Editing Dice rolls
Jump to navigation
Jump to search
The edit can be undone. Please check the comparison below to verify that this is what you want to do, and then publish the changes below to finish undoing the edit.
Latest revision | Your text | ||
Line 321: | Line 321: | ||
{{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<nowiki>}</nowiki> }} | ||
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<nowiki>}</nowiki> }} | ||
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<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) }} |