Calculate the probability of rolling a 3 on the first die and a 4 on the second die

When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b):

With the sample space now identified, formal probability theory requires that we identify the possible events. These are always subsets of the sample space, and must form a sigma-algebra. In an example such as this, where the sample space is finite because it has only 36 different outcomes, it is perhaps easiest to simply declare ALL subsets of the sample space to be possible events. That will be a sigma-algebra and avoids what might otherwise be an annoying technical difficulty. We make that declaration with this example of two dice.

With the above declaration, the outcomes where the sum of the two dice is equal to 5 form an event. If we call this event E, we have

E={(1,4),(2,3),(3,2),(4,1)}.

Consider next the probability of E, P(E). Here we need more information. If the two dice are fair and independent , each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their probabilities must equal 1, each singleton event {(a,b)} is assigned probability equal to 1/36. Because E is composed of 4 such distinct singleton events, P(E)=4/36= 1/9.

In general, when the two dice are fair and independent, the probability of any event is the number of elements in the event divided by 36.

What if the dice aren't fair, or aren't independent of each other? Then each outcome {(a,b)} is assigned a probability (a number in [0,1]) whose sum over all 36 outcomes is equal to 1. These probabilities aren't all equal, and must be estimated by experiment or inferred from other hypotheses about how the dice are related and and how likely each number is on each of the dice. Then the probability of an event such as E is the sum of the probabilities of the singleton events {(a,b)} that make up E.

In 1654, Chevalier de Mere, a French gambler, wrote to Pierre Fermat and Blaise Pascal, two of France's mathematical giants, with a number of problems concerning the odds of particular combinations of numbers occurring, when several dice are rolled. This event is considered to be the birth of probability theory.

Let's investigate a simple question that Chevalier de Mere could have asked. Suppose we roll two dice. We can get a sum of 4 in two different combinations: (1,3) and (2,2). We can get a sum of 5 in two different combinations also: (1,4) and (2,3). Why is it that in de Mere's practice 5 appears more often than 4?

The answer is the following: the combinations (1,3) and (2,2) are not equiprobable. We have a probability of 1/6 that the first die rolls 2, and a probability of 1/6 that the second die rolls 2, thus making a combination (2,2) with the probability 1/36. By a similar argument we see that the probability that the first die rolls 1 and the second die rolls 3 is 1/36. The probability that the first die rolls 3 and the second die rolls 1 is also 1/36. Hence, the combination (1,3) is rolled with probability 2/36 = 1/18.

In the table below, the numbers in the left column show what is rolled on the first die and the numbers in the top row show what is rolled on the second die. We will color in blue the cells corresponding to the sum of 4, and in pink the cells corresponding to the sum of 5.

Probabilities for Two Dice

Now we can see that the sum 4 will be rolled with probability 3/36 = 1/12, and the sum 5 with probability 4/36 = 1/9.

Below you can check our random "roll of dice" generator. It will count for you the total number of rolls and the total for each sum. To set the count back to 0, press "Start Over" button.

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