantage, we are looking for the probability that both dice don’t fail. 100% minus 50% is still 50%, and this is the chance of failure for an individual die. So we multiply 50% by 50% to find the chance that both dice fail, which is 25%. Then we subtract this from 100% so we can tell that the chance of at least one die succeeding is 75%.

Notice that the chance of success either way is modified by 25%, which is 5 pips on a twenty-sided die! However, this does not hold true for all chances of success. Consider the more likely case that the base chance of success is 60%. With disadvantage, that is reduced to 36% (60% times 60%). With advantage—and I’ll spare you the math this time—the chance of success is increased to 84%.

Chances of success that are greater than 50% are exponentially “safer” when disadvantaged than chances of success that are less than 50%. Having even just a 40% base chance of success results in a disadvantaged chance of 16%. This is probably one reason why Fifth Edition has you make characters with such high ability bonuses.

Arbitrary Dice

What if you decided that X and Y had totally different base chances of success? The formulas for advantage and disadvantage remain the same, except you replace the probabilities of A and B with their own instead of assuming that they are equal.

Let’s suppose that the probability of X equals 60% and that of Y equals 40%. For a situation where both tasks must succeed, the overall chance of success is 24% (60% times 40%). For a situation where only one task must succeed in order to succeed overall, the overall chance of success is 76%.

Consider that relationship between rolls when you ask players to roll for a bunch of dice. If everyone has to succeed individually (meaning that no one fails), the chance of overall success is extremely low. If just one player needs to pass for everyone to succeed, the task is that much easier.

Helping Hands

I wrote about similar math in the context of the original D&D, where characters exploring a dungeon can attempt a task together in groups of three in order to get a better overall chance of success (though the characters tended to have the same individual chance of success) [1]. Imagine if you replaced groups of three with pairs of two—that’s literally advantage!

Which means that the “helping” action granting advantage in Fifth Edition is actually an extremely apt and intuitive way of handling the situation. Remember that what it really means is that you roll two dice and, if one passes, you succeed. Isn’t that actually quite nice, especially since (presumably) both dice actually have the best possible chance of success if the “main person” doing the action is the most fit for the task?

I think I would ideally handle things more like OD&D with its same chance of success per die (and smaller dice, six-sided instead of twenty-sided), but also restricting “help” to two characters instead of three. That’s just because I like the idea of zooming into two characters and seeing how they interact with each other. I think it’s cutesy.

Alternatively, similar to my post about thief skills, I’d let everyone define character abilities in terms of dice (intelligence, strength, etc.) and ask for rolls greater than or equal to 4 or something like that [2]. You might want two d8 strength characters to push open the big door, while two d8 intelligence characters look for a hidden lever. I’m just also hesitant to impose such a strict scheme for this context, while not knowing how well it would apply to others. But I think 24XX has that kind of dice math as well as a similar helping rule [3]?

Natural Twenties & Ones

My friend and Fifth Edition enjoyer Jay Dragon pointed out that this model does not account for rolling natural twenties and ones, so I will do that here under a new subheading! The chance of rolling a natural twenty or one is independent of the likelihood of succeeding at an ability check itself. Regardless of whether your chance of success is 25% or 75%, your chance of rolling a natural twenty or a natural one is always 5% each.

The only difference when introducing multiple dice into the equation is that, depending on whether we want the best or worst roll, the chance of either event (rolling a natural twenty or rolling a natural one) changes. We typically view this in terms of the likelihood of getting the highest or lowest number when picking the best or worst of two rolls. When rolling with advantage, the chance of rolling a natural twenty is 9.75% while the chance of rolling a natural one is 0.25%. These chances are flipped when rolling with disadvantage.

What’s interesting is that these values are also subject to the same boolean conditions as the ability checks, except it is always an independent 5% chance regardless of the ability check’s own likelihood. Getting a natural twenty while having advantage means at least one of your dice rolled a twenty, while getting a natural one means that both of your dice must have rolled a one. The reverse is true for rolling with disadvantage: if you rolled a natural twenty, it’s because both of your dice turned up as twenties.

Even odder is that regardless of having advantage, disadvantage, or neither, the total chance of rolling either a natural twenty or a natural one is always 10% (either 5% plus 5%, or 9.75% plus 0.25%).

If we wanted to, we could construct a probability matrix with four events for each of the three modes (advantage, disadvantage, neither): success, critical success, failure, and critical failure. Just subtract the critical success rate from the overall success rate to find the rate for uncritical success, and do the same for failure. Model complete!


I hope this is an interesting and helpful new way to understand how advantage and disadvantage impact the dice we roll. I think it also helps to see how we implicitly use the same “logic” as advantage or disadvantage when we try to figure out how characters are acting in a group. Do too many cooks spoil the broth, or is it the more the merrier? Do certain restrictions (e.g. groups of three in OD&D, or only getting advantage from one helper in Fifth Edition) represent reasonable restrictions?


[1] B., Marcia. 2021-08-11. “Time, Movement, and Action Economy in Dungeon Games”, Traverse Fantasy.

[2] B., Marcia. 2012-07-05. “On Thieves: A Trifunctional Analysis of OD&D“, Traverse Fantasy.

[3] Tocci, Jason. 2020. 24XX.


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