# Isn't rolling with advantage unordered with replacement?

Conventional wisdom for rolling with advantage (roll 2 die take the higher) is say the probability of getting a 20 is - $$39 / 400 = 0.0975$$ Where 400 is total number of rolls from $$20^2$$ and 39 is the total ways one can get a 20 from 2 dice.

As far as I understand the above means that order matters. But when one rolls with advantage, they don't care which die is a 20 just that a 20 occurs. So shouldn't the calculation be total combinations = $$\binom{n-1+k}k$$ = $$\binom{21}2$$ = 210. And the number of ways to roll a 20 given order doesn't matter is 20. Therefore the probability of a 20 occuring is $$20/210 = 0.0952$$.

Edit: In this problem I'm considering two 20 sided dice being rolled simultaneously.

• When you're rolling physical dice in the real world, the universe "cares" about which die is which. That's why rolling a total of 3 on two dice is twice as likely as rolling a total of 2, even though there is only one combination of dice (1 and a 2 in either order) that can produce the total 3. Commented Mar 6 at 3:23
• @peterwhy I've edited: 2 d20 Commented Mar 6 at 3:24
• @DavidK you're discussing summing the two dice values which is a different scenario to the one I'm proposing. Commented Mar 6 at 3:29
• The universe cares just as little about your interpretation of rolling with advantage as it cares about the fact that there's only one "combination" with total 3. If you lived in a different universe where dice interacted quantum-mechanically you might have a good idea. In this universe it's a bad idea. Commented Mar 6 at 3:33
• Does this answer your question? Rolling $2$ dice: NOT using $36$ as the base? Commented Mar 6 at 3:33

Regardless of whether you roll the dice together or separately, rolling a d20 with advantage (aka rolling two d20s) is two independent events - what you roll on one dice does not affect what you roll on the other. Each of those events has 20 outcomes, each of which has a probability of $$\frac{1}{20}$$. As the two events are independent, we determine the probability of any single outcome as the product of the probability of both events. That simply doesn't change regardless of whether the order of those events is disregarded - you still have two events whose probability needs to be multiplied together.

With that said, you're not entirely wrong here - rolling with advantage can be modelled as a binomial distribution, and binomial distributions famously don't care about order, but you've made two errors with your reasoning. Firstly, we don't use binomial coefficients to determine the total number of outcomes - we use them to determine the specific number of outcomes for a given number of successes and failures (this then becomes a multiplier once the probability of that given number of successes and failures). Secondly, the binomial that you're using is incorrect - you're using the negative binomial coefficient, which is not appropriate here. In terms of dice probability, the negative binomial coefficient is more typically used for dice rolls where a given outcome allow us to reroll, or more generally, where the number of dice we roll is dependent on the outcomes of those rolls. In the case we're looking at, we know the number of dice we're rolling - it's always two, and no outcome on the dice allow us to reroll either of these dice, so the negative binomial coefficient simply doesn't apply here.

Let's go through the math of rolling with advantage using the binomial to demonstrate where you're going wrong here. The probability of a given binomial outcome is:

$$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$

We know that the probability of rolling a 20 on a d20 is $$\frac{1}{20}$$ (so $$p= \frac{1}{20}$$), we have 2 dice (so $$n= 2$$), and we're looking for at least 1 success, so we need to calculate the probabilities for both 1 success and 2 successes, then add them together (so, $$k=\{1,2\}$$). We can plug each of those in to get:

\begin{align*}P(X\geq1) =& \binom{2}{1}\left(\frac{1}{20}\right)^1\left(1-\frac{1}{20}\right)^{2-1}+\binom{2}{2}\left(\frac{1}{20}\right)^2\left(1-\frac{1}{20}\right)^{2-2}\\ =& 2\left(\frac{1}{20}\right)^1\left(\frac{19}{20}\right)^{1}+1\left(\frac{1}{20}\right)^2\left(1-\frac{1}{20}\right)^{0}\\ =& 2\left(\frac{1}{20}\right)\left(\frac{19}{20}\right)+\frac{1}{400}\\ =& 2\left(\frac{19}{400}\right)+\frac{1}{400}\\ =& \frac{38}{400}+\frac{1}{400}\\ =& \frac{39}{400} \end{align*}

In fact, we can even double-check this - We can consider that having a 20 appear on either dice is the opposite of no dice having a 20 on the roll. So, we can calculate that out:

\begin{align*}P(X=0)&=\binom{2}{0}\left(\frac{1}{20}\right)^0\left(\frac{19}{20}\right)^2\\ &=1\left(\frac{19}{20}\right)^2=\frac{361}{400} \end{align*}

And since our dice roll is the very opposite of this number, we determine the probability we actually want by subtracting this figure from 1:

$$P(X\geq1) =1-P(X=0)=1-\frac{361}{400}=\frac{39}{400}$$

So, in conclusion:

1. Your first error is that we don't determine the number of possible outcomes through binomial coefficients - we determine them by multiplying the total possible outcomes of each independent event. In cases such as these, that would always be the number of sides of our die times the number of dice thrown.
2. Your second error is using the negative binomial coefficient. Negative binomials aren't generally used with binomial distributions, they are instead used with geometric distributions, ie situations where the results of our dice roll directly affect the number of the dice we throw.
3. As a bonus, even if we were dealing with a geometric distribution, we still wouldn't be using the negative binomial coefficient to determine the number of possible outcomes. In that case, we'd instead be using the initial number of dice, plus the number of additional dice thrown, multiplied by the number of sides on our dice. The negative binomial coefficient instead represents the number of ways we can assign $$k$$ successes to $$n$$ dice - similar to our binomial example, it would reflect the number of ways a particular outcome can occur, not the total number of outcomes.
• Thanks for the detailed answer! I need to look further into the negative binomial idea as I'm not familiar with that but overall your explanation makes a lot of sense. Commented Mar 15 at 0:34
• Negative Binomials are also known as "multisets", "stars and bars" and "combinations with replacements", if that helps your research! The Negative Binomial distribution is more commonly referred to as a "geometric" distribution. Here's a link to a very good answer on this site that shows how you might use geometric distributions to model a type of die roll (exploding dice pools, in specific): math.stackexchange.com/a/1649514/1217632 Commented Mar 16 at 2:35