Dice rolls: Difference between revisions

{{math_block|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=\sum_{i=1}^n i = \frac{1}{2}n(n+1) }}, we get  

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

Applying that to the formula of an average of a die D{{math|n}} we get

{{math_block|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=\sum_{i=1}^n i^2 = \frac{1}{6}n(n + 1)(2n + 1)}}, which gives

To know what bonus having advantage gives to our roll, we calculate

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