# How does rolling two dice and taking the highest affect the average outcome?

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

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

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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:

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 = .5x - .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:

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

Data:

Advantage
#     %
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

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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 – Dale M Jul 23 '14 at 5:06
@DaleM True for attack rolls, but ability checks don't have critical success or automatic failure. – mattdm Sep 4 '14 at 1:51
Advantage => 50% chance of rolling 15+. Disadvantage => 50% chance of rolling 7- I will use this to pitch my players on the difference :) – Gates VP Dec 29 '14 at 6:50
@DaleM I have seen a couple rare situations where a character had enough conditional bonuses to his attack roll to hit on a 2, that isn't necessarily true. – Patrick vD Feb 19 at 1:37

The math is straightforward

With an advantage you are looking for best of two results. To figure out your odds you need to multiply the chance of FAILURE together to find out the new chance of failure. For example if you need 11+ to hit rolling two dice and taking the best means instead of a 50% of failing you have only a 25% chance of failing (.5 times .5).

For a disadvantage where you take the worst of two dice roll you need to multiply the chances of SUCCESS to find out the new odds. For example if you need a 11+ to hit your chance success drops from 50% to 25% (.5 time .5).

Advantage 16+ to hit, goes from 25% chance of success to roughly 43% chance of success. (.75 time .75)

Disadvantage 16+ to hit, goes from 25% chance of success to roughly a 6% chance of success (.25 times .25)

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.

An interesting property of the system is that there always a chance of success and always a chance of failure. Unlike a modifier systems where enough modifiers can mean auto success or auto failure. (Unless you have a 20 is an automatic success and 1 a automatic failure)

A useful application of knowing the odds of rolling two dice is that you can just convert it to a straight bonus when rolling for a large number of NPCs. A bunch of goblins with an advantage from surprise that need 13+ to hit the players you can just apply a +4 (or +5 if you round up) bonus instead of rolling the second dice. This is because they have a 60% chance of failure on 13+. Taking .6 times .6 yields .36 a drop of 24%. Not quite a +5 bonus on a d20 dice.

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very interesting different way of looking at it. Only issue I see with allowing for a straight bonus for a group of NPCs is the reduction in critical successes that would proceed from there (unless the critical range is then expanded as well to 19-20). – wax eagle May 30 '12 at 12:56
considering this is a straight repost from your blog, they should probably link to each other. – wax eagle May 30 '12 at 14:25

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.)

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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.

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

The average result from the disadvantaged roll was 7.

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

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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. – nitsua60 Apr 6 at 22:17
Noted, I'll keep that in mind next time. :) – pizoelectric Apr 26 at 23:43

# Probability

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

# 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.

# Benefit

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.

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.

# 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

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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. – nitsua60 Jun 2 at 13:57
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? – nitsua60 Jun 2 at 15:09

## protected by Oblivious SageFeb 18 at 14:47

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