Odds of rolling the same results twice on advantage rolls

I know that this has probably been answered before, but I couldn't find any specific threads on it.

This happened in a tabletop game recently, where all dice rolls are rolled with advantage anyways because of how the VTT system works.

So what are the odds that you get the same results twice in a row like this? Is it 1/400 or 1/160000, assuming order matters or not?

• @mprogrammer I don't think the system matters for dice percentage Feb 6 '20 at 0:22
• To clarify, are you interested in getting the same high number twice, or the same unordered pair twice? Feb 6 '20 at 9:59
• You may want to review the accepted answer in light of some new answers. Typically it is good to wait a day or so before accepting an answer in case more answers come in Feb 6 '20 at 12:09
• What is the exact question? Are you asking if you have just rolled d20 what are your odds of getting the same 2 d20 the second time? And do they need to be in the same order? Or are you asking before your roll any dice, what are your odds of getting the same result value with advantage on two consecutive rolls of two d20? Feb 6 '20 at 19:12
• Sorry I’ve been at work, the question was more what are the odds you get the same pair of rolls again, it doesn’t matter the order displayed in the screenshot. Could be 1, 2 and 1, 2 or 2, 1 and 2, 1 Feb 7 '20 at 0:50

The other answers have made a mathematical error. They neglect the possibility of rolling pairs. You can roll a 6 and a 9 as 6,9 or as 9,6, but there is only one way to roll a pair of 6s. This slightly lowers the probability of rolling equivalent results twice.

The probability of rolling a pair is 1/20 (the probability that the second die lands on the 1 face out of 20 that matches the first), and there is only 1 outcome out of 400 possibilities where the second set of dice match the first if the first is a pair.

The probability of not rolling a pair is 19/20, and there are 2 outcomes out of 400 possibilities where the second set of dice match the first if the first is not a pair.

The probability of rolling 2 sets of 2 dice and getting the same results up to order is thus 1/20*1/400 + 19/20*2/400 = 39/8000, or 0.004875, slightly less than the 1/200 number the other answers give.

• +1, this is the correct answer. FWIW, here's an AnyDice program to compute the same result by brute force enumeration of all the possible rolls. Feb 6 '20 at 11:33
• your reasoning smells flawed - the pair has the same individual chance as any other result ... since it still is made from 2 dice (essentially a pair of 6 as in your example can be made 2 ways (by giving the dice an "order" - 6[1],6[2] and 6[2],6[1] thus the same probability as the others Feb 6 '20 at 12:26
• @eagle275 No, this reasoning is correct and the other reasonings are flawed. Consider what happens when you generalize this system to dice with any number of sides. For a one-sided die, the other answers would yield a probability of 200%, which is of course impossible. Feb 6 '20 at 13:24
• For a two-sided die, the other answers would yield a 50% chance, even though the correct chance is actually 37.5%. You can check this by hand. Feb 6 '20 at 13:39
• @eagle275: Incidentally, that's not right. If we're counting unordered results, rolling e.g. two sixes is half as likely as rolling a six and a five. That's precisely because there are two ways of getting a six and a five when rolling the dice one after another (either 6 on the first die and 5 on the second, or vice versa) but only one way of getting two sixes. Feb 6 '20 at 13:42

Assuming order doesn't matter, as in your photo...

\begin{align} P(\text{Same Advantage Rolls}) &= P(\text{Same Roll #1}) \times P(\text{Same Roll #2}) \times 2!\\ &= \frac{1}{20} \times \frac{1}{20} \times 2!\\ &= \frac{1}{400} \times 2\\ &= \frac{1}{200}\\ \end{align}

This can be generalized to $$\\left( \frac{1}{20} \right)^n \times n!\$$, where n is the number of dice. This will be true for disadvantage as well.

Edit: Slightly incorrect, as I didn't account for doubles. See most upvoted post

• Bonus points for the use of factorials! Feb 6 '20 at 1:04
• Correct but clearly OP doesn't know how to calculate probabilities, so I find this answer a little sparse for their needs. Feb 6 '20 at 2:50
• Reasons I hate statistics: for some reason I feel like this is missing something, but I coded up a quick demo and sure enough ~0.5% consistently when using a sample size of 1 million Feb 6 '20 at 3:05
• Not quite. You've computed the probability conditional on the dice showing distinct numbers. If the dice roll the same number, then switching the order doesn't produce a distinct outcome. Feb 6 '20 at 9:17
• @user2357112supportsMonica: It's not even the correct conditional probability, since conditioning on the dice in each pair being different would reduce the number of possible outcomes for each (ordered) roll from 20 × 20 = 400 to 20 × 19 = 380. With this correction, the conditional probability works out to 1 / 190 or approximately 0.526%. Feb 6 '20 at 11:41

Odds of getting those exact same two numbers in a row (maybe in different orders, maybe not)?

4 possibilities ( 6-21 or 21-6 on first roll, 6-21 or 21-6 on second ) out of 20 x 20 x 20 x 20 (160,000) possible rolls = 1/40,000

Odds of just getting the same two numbers in a row, disregarding order?

Two chances of matching (same order, different order) out of 400 possible outcomes = 1/200

Or are you asking what are the odds two rolls with advantage have the same end result for the roll?

No idea. But now I want to know.

Edit: I went and figured it out: I'll spare you the math, but there's roughly a 1/15 chance of getting the same end result twice when you have advantage

Basically SUM(from 1-20, (odds of getting any given number with advantage) ^2) = 0.066625 or ~ 1/15

• The question you posed in your 5th line of your answer is answered by looking at a Max of Uniform distribution $X \sim \max(U_1, U_2)$ Feb 6 '20 at 1:58
• Your first part is not correct. The first roll does not matter. Feb 6 '20 at 2:52
• I'm also unsure of what your edit is getting at. The probability of rolling the same number twice on a d20 is 1/20. Feb 6 '20 at 2:56
• @lucasvw Incorrect. When a number is rolled with advantage (or disadvantage) the probability of getting the same outcome changes from a simple 1/20 Feb 6 '20 at 3:08
• @lucasvw Advantage dice work differently. They have 400 sides, and 1 has "1", 3 have "2", 5 have "3", ... 31 have "16", 35 have "18" and 39 have "20". Roll two of those and what are the chances of the face numbers matching? Feb 6 '20 at 9:49

The current accepted answer is part right in mentioning pairs not being counted twice, but there is a flaw in the original assumption/setup so I am giving my own anawer.

Rolling pairs needs to be kept in mind, but so do duplicate rolls is the key thing to remember. If I understand the question correctly it makes no discernable difference between rolling a 1 then rolling a 16, or rolling a 16 and then rolling a 1. If this is the case, the end result of the two dice rolls is what matters (since you have to roll two dice no matter what) and therefore such rolls should not be counted twice in the subsequent calculations. If the end result of rolling two dice is compared with the end result of two more dice rolled that changes the probability calculation.

There are 210 distinct, or unique, ways to roll two unordered d20s. 210 is gotten via the triangular number calculation of 20 (not counting pairs twice, and not count a 1 and 16 twice). So that means a 1/210 chance for any particular outcome of a single roll of 2d20s. That is a 0.476% chance.

Take this number times itself and you get 1/44100, which is 0.0023%.

Edit: Original answer was completely and entirely wrong in regards to the OP's question. I made a gross mistake via mental detour in my thought process and answered a question that wasn't there. I wasn't sure whether to repost an answer or edit this.

• Although there are 210 distinct unordered results for 2d20, they are not equiprobable.
– amd
Feb 7 '20 at 6:19

Edit: I'm unclear what you mean. Maybe you're asking "what are the chances of rolling a 1 and a 16 (as components of your advantaged 16) twice in a row?" Well, the first one doesn't count, for reasons I describe further in. However, the second one to match it, your two dice have 400 possibilities, and two meet your criteria: 1/16, or 16/1. So the simple answer is 2 in 400, or 1 in 200.

That applies in your case, but doesn't apply globally, as user2357112 discuses in an answer.

What are the chances of rolling 16 twice? 1 in 166.49.
160,000 total possibilities; 961 possibilities result in advantage rolls being both 16.

But hold on. Why is 16 special? 16 is only special because the first roll was 16. Hence we have a fallacy. You only consider the second one special because it matched the first. But this condition could have happened with any first number!!!

Get it? The chance of the first number being a number is 100%. So the 16 is not special at all; it had to be some number. The only chance here is the chance of the second number matching the first. That depends on the second number; for instance in advantage, you only have a 1 in 400 chance of rolling an advantaged 1, but a 39 in 400 chance of rolling an advantaged 20. In this case, you have a 31 in 400 chance (1 in 12.9) of rolling an advantaged 16.

Not terribly remarkable, and even less remarkable for an advantaged 20 (almost 1 in 10). Higher numbers are more probable; that's what advantage is all about.

However if we let go of the 16, since it isn't special, let's look at your core question. What are the chances of rolling two consecutive same-numbers? Without advantage, that's easy: 1 in 20. (as said, the first number is anything; the second simply matches it.) With advantage, it gets more complicated, but also more likely, since high numbers are more probable than low ones.

The answer is 1 in 15.00938.
160,000 total possibilities; 10660 result in advantaged rolls being equal.

• This answer is founded on the premise that the question is about rolling the same high number twice (what would usually be considered the "result" of an advantage roll), but from the example given, it seems like the question is intended to be about getting the same unordered pair twice. The question is unclear, and could use clarification. Feb 6 '20 at 9:53
• @user2357112 it is vague. I'll cover it both ways. Feb 6 '20 at 10:04