5,797
editsMore actions
Fixed errant <nowiki> tags that got accidentally inserted
(Significant overhaul of most sections. Cleaned up a bunch of redundant info, fixed typos, added additional info about AC, etc.) |
(Fixed errant <nowiki> tags that got accidentally inserted) |
||
| Line 318: | Line 318: | ||
{{math_block|1=\mathbb{E}[\text{D}n] = \sum_{i=1}^n (i \cdot P(i))}} | {{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 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 | ||
{{math_block|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 | {{math_block|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 | 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_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}^n i}} | {{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 | 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^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}{2n} = \frac{2n}{3} + \frac{1}{2} - \frac{1}{6n | {{math_block|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}{2n} = \frac{2n}{3} + \frac{1}{2} - \frac{1}{6n} }} | ||
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) }} | ||