What are the possible outcomes of rolling three dice simultaneously?

Throwing coins and dice or blindly removing balls from a box are some of the simplest experiments we can carry out to test our understanding of the different concepts related to statistics. They are easy experiments to perform, that anyone can do at home, they give clear and unambiguous results, and these can be easily converted into numerical data.

In the case of dice throwing, there is also a clear relationship between them and games of chance, which makes the application of statistics more tangible in something that is part of the daily life of many people or, at least, something with what almost all of us have come across at least once in their lives.

Rolling three dice simultaneously can produce different types of results that we can interpret in different ways. We can be interested in the individual results themselves, or we can be interested in the value of the sum or in the number of even or odd results that come up between the dice, etc. Of the three, the most common is to be interested in the result of the sum of the values ​​of the three dice. In the following sections, we will explore how to calculate the probability of occurrence of each of the sums when rolling three dice at the same time.

The sample space for rolling three dice

Rolling a single cube die is a simple experiment that has only six possible outcomes. That is, it is an experiment whose sample space is formed by the results S _{1 given} = {1; 2; 3; 4; 5; 6}.

When rolling two dice simultaneously, it can be assumed that the result of each die is independent of the other, so that each one can result in any of the six previous results. ^{This brings as a consequence that 6 2} = 36 possible results can be given corresponding to all the possible combinations between the 6 values ​​of one die and the 6 values ​​of the other.

In this case, we will have a sample space of S _{2 given} = {11; 12; 13; 14; fifteen; 16; twenty-one; 22; 23; 24; 25; 26; …; 61; 62; 63; 64; 65; 66}. Of these 36 results, the number of unique combinations (without considering the order) can be calculated by means of a combinatorics with repetition in which groups of n = 2 are taken (the two dice that are thrown) with m = 6 possible results. :

These 21 results correspond to {11; 12; 13; 14; fifteen; 16; 22; 23; 24; 25; 26; 33; 3. 4; 35; 36; 44; Four. Five; 46; 55; 56; 66}. The probability of each of these results corresponds to 1/36 multiplied by the number of different permutations that can be created with the digits of each number (1 if the number is repeated, as in 11, 22, etc., and 2 if the number is not repeated, since we can have 12 or 21, 13 or 31, etc.)

In the case of rolling 3 dice, the total number of possible outcomes in the sample space is given by 6 ³ = 216. These outcomes are S _{3 dice} = {111; 112; 113; 114; 115; 116; 121; …; 126; 131; …; 136; …; 166; 211; 212; …; 656; 666}. In this case, the probability of any individual outcome must be 1/216.

Probability of individual results when rolling three dice

Now that we have well defined the sample space of all the possible results of the throw of 3 dice, let’s see how to calculate the probability of each of the different results that can be obtained.

In the case of rolling three dice, considering that the order in which the results come up is irrelevant, many of the 216 results will actually be repeated. The total number of unique results can be calculated again as a combination of groups of 3 with 6 options each and with the possibility of repetitions, that is:

Among these 56 results, those consisting of three equal numbers (let’s call them AAA) only occur once. On the other hand, those with two identical figures and one different one (AAB) are repeated 3 times each (corresponding to the permutations AAB, ABA and BAA). Finally, those who have three different figures (ABC) will appear 3! = 6 times (ABC, ACB, BAC, BCA, CAB and CBA).

From this information and the total number of possible outcomes (216), we can calculate the probability of each outcome as

Depending on the result, it has 1, 2 or 3 different figures. The 56 possible outcomes and their probabilities are shown in the following table:

Probability of the sum when rolling three dice

As mentioned before, when rolling the dice, a more important result than the particular number each heads lands on is the sum of the dice. In the experiment in which three dice are rolled and the sum is obtained, the sample space is made up of all possible sums between three numbers from 1 to 6.

The smallest value that can result from this sum is the one obtained when the three dice land on 1, obtaining a sum of 1+1+1 = 3, while the maximum value corresponds to 6+6+6 = 18, with the possibility of obtaining any of the intermediate sums. Therefore, the sample space of this experiment corresponds to:

S = {3; 4; 5; 6; 7; 8; 9; 10; eleven; 12; 13; 14; fifteen; 16; 17; 18}

The last column of the table shows the total number of results that each sum gives, including equivalent results (from all permutations of each unique combination). For example, for the sum of 15, the roll of the dice must be 366, 356, or 555. But there are 3 permutations of 366 (366, 636, and 663) and 6 permutations of 356 (356, 365, 536, 563, 635 and 653) and a single one out of 555, so the total number of possible outcomes that equal 15 is 10.

With the previous table we can practice calculating the probability of each sum for the throw of three dice in two different ways. These are detailed below.

Strategy 1: Using the probability of each unique outcome

The first strategy consists of adding the probability of all the unique results that each sum can give. This involves using the unique outcomes from the third column and the respective probability of each outcome presented above.

Example

Suppose we want to calculate the probability that the sum of the three dice is 11 (that is, P(11)). In this case, there are 6 unique combinations (regardless of order) that give a sum of 11. These results are (according to the third column of the table above): {146; 155; 236; 245; 335; 344}.

The probability of each result is determined based on the total number of possible permutations in each case, as explained in the previous section. In this case:

Therefore, the probability that the result of the sum is 11 will be:

Similarly, if we wanted the probability that the sum is 16, the result would be the sum of the probabilities of 466 and 556, which are both equal to 1/72, so the probability would be:

Strategy 2: Using the total number of results corresponding to each sum

In this case, a simpler path is taken, as long as there is a list of all the possible results for each summation, including the permutations. Then the probability of each sum is simply the total number of outcomes for the sum divided by the total number of possible outcomes (216).

Example

In the case of the sum = 11, the total number of possible results that give said sum is 27 (see the third column of the previous table), so the probability that the sum of 11 will be:

As you can see, the result is the same as before and it is very simple if we have a table like the previous one already built. However, for more complex cases where there are more possible outcomes (such as rolling 4, 5, or 4 dice), this strategy may be less convenient and the former more practical.

References

Graffe, S. (2021, September 21). What is the probability that when you roll three dice, you get a sum of 7? Quora. https://en.quora.com/What%C3%A9-probabilidad-hay-que-al-lanzar-tres-dados-salga-una-sumatoria-de-7

Montagud Rubio, N. (2022, March 17). Counting techniques: types, how to use them and examples . Psychology and Mind. https://psicologiaymente.com/miscelanea/tecnicas-de-conteo

Naps. (2017, November 16). Counting Techniques in Probability and Statistics . Naps Technology and education. https://naps.com.mx/blog/tecnicas-de-conteo-en-probabilidad-y-estadistica/

Valdés Gómez, J. (2016, November 23). Combinations with repetition . Youtube. https://www.youtube.com/watch?v=WqHZx64RW-Q