# Odds and Evens

## Problem

*Odds and Evens printable sheet*

Here is a set of numbered balls used for a game:

To play the game, the balls are mixed up and two balls are randomly picked out together. For example:

The numbers on the balls are added together: $4 + 5 = 9$

**If the total is even, you win.**

If the total is odd, you lose.

How can you decide whether the game is fair?

If the total is odd, you lose.

How can you decide whether the game is fair?

*You might like to experiment with the interactivity below.*

Here are three more sets of balls:

Which set would you choose to play with, to maximise your chances of winning?

Click on the links below to explore each set using the interactivity.

**What proportion of the time would you expect to win each game?**

*You may wish to look at the problem Odds and Evens Made Fair to explore whether it is possible to change the number of balls to make the game fair.*

## Getting Started

Try to list all the possible combinations for set $A$.

Try to be systematic so you don't miss any combination.

How can I end up with an even total?

By adding two even numbers? An even and an odd? Two odds?

## Student Solutions

Alfie from Tain Royal Academy in Scotland used the interactvity to test whether the game is fair with set A:

I'm going to take a total sum of wins for every hundred rolls up to 500 rolls:

100 rolls -42 wins (relatively fair random results, the closer to 50 it is the more fair it is)

200 rolls- 88 wins (still 12 wins from exactly fair)

300 rolls- 133 wins (a bit further than last time)

400 rolls- 169 wins (straying further from 50/50 split)

500 rolls- 207 wins (almost 50 wins from fair)

Now I'll jump to 1000 rolls- 416 wins (still 84 wins from an even 500)

From this I can conclude that from 100 rolls the results are slowly dropping from the even split. From here I can say with almost certainty that it is not quite fair”ƒ

Scout from Jmes Allen Girls School used the same strategy for sets B, C and D:

Set B

The mean of winning when I played 100 times:

0.570, 0.500, 0.640, 0.600, 0.560

Overall mean: 0.574

Set C

The mean of winning when I played 100 times:

0.460, 0.470, 0.420, 0.480, 0.440

Overall mean: 0.454

Set D:

The mean of winning when I played 100 times:

0.690, 0.670, 0.590, 0.630, 0.670

Overall mean: 0.650

I would probably play with D to make sure I had the most chance of winning.

Scout also looked at set A in more detail:

I worked out the many possibilities for Set A:

Please note: W=win L=lose and each combination is in inverse

*(e.g. 2 + 3 and 3 + 2 are both listed here)*

2+3=5=L

2+4=6=W

2+5=7=L

2+6=8=W

We notice a pattern as we go systematically through them, as we increase the last number we are adding by 1, it goes through a multitude of odd, even, odd, even etc.

3+2=5=L

3+4=7=L

3+5=8=W

3+6=9=L

Here, because 3 is an odd number there are three loses and a win instead of an equal number of each set.

4+2=6=W

4+3=7=L

4+4=8=W

4+5=9=L

An equal number on the sets. It seems like on an even is equal. This makes sense as an odd add an even is an odd and an odd add an even is an odd too. This means it is higher probability of you getting an odd result.

5+2=7=L

5+3=8=W

5+4=9=L

5+6=11=L

Already, we can see the probability is higher to get an odd result than an even one by a long shot.

6+2=8=W

6+3=9=L

6+4=10=W

6+5=11=L

Again, an equal number.

Daniele from Southlands International School, Rome in Italy, Ali from Garden International School in Malaysia and Mahdi from Mahatma Ghandi International School in India also listed the probabilities, and counted up the number of winning combinations. Ali wrote:

The game is not fair because with the set of different 5 balls there are 10 possible number combinations

*(Ali does not list the combinations that Scout described as "inverse", so has 10 instead of 20)*

When I checked there are 6 answers which are odd and 4 answers which are even. Therefore, there is a 60% chance you will get an odd number and a 40% chance you will get an even number.

Daniele also made lists of all of the possibilities for sets B, C and D:

If I want to maximise my chances of winning, I would choose set D.

Set 8B from Manchester High School for Girls used tables to show all of the possible combinations for all of the sets:

From the table we have worked out all the possible outcomes of the experiment. We have ignored the diagonals as we are choosing numbers without replacement.

One group noticed that we only needed to use half of the table to work out the probabilities as the table is mirrored about the diagonal. Using a similar table for set B, we can see that there are 6 evens out of a possible 10. Therefore, the probability of winning is 60%.

Using a similar method, we found out what all the outcomes were using set C. From this we can see that there are 14 even outcomes and 16 odd outcomes. Therefore, we have 14 winning outcomes out of a possible 30 and so the probability of winning is 46.7%.

Below is a table representing the sample space diagram for set D. From here we can see that there are 20 even outcomes and 10 odd outcomes. Therefore, we have 20 winning outcomes from a possible 30 and so the probability of winning is 66.7%.

Overall, as set D gives us the greatest probability of winning, we would choose to play with set D to maximise the chances of winning.

Sanika P from PSBBMS, OMR in India used tree diagrams to represent the possibilities for each set:

Set A

Probability of Odd-Odd combination $ = \frac25\times\frac14 = \frac2/{20}$

Probability of Even-Even combination $ = \frac35\times\frac24 = \frac6{20}$

Total chance of winning $ = \frac2{20}+\frac6{20} = \frac8{20} = 0.4 = 40\%$

Set B

Probability of Odd-Odd combination $ = \frac45\times\frac34 = \frac35$

Probability of Even-Even combination $ = \frac15\times0 = 0$

Total chance of winning $ = \frac35+0 = \frac35 = 0.6 = 60\%$

Set C

Probability of Odd-Odd combination $ = \frac26\times\frac15 = \frac2{30}$

Probability of Even-Even combination $ = \frac46\times35 = \frac{12}{30}$

Total chance of winning $ = \frac2{30} + \frac{12}{30} = \frac{14}{30} = 0.467 = 46.7\%$

Set D

Probability of Odd-Odd combination $ = \frac56\times\frac45 = \frac{20}{30} = \frac23$

Probability of Even-Even combination $ = \frac16\times0 = 0$

Total chance of winning $ = \frac23+0 = \frac23 = 0.667 = 66.7\%$

Maria Ester from Italy used a similar method but didn't draw the whole trees. This is Maria Ester's work:

Tom and Hussein from Wilson's School and Daniele found the same set that produced a fair game:

To produce a fair set of numbers from what we found out you would need 4 balls, one odd and three even, or the other way round, one even and three odd.

Daniele also found that these sets would give a fair game:

Mahdi used the notation $\left(\begin{matrix}n\\2\\ \end{matrix}\right)$, where $\left(\begin{matrix}n\\2\\ \end{matrix}\right)$ is the number of ways of choosing $2$ balls from a set of $n$ balls.

For any general set containing $o$ odd numbers and $e$ even numbers

Total number of pairs formed $ = \left(\begin{matrix}o+e\\2\\ \end{matrix}\right)$

Total favourable pairs can be formed by choosing pairs of odd numbers and pairs of even numbers

Number of favourable outcomes $ = \left(\begin{matrix}o\\2\\ \end{matrix}\right) + \left(\begin{matrix}e\\2\\ \end{matrix}\right)$

Probability $= \dfrac{\left(\begin{matrix}o\\2\\ \end{matrix}\right)+\left(\begin{matrix}e\\2\\ \end{matrix}\right)}{\left(\begin{matrix}o+e\\2\\ \end{matrix}\right)}$

Let us verify the general strategy for set A, set B, set C and set D

For set A, 2 odd, 3 even: $= \dfrac{\left(\begin{matrix}2\\2\\ \end{matrix}\right)+\left(\begin{matrix}3\\2\\ \end{matrix}\right)}{\left(\begin{matrix}5\\2\\ \end{matrix}\right)} = \dfrac{1+3}{10} = \dfrac4{10}$

For set B, 4 odd, 1 even: $= \dfrac{\left(\begin{matrix}4\\2\\ \end{matrix}\right)+\left(\begin{matrix}1\\2\\ \end{matrix}\right)}{\left(\begin{matrix}5\\2\\ \end{matrix}\right)} = \dfrac{6+0}{10} = \dfrac6{10}$

For set C, 2 odd, 4 even: $= \dfrac{\left(\begin{matrix}2\\2\\ \end{matrix}\right)+\left(\begin{matrix}4\\2\\ \end{matrix}\right)}{\left(\begin{matrix}6\\2\\ \end{matrix}\right)} = \dfrac{1+6}{15} = \dfrac7{15}$

For set D, 5 odd, 1 even: $= \dfrac{\left(\begin{matrix}5\\2\\ \end{matrix}\right)+\left(\begin{matrix}1\\2\\ \end{matrix}\right)}{\left(\begin{matrix}6\\2\\ \end{matrix}\right)} = \dfrac{10+0}{15} = \dfrac{10}{15}$

## Teachers' Resources

### Why do this problem?

This problem offers an opportunity to explore and discuss two types of probability: experimental and theoretical. The simulation generates lots of experimental data quickly, freeing time to focus on predictions, analysis and justifications. Calculating the theoretical probabilities provides a motivation for using sample space diagrams or perhaps tree diagrams.

### Possible approach

You may wish to use the start of What Numbers Can We Make? as a preliminary activity to get students thinking about the effect of combining odd and even numbers.

*The notes that follow are based on a lesson with a group of 14 year old students. The first part forms a single lesson on probability and sample space diagrams or tree diagrams. The second part, a collaborative task that leads to interesting and unexpected results, can be found in the extension task Odds and Evens Made Fair.*

**think on their own, without writing,**about which of the four sets they would choose to play with, to maximise their chances of winning. Once they have had a short time to reflect, ask them to discuss in pairs their choice, and to justify their decisions (again, without writing). There is often disagreement about which set offers the best chance of winning, so bring the class together to compare ideas before setting them the task of calculating the probabilities - discourage them from using inefficient listing methods. You may wish to use the interactivities for sets B, C and D, to compare the theoretical and experimental probabilities.

### Key questions

How can you decide if a game is fair?

### Possible support

Flippin' Discs could be used as a simpler context for exploring theoretical and experimental probability.

### Possible extension

The problem In a Box offers another context for exploring exactly the same underlying mathematical structure, and could be used as a follow-up problem a few weeks after working on this one.