Expected value

In probability theory, the expected value or expectation of a random variable, intuitively, is the long-run average value of repetitions of the same experiment it represents. In other words, the law of large numbers states that the arithmetic mean of the values almost surely converges to the expected value as the number of repetitions approaches infinity.

Although the definition of the drop rate is based on that of the expected value, the two are not the same. The drop rate of an item (for example, a Pro Kit card) is its expected value in relation to all items (for example, the content of a Pro Kit Box).
 * The expected value is a number (the average amount of a card in a box).
 * The drop rate is a ratio, expressed in percent (as provided in the box info).

Definition
Let $$X$$ be a random variable with a finite number of finite outcomes $$x_1$$, $$x_2$$, ..., $$x_k$$ occurring with probabilities $$p_1$$, $$p_2$$, ..., $$p_k$$, respectively. The expected value of $$X$$ is defined as


 * $$\operatorname{E}[X] = \sum_{i=1}^k x_i\, p_i = x_1p_1 + x_2p_2 + \ldots + x_kp_k$$.

Since all probabilities $$p_i$$ add up to 1 ($$p_1 + p_2 + \ldots + p_k = 1$$), the expected value is the weighted average, with $$p_i$$’s being the weights. In the special case of all probabilities being the same ($$p_1 = p_2 = \ldots = p_k$$), the weighted average turns into the simple average.

In this case (all outcomes of $$X$$ have the same probability $$p$$), the general formula becomes


 * $$\operatorname{E}[X] = \sum_{i=1}^k x_i\, p = x_1p + x_2p + \ldots + x_kp$$.

Simple average
If $$X$$ represents the outcome of a roll of a fair six-sided die, the possible values for $$X$$ are 1, 2, 3, 4, 5, and 6, all equally likely (each having the probability of $$\tfrac16$$). The expectation of $$X$$ is


 * $$\operatorname{E}[X] = 1\cdot\frac16 + 2\cdot\frac16 + 3\cdot\frac16 + 4\cdot\frac16 + 5\cdot\frac16 + 6\cdot\frac16 = 3.5$$.