What is the probability that when two dice are rolled the sum of the numbers on the two dice is 5?

One popular way to study probability is to roll dice. A standard die has six sides printed with little dots numbering 1, 2, 3, 4, 5, and 6. If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on. But what happens if we add another die? What are the probabilities for rolling two dice?

To correctly determine the probability of a dice roll, we need to know two things:

• The size of the sample space or the set of total possible outcomes
• How often an event occurs

In probability, an event is a certain subset of the sample space. For example, when only one die is rolled, as in the example above, the sample space is equal to all of the values on the die, or the set (1, 2, 3, 4, 5, 6). Since the die is fair, each number in the set occurs only once. In other words, the frequency of each number is 1. To determine the probability of rolling any one of the numbers on the die, we divide the event frequency (1) by the size of the sample space (6), resulting in a probability of 1/6.

Rolling two fair dice more than doubles the difficulty of calculating probabilities. This is because rolling one die is independent of rolling a second one. One roll has no effect on the other. When dealing with independent events we use the multiplication rule. The use of a tree diagram demonstrates that there are 6 x 6 = 36 possible outcomes from rolling two dice.

Suppose that the first die we roll comes up as a 1. The other die roll could be a 1, 2, 3, 4, 5, or 6. Now suppose that the first die is a 2. The other die roll again could be a 1, 2, 3, 4, 5, or 6. We have already found 12 potential outcomes, and have yet to exhaust all of the possibilities of the first die.

The possible outcomes of rolling two dice are represented in the table below. Note that the number of total possible outcomes is equal to the sample space of the first die (6) multiplied by the sample space of the second die (6), which is 36.

The same principle applies if we are working on problems involving three dice. We multiply and see that there are 6 x 6 x 6 = 216 possible outcomes. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. For two dice, there are 62 possible outcomes. For three dice, there are 63 possible outcomes. In general, if we roll n dice, then there are a total of 6n possible outcomes.

With this knowledge, we can solve all sorts of probability problems:

1. Two six-sided dice are rolled. What is the probability that the sum of the two dice is seven?

The easiest way to solve this problem is to consult the table above. You will notice that in each row there is one dice roll where the sum of the two dice is equal to seven. Since there are six rows, there are six possible outcomes where the sum of the two dice is equal to seven. The number of total possible outcomes remains 36. Again, we find the probability by dividing the event frequency (6) by the size of the sample space (36), resulting in a probability of 1/6.

2. Two six-sided dice are rolled. What is the probability that the sum of the two dice is three?

In the previous problem, you may have noticed that the cells where the sum of the two dice is equal to seven form a diagonal. The same is true here, except in this case there are only two cells where the sum of the dice is three. That is because there are only two ways to get this outcome. You must roll a 1 and a 2 or you must roll a 2 and a 1. The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18.

3. Two six-sided dice are rolled. What is the probability that the numbers on the dice are different?

Again, we can easily solve this problem by consulting the table above. You will notice that the cells where the numbers on the dice are the same form a diagonal. There are only six of them, and once we cross them out we have the remaining cells in which the numbers on the dice are different. We can take the number of combinations (30) and divide it by the size of the sample space (36), resulting in a probability of 5/6.

Neither method is correct.

• In the first method, your outcomes are not equally likely

• In the second method, you're counting some outcomes twice.

In order to use the formula $$ \frac{\text{favorable outcomes}}{\text{total outcomes}}, $$ you need your events to be equally likely. In the first method, the outcomes $(1,1)$ and $(1,2)$ are not equally likely. The first method requires both dice to have a value $1$ while the second method has two situations for the dice.

To make this clearer, suppose that the dice are red and blue. Then, $(1,1)$ means that both the red die and the blue die show $1$. On the other hand, in the first method, $(1,2)$ represents the two possibilities ($1$-Red and $2$-Blue) or ($2$-Red and $1$-Blue). Since there are two possible ways to get a $1$ and a $2$, this $(1,2)$ has double the chances of occurring when compared to $(1,1)$.

For the second formulation, you're double counting the pairs of the form $(1,1)$. In this case, you're trying to describe $(1,1)$ for $1$-Red and $1$-Blue as well as $(1,1)$ for $1$-Blue and $1$-Red, but these are exactly the same situation.

Therefore, in the second case, you shouldn't duplicate the pairs that are identical under reversing the coordinates.

To calculate the probability correctly, the list should be $$ (1,1),(1,2),(1,3),(2,1),(2,2),(3,1). $$ Or, in other words, for the red and blue dice, $$ (1R,1B),(1R,2B),(1R,3B),(2R,1B),(2R,2B),(3R,1B). $$ Since there are $6$ possibilities for the red die and $6$ possibilities for the blue die, this results in $36$ total possible outcomes. Putting this all together, the probability is $6/36=1/6$.