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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. | 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: | In the case of an {{math|n}}-sided die, D{{math|n}}, this becomes: | ||
{{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= | 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} }} | |||
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} }} | |||
Applying that to the formula of an average of a die Dx we get | Applying that to the formula of an average of a die Dx 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= | 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}{n} = \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{x + 1}{2} = \frac{1}{6}\left(n - \frac{1}{n}\right) }} | |||
When we apply this elegant expression to a D20 we get that '''having advantage is equivalent to an average bonus of +3.325'''. | When we apply this elegant expression to a D20 we get that '''having advantage is equivalent to an average bonus of +3.325'''. | ||