You can calculate this with a basic rule of probability: Next, you must determine the probability that each combination appears. As a result, each element of value will refer to the elements of Var1 and Var2 that appear in the same row. R will match up the elements in each vector before adding them together. This will be the sum of the two dice, which you can calculate using R’s element-wise execution: rolls $value <- rolls $Var1 + rolls $Var2 You can determine the value of each roll once you’ve made your list of outcomes. Each combination will contain exactly one element from each vector. id will always return a data frame that contains each possible combination of n elements from the n vectors. For example, you could list every combination of rolling three dice with id(die, die, die) and every combination of rolling four dice with id(die, die, die, die), and so on. You can use id with more than two vectors if you like. This will capture all 36 possible combinations of values: rolls To do so, run id on two copies of die: rolls <- id(die, die)Įid will return a data frame that contains every way to pair an element from the first die vector with an element from the second die vector. For example, you can list every combination of two dice. The id function in R provides a quick way to write out every combination of the elements in n vectors. Listing out these combinations can be tedious, but R has a function that can help. Or, you may roll (1, 2), one on the first die and two on the second. For example, you might roll (1, 1), which notates one on the first die and one on the second die. A total of 36 different outcomes can appear when you roll two dice. Let’s do this step by step.įirst, list out all of the possible outcomes. For example, you could calculate the expected value of rolling a pair of weighted dice. You can use these steps to calculate more sophisticated expected values. The expected value was then just the sum of the values in step 2 multiplied by the probabilities in step 3. Calculated the probability that each outcome occurred.
Determined the value of each outcome (here just the value of the die).Listed out all of the possible outcomes.Notice that we did the same three things to calculate both of these expected values. If you rolled a loaded die an infinite number of times, the average outcome would be 4.125, which is higher than what you would expect from a fair die. Hence, the expected value of a loaded die does not equal the average value of its outcomes. The expected value of a random event is a type of weighted average it is the sum of each possible outcome of the event, weighted by the probability that each outcome occurs:Į(x) = \sum_\\
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