# Odd Dice

These strange dice are rolled. What is the probability that the sum obtained is an odd number?

Three fair, six-sided dice are numbered as follows:

A: 1, 1, 1, 2, 2, 2

B: 3, 3, 4, 4, 5, 5

C: 6, 7, 7, 8, 8, 8

The three dice are rolled once. What is the probability that the sum obtained is an odd number?

*This problem is taken from the World Mathematics Championships*

To find out if a sum of three numbers is odd or even, you don't actually need to know the numbers. You only need to know if they are odd or even.

This means that we can replace the numbers on the dice with 'odd' or 'even' (or O and E) without losing any information.

A: 1, 1, 1, 2, 2, 2 becomes O, O, O, E, E, E

B: 3, 3, 4, 4, 5, 5 becomes O, O, E, E, O, O

C: 6, 7, 7, 8, 8, 8 becomes E, O, O, E, E, E

**Using a sample space diagram**

All possible outcomes can be shown in two tables like the ones below. Two tables are necessary because the outcomes of only two dice can be shown on each table. They can be split only by the outcome on A because it is equally likely to be odd or even.

Image

Some of the rows and columns are repeated. A simpler version of the tables is shown below, with fewer rows and columns - but the ratio between the odd and even outcomes on each dice is preserved.

Image

Using the second pair of tables, there are 18 possible outcomes. 9 of them are 'O'. So the probability of getting an odd sum is $\frac12$.

**Using a tree diagram**

The tree diagram below shows the possible outcomes.

The outcomes where the sum is an odd number are indicated with a green arrow.

Image

So the total probability is: $$\left(\frac12\times\frac23\times\frac13\right)+\left(\frac12\times\frac13\times\frac23\right)+\left(\frac12\times\frac23\times\frac23\right)+\left(\frac12\times\frac13\times\frac13\right)\\

= 2\times\left(\frac12\times\frac23\times\frac13\right)+\left(\frac12\times\frac23\times\frac23\right)+\left(\frac12\times\frac13\times\frac13\right)\\

= 2\times\frac{1\times2\times1}{2\times3\times3}+\frac{1\times2\times2}{2\times3\times3}+\frac{1\times1\times1}{2\times3\times3}\\

=\frac{2\times2}{18}+\frac{4}{18}+\frac{1}{18}\\

=\frac{4+4+1}{18}\\

=\frac9{18}\\

=\frac12$$

**Using combinations of products**

To add three numbers and get an odd number, the numbers must have all been odd, or two evens and an odd.

The probability that all of the numbers are odd is $\dfrac12\times\dfrac23\times\dfrac13=\dfrac{1}{9}$.

There are three ways of getting two even numbers and an odd number. The odd number could be on any of the three dice.

Die A odd: $\dfrac12\times\dfrac13\times\dfrac23=\dfrac19$

Die B odd: $\dfrac12\times\dfrac23\times\dfrac23=\dfrac29$

Die C odd: $\dfrac12\times\dfrac13\times\dfrac13=\dfrac1{18}$

So the probability of getting an odd sum is: $$\begin{split}\frac19+\frac19+\frac29+\frac1{18}&=\frac2{18}+\frac2{18}+\frac4{18}+\frac1{18}\\&=\frac9{18}=\frac12\end{split}$$