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{{math_block|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_block|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} }} | ||
Applying that to the formula of an average of a die D{{math|n}} we get | 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}^ | {{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}^i 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 | 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 | ||
{{math_block|1= \mathbb{E}[\text{D}n \text{ with advantage}] = \frac{2}{n^ | {{math_block|1= \mathbb{E}[\text{D}n \text{ with advantage}] = \frac{2}{n^3}\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}{2n} = \frac{2n}{3} + \frac{1}{2} - \frac{1}{6n} + 3 + n^3{3<nowiki>}</nowiki> }} | ||
To know what bonus having advantage gives to our roll, we calculate | 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) }} | {{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) }} | ||