D&D 5E has an "advantage" concept where instead of rolling 1d20, you roll 2d20 and take the higher. Likewise, disadvantage means rolling 2d20 and taking the lower.

How does this affect the expected average outcome of the roll?


All this does is linearly adjust the normally-flat 5% probability for each number to occur. What results is a increased or decreased probability of any number above or below average to occur, positively for advantage and negatively for disadvantage. See this AnyDice function set, which yields the following:

Probability of x Black is d20, orange is highest of 2d20, blue is lowest of 2d20.

Since the probability of achieving any given number is a linear function, we can use linear regression (via Wolfram Alpha and our sample data from AnyDice to eventually solve for probability of x = 0.5x - 0.25 - multiply by 100, and there's your percent chance that you'll roll any particular number.

Additionally, what you're likely looking for is the probability that at least a particular number will be rolled, using either advantage or disadvantage. AnyDice, again, is king:

Probability of at least x Black is d20, orange is highest of 2d20, blue is lowest of 2d20.


#     %
1     100
2     99.75
3     99
4     97.75
5     96
6     93.75
7     91
8     87.75
9     84
10    79.75
11    75
12    69.75
13    64
14    57.75
15    51
16    43.75
17    36
18    27.75
19    19
20    9.75

#     %
1     100
2     90.25
3     81
4     72.25
5     64
6     56.25
7     49
8     42.25
9     36
10    30.25
11    25
12    20.25
13    16
14    12.25
15    9
16    6.25
17    4
18    2.25
19    1
20    0.25
  • 1
    \$\begingroup\$ Of course, unless you are a halfling (or have another similar luck feature) a 1 is always a failure so your chance of success on a DC1 is the same as DC2 \$\endgroup\$ – Dale M Jul 23 '14 at 5:06
  • 22
    \$\begingroup\$ @DaleM True for attack rolls, but ability checks don't have critical success or automatic failure. \$\endgroup\$ – mattdm Sep 4 '14 at 1:51
  • 10
    \$\begingroup\$ Advantage => 50% chance of rolling 15+. Disadvantage => 50% chance of rolling 7- I will use this to pitch my players on the difference :) \$\endgroup\$ – Gates VP Dec 29 '14 at 6:50

The math is straightforward





The general rule of thumb that in the mid range of the d20 (from success on a 9+ to 12+) advantage grant roughly a equivalent to a +5 bonus and disadvantage a -5 penalty. The increase and decrease in odds tappers off when your odds of success approach 1 or 20. For example a advantage on a 19+ your chance of failure goes from 90% to 81% not quite a +2 bonus on a d20.




The mean result goes from 10.5 to 7.175 for disadvantage and to 13.825 for advantage. The odds go from a flat 5% for each of 1 through 20 to (disadvantage results shown; reverse the first column for advantage results):

 1 39 9.75% 
 2 37 9.25% 
 3 35 8.75% 
 4 33 8.25% 
 5 31 7.75% 
 6 29 7.25% 
 7 27 6.75% 
 8 25 6.25% 
 9 23 5.75% 
10 21 5.25% 
11 19 4.75% 
12 17 4.25% 
13 15 3.75% 
14 13 3.25% 
15 11 2.75% 
16 9  2.25% 
17 7  1.75% 
18 5  1.25% 
19 3  0.75% 
20 1  0.25% 

(Middle column is how many of the 400 combinations of two numbers from 1-20 yield the result given in the first column.)


I just wanted to add a more generalized answer to this question that will give you a formula for computing your odd of success with advantage and disadvantage rather than looking up the value in a table. I am going to do my best to make this clear to anyone with any math background, so let me know in the comments if any of the steps don't make sense.


With advantage when you need to roll at least \$n\$ to succeed on your check (i.e. check - mod = \$n\$), you succeed if any one of your two dice rolls a value of \$n\$ or greater. Conversely, you fail when both of your two dice roll a value of \$n-1\$ or less. Since these are the only two options, you succeed or you fail, the probability of one of these two things happening is \$1\$, so we can say:

$$ P(success) + P(failure) = 1 $$

Where \$P(x)\$ indicates the probability of event \$x\$ occurring. We can re arrange this to get:

$$ P(success) = 1 - P(failure) $$

So now we know that we can find the value we want using the probability of failure, which we previously defined as:

$$ P(failure) = P(\text{both dice }\leq n-1) $$ the probability that both dice roll a value of \$n-1\$ or less. For one dice, we know that there are \$n-1\$ ways that you can roll \$n-1\$ or less (e.g. if \$n-1 = 5\$ you could roll \$1, 2, 3, 4, \text{or }5\$ so there are \$5\$ possible ways to do it). There are \$20\$ total possible ways to roll the dice. so the probability of one dice rolling \$n-1\$ or less is the number of ways to roll \$n-1\$ divided by the total number of ways to roll the dice or:

$$ P(\text{one die } \leq n-1) = \frac{n-1}{20} $$

Since both dice are the same, their probability of rolling \$n-1\$ is the same, so we know the probabilities for both dice. The two dice rolls are independent of one another, meaning that the number you roll one one die doesn't effect the number you roll on the other one. In other words, if you roll a 5 on the first die, the odds of rolling a 7 on the other one don't change. When two events are independent, we can find the probability of both events happening by multiplying their probabilities. In other words:

$$ P(\text{both dice }\leq n-1) = P(\text{one die }\leq n-1) \times P(\text{one die }\leq n-1)\\ P(\text{both dice }\leq n-1) = \frac{n-1}{20} \times \frac{n-1}{20}\\ P(\text{both dice }\leq n-1) = \Big( \frac{n-1}{20}\Big)^2 $$

Substituting this into our original equation we get:

$$ P(success) = 1 - \Big( \frac{n-1}{20}\Big)^2 $$


Now let define what it means to succeed with disadvantage in the same way we defined what it meant to succeed with advantage. For disadvantage where you need to roll at least \$n\$ to succeed, both dice must roll a value of \$n\$ or greater. In other words, if we need to roll at least an \$18\$ to succeed, both dice must roll either \$18, 19, \text{or } 20\$. The total number of ways to roll at least \$n\$ on a 20 sided die are:

$$ \{\text{# of ways to roll }\geq n\} = \{\text{total # of ways to roll}\} - \{\text{# of ways to roll }\leq n-1\}\\ \{\text{# of ways to roll }\geq n\} = 20 - (n-1) = 21 - n $$

We can create a probability from this by dividing by the total number of ways to roll the die giving us:

$$ P(\text{one die }\geq n) = \frac{21 - n}{20} $$

As before, the dice rolls are independent, so we can get the probabilities of both dice being greater than or equal to \$n\$ is:

$$ P(success) = \Big( \frac{21-n}{20}\Big)^2 $$


Since we worked through the math, we can also see how we can easily change this formula to get new probabilities. For example, if we make a house rule of "super advantage" where you roll 3 dice instead of 2, we simply multiply our \$P(failure)\$ by one more die \$\frac{n-1}{20}\$ changing the \$^2\$ to \$^3\$. We can therefore generalize the formula to be:


Where \$m\$ is the number of dice. Similarly, the probabilities for "super disadvantage" would be:

$$ P(success) = \Big( \frac{21-n}{20}\Big)^m $$

Going further, if we want to we could also sub out the \$20\$ in the denominator for another number if you wanted to look at the odds for other dice. For example, you are a GM and after character creation, one player comes to you and wants to re-roll their stats. They say they rolled all 1s and 2s on their 4d6s for a stat and they feel this was so unlikely that it will make the game be unbalanced for their character. Lets help the GM figure out if the player is right or not. In other words, we want to know \$P(\text{all dice }\leq 2)\$. This is the same as our "failure" condition for advantage except with 4 6-sided dice instead of 2 20-sided dice. So we can sub the 2 out for 4 and the 20 out for 6 and get:

$$ P(\text{all dice }\leq \text{max roll}) = \Big( \frac{\text{max roll}}{\text{# of sides}}\Big)^\text{# of dice}\\ P(\text{all dice }\leq 2) = \Big( \frac{2}{6}\Big)^4 = 0.01234 $$

So there is a 1.234% chance of this happening (i.e. 1 in 81 stats rolled up will be this low). Since characters have to roll 6 stats per game, the DM decides this isn't actually as unlikely as the player thinks and tells them to keep the stat block.



The answers provided effectively cover the probability for every result, 1 through 20, for advantage/disadvantage with 2d20. For completeness, the probabilities follow:

Probability Table

Expected Value

When rolling 2d20, and keeping the Maximum value from each of the 400 permutations, the expected value is 13.825. By contrast, the expected value when you keep the Minimum value is 7.175. The departure from the average of a single d20 is 3.325
Yes, the two average values sum to 21.


Unaddressed is the inherent benefit, or detriment, on the outcome expected rolling 2d20. To minimize duplication of effort, the following analysis assumes the roll is performed with advantage.

By definition, rolling with advantage is the act of rolling 2d20, and taking the higher value; the lower die, or one die if they have the same value, is disfavored in comparison to the other. The order in which the dice are rolled is immaterial. Instead, focus on the values they are capable of producing, e.g. of the 400 permutations, there are 39 opportunities to receive a 20 as the favored result. In rolling two dice, benefit for rolling a 1 and a 20, or a 20 and a 1, is still 19. The 1 is the disfavored value, and discounted by the procedure for rolling with advantage.

However, it would be a statistical error to assume that one die will always be disfavored and focus on the cases where the value of the other die is greater or equal to the value of the disfavored die. Doing so negates 190 cases where a benefit would still be gained from rolling 2d20 instead of a single die. This is because for each result where the die values aren't equal, there are 2 cases in which it can occur. In total, there are 20 cases where the values are equal, 190 where A < B, and 190 where A > B. Weighted Average Mistake

To correctly analyze the benefit of rolling 2d20, each of the 400 cases must be examined. For each resultant PAIR, the benefit demonstrated by the roll is the absolute difference between the dice, e.g. the values of the two dice are equal, the benefit is zero. Through this, the disfavored value is presumed to be the result we would have gotten in rolling 1 die, while the difference between it and the favored die is the benefit gained. The average of every Benefit is 6.650.

Benefit Chart

Why +5 Modifier

The PHB provides a short cut for applying advantage via a +5 modifier to supplant the roll. Coincidently

6.650 - (6.650-3.325)/2 = 4.9875 ~ 5

  • 1
    \$\begingroup\$ Your "benefit" strikes me as a strange statistic. Viewed one way: it reads the rolling of a 20 followed by a 1 as a benefit of 19, even though in plain language the second roll didn't benefit the roller at all. But we don't want to chain ourselves to thinking that the rolls are sequential, so let's imagine simultaneous rolls of 1 and 20: the claimed benefit is 19. But had I rolled 1 die and received one of those two results there'd be a 50% chance of getting a result 19 better than 1, and a 50% chance of getting a result 0 better than 1. It's a weird construct; I'm not quite sure what it adds. \$\endgroup\$ – nitsua60 Jun 2 '16 at 13:57
  • 4
    \$\begingroup\$ So I think this latest edit has clarified my... dissatisfaction? The "benefit" statistic is an accurate expression of "if I assume the lower of the two would have been my result, then how much better off am I taking the higher?" But why start off assuming the lower roll would have been the result? \$\endgroup\$ – nitsua60 Jun 2 '16 at 15:09

I actually made an ipython notebook for this:

To start, I simply rolled a random d20 1000 times.

The average 1d20 result for this series was 10.

For this graph, I rolled 2d20 1000 times and threw out the lower result.

The average result from an advantaged 2d20 roll was 13.

enter image description here

The last graph is a 2d20 1000 times disadvantaged roll.

The average result from the disadvantaged roll was 7.

enter image description here

So you can see here that there is a general +- 3 bias for advantaged or disadvantaged rolls.

  • 19
    \$\begingroup\$ I haven't downvoted, but I certainly can't upvote as it is. Your plots strike me as the least useful way one could have presented the simulated data. A frequency tabulation, a histogram, or even just presenting the mean and deviation of each dataset... I think any would have been better. \$\endgroup\$ – nitsua60 Apr 6 '16 at 22:17
  • \$\begingroup\$ Noted, I'll keep that in mind next time. :) \$\endgroup\$ – pizoelectric Apr 26 '16 at 23:43
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    \$\begingroup\$ Surely the average 1d20 result was 10.5! If it wasn't then you need to simulate more rolls or - and this is infinitely preferable - use probability theory to calculate the probabilities. You could then use simulated rolls to corroborate your results. \$\endgroup\$ – Clearly Toughpick Mar 4 '17 at 6:51

Play around with the numbers in Excel

This was not intuitive to me at first, so I created an Excel spreadsheet to help me see how it worked with simulated rolls.

You can change the number of rolls and change the type of die (d20, d12, d33--knock yourself out), and watch how the rolls change.

It gives you a nifty chart, like this: enter image description here

...or this: enter image description here

Find the spreadsheet here. Enjoy!


Effectively the trick is to find the percentage chance to hit under advantage. Subtract your percentage to hit normally. Divide by 0.05 (5%). Round down. This will provide the effective bonus that advantage provides.

Why divide by 5%? Because the standard d20 is 20 outcomes and 100/20 = 5. So if you want to know how many effective die results the bonus is helping you you have to divide by 5. (Interesting fact: the die size would affect your results in different games - but since the standard is D20 that's a moot point.)

In case you're wondering, I took from this article and computed the effective bonus advantage provides. As you can see, it provides more of a bonus the closer to the middle of the to hit range you needed. But less of a bonus at the extremes. In games term, if you were really good - or really terrible - at hitting the AC/DC before then advantage can't really help you. However if you're just an average joe, you get the most benefit from it.

In other words, for having advantage, the game rewards you. However, the effective reward is not a flat bonus. Instead, the reward is bell curved around your original chance of success. Disadvantage works the same, except the reward is instead a penalty.

RLL NORMAL  ADV   ADV-Normal  Effective Bonus
20  0.050   0.098 0.48        +0
19  0.100   0.191 0.91        +1
18  0.150   0.278 .128        +2
17  0.200   0.359 .159        +3
16  0.250   0.437 .187        +3
15  0.300   0.510 .210        +4
14  0.350   0.576 .226        +4
13  0.400   0.639 .239        +4
12  0.450   0.698 .248        +4
11  0.500   0.751 .251        +5
10  0.550   0.798 .248        +4
9   0.600   0.840 .240        +4
8   0.650   0.877 .227        +4
7   0.700   0.910 .210        +4
6   0.750   0.938 .188        +3
5   0.800   0.960 .160        +3
4   0.850   0.978 .128        +2
3   0.900   0.990 .090        +1
2   0.950   0.998 .048        +0
1   1.000   1.000 .000        +0

The average expected outcome is 1 out of 20.

Rolling twice makes the expected outcome as 1 out of 10, which is 2 in 20 simplified, instead of 1 in 20 as with a regular roll.

This is the same for both advantage and disadvantage with the difference being taking the lower number instead of the larger number. But otherwise it is the same.

Without overcomplicating this I'll keep it short with this example; you have a better chance of find a Willie Wonka golden ticket if you eat 2 bars rather than 1 bar, and the players know this when they roll with advantage or disadvantage. Which is why they like advantage and dont like disadvantage.

Hope that helps clear up any confusion, without going scientific on you.

  • 3
    \$\begingroup\$ The downvotes here are probably reacting to misused terms and statistical errors. For an example of one, the average expected outcome of a d20 roll is 10.5. \$\endgroup\$ – SevenSidedDie Aug 14 '18 at 0:41

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