Advantage: Difference between revisions

For this we first need to clarify the notations used below: D{{math|n}} represents an {{math|n}}-sided die, {{math|P(i)}} is the probability that a variable has value {{math|a}}, {{math|\mathbb{E} }} denotes the average or expected value of a roll, the subscript "adv" means "with advantage", and {{math|1=\textstyle\sum_{i=a}^b x_i}} denotes the sum of a series of numbers {{math|x}} over an index {{math|i}} with {{math|i}} going from {{math|a}} through {{math|b}}.

The formula to calculate the expected value, {{math|\mathbb{E}[x]}}, of a variable {{math|x}} is equal to the sum of every possible value of {{math|x}} multiplied by the chance for {{math|x}} to have that value.

In the case of an {{math|n}}-sided die, D{{math|n}}, this becomes:

* {{math|1=\mathbb{E}[\text{D}n] = \sum_{i=1}^n (i \cdot P(i))}}

For a regular dice roll the probability distribution is uniform, which means {{math|1=P(i) = 1/n}} for any {{math|i}}, and using {{math|1=\textstyle\sum_{i=1}^n i = \frac{n(n+1)}{2} }}, we get  

* {{math|1=\mathbb{E}[\text{D}n] = \sum_{i=1}^n(i \cdot P(i)) = \frac{1}{n}\left(\frac{n(n+1)}{2}\right) = \frac{n+1}{2} }}

For a dice roll with advantage the chance to roll the number {{math|i}} is equal to the chance that the first die rolls {{math|i}} multiplied by the chance that the second die rolls {{math|i}} or less, multiplied by 2 (because the 2 dice are interchangeable), minus the chance of both dice rolling {{math|i}} (because we counted that possibility twice by multiplying by 2). This gives

* {{math|1=P_\text{adv}(i) = 2P(i)\sum_{j=1}^i P(j) - P(i)^2 = 2\frac{1}{n} \cdot \frac{i}{n} - \frac{1}{n^2} = \frac{2i - 1}{n^2} }}

* {{math|1=\mathbb{E}[\text{D}n \text{ with advantage}] = \sum_{i=1}^n i \cdot\frac{2i - 1}{n^2} = \frac{2}{n^2} \cdot \sum_{i=1}^n i^2 - \frac{1}{n^2} \cdot \sum_{i=1}^n i}}

Here we can use that the sum of squares is {{math|1=\textstyle\sum_{i=1}^n i^2 = n(n + 1)(2n + 1)/6}}, which gives

* {{math|1= \mathbb{E}[\text{D}n \text{ with advantage}] = \frac{2}{n^2}\left(\frac{n(n+1)(2n+1)}{6}\right) - \frac{1}{n^2}\left(\frac{n(n+1)}{2}\right) = \frac{2n}{3} + 1 + \frac{1}{3n} - \frac{1}{2} - \frac{1}{n} = \frac{2n}{3} + \frac{1}{2} - \frac{1}{6n} }}

* {{math|1= \mathbb{E}[\text{D}n \text{ with advantage}] - \mathbb{E}[\text{D}n] = \frac{2n}{3} + \frac{1}{2} - \frac{1}{6n} - \frac{x + 1}{2} = \frac{1}{6}\left(n - \frac{1}{n}\right) }}

Because of symmetry, having disadvantage instead of advantage means we can simply make the permutation of {{math|\{1, \dots, n\} \to \{n, \dots, 1\} }} for the values of dice rolls and all the calculations will remain the same. Therefore the size of the bonus of advantage is equal to the size of the penalty of disadvantage.