Probability - Throwing Dice

Throwing a single dice

A number of interesting, and simple, questions can be framed on throwing a single dice. As you may be aware, a dice has six faces numbered one to six.

A dice

A dice has six faces numbered one to six

When you throw a single dice once, either one of the six faces can come up. Thus when you throw a dice the total number of possible outcomes is six. The probability of, say, number 1 coming up is given by the formula for theoretical probability:

`P(1) = "number of favorable outcomes"/"total number of possible outcomes"`

Number of favorable outcomes is equal to one because there is only one face of the dice that can show number 1 on it. Total number of possible outcomes is six because there are six faces on a dice and any one of the six faces can show up. Hence the probability of the number one showing up on throwing a dice is given by:

`P(1) = 1/6`

Thus, when you throw a dice, the probability of any one of the six numbers showing up is equal to 1/6.

Different types of questions based on this are explained in: "Probability - Throwing One Dice"

Throwing Two Dice

Throwing two dice or throwing a single dice two times, both are the same things. The probabilities of different outcomes are same in both cases. Here we shall consider throwing two dice for simplicity.

On throwing two dice, you can get many pairs of two numbers on the two dice. For example, 1 on the first dice and 3 on the second. How many different pairs as such are possible? A total of 36 combinations are possible on throw of two dice. This can be understood as follows:

Suppose you get the number 1 when you throw the first dice. Now you throw the second dice. You can get any one of the six numbers on the second dice. Thus, for each number on the first dice you can get six numbers on the second dice. Since there are six numbers on the first dice, hence total number of possible pairs of numbers you can get are 6 + 6 + 6 + 6 + 6 + 6 = 36.

These possible pairs are shown below:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Thus we have one important information already for solving the questions based on throw of two dice: total number of possible outcomes is equal to 36.

Different types of questions based on probability when throwing two dice are explained: "Throwing Two Dice"

Throwing Three Dice

If you throw three dice computation of the probabilities become a bit more tedious since the sample space, that is, the total number of possible combinations of numbers on the three dice, increases tremendously (it becomes 6 * 6 * 6 = 216). Hence, rather than using the theoretical probability approach that involves counting the number of favorable outcomes, here we use another approach that involves the binomial theorem. So if you are not familiar with the binomial theorem please read this post: Application of Binomial Theorem in Probability