# Mathsland National Lottery

## Problem

*Mathsland National Lottery printable sheet*

In the Mathsland National Lottery, six balls painted with the numbers 1 to 6 are placed in a bag.

To enter, you choose three numbers.

To win, your numbers must match (in any order) the three numbers that are drawn from the bag.

**What is the probability of winning?**

The Mathsland Lottery organisers decide to change to a simpler system where you just choose two numbers from the six, and then two numbers are drawn from the bag.

To win, both of your numbers must match.

**How will this affect your chance of winning?**

The two-number lottery is very unpopular, so the organisers move to a four-number (six ball) lottery instead.

**How will this affect your chance of winning?**

In their effort to find the most popular lottery, the organisers also trial a one-number and a five-number (six ball) lottery.

**What can you say about the chances of winning these?**

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The Mathsland Lottery organisers decide to move to a 10-ball lottery, but they can't decide how many numbers should be drawn.

The organisers want to make the chance of winning as low as possible! Can you advise them?

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When the UK National Lottery was first introduced it allowed you to choose six numbers from 49. What is the probability of winning the jackpot in the UK lottery?

## Getting Started

What is the probability that the first number drawn matches one of your choices?

What is the probability that the second number drawn also matches one of your choices?

What is the probability that the first two numbers drawn match two of the numbers you have chosen?

Does it help to draw a tree diagram?

## Student Solutions

We hope that you enjoyed exploring this problem.

Here are some of the solutions which we have previously received from students:

Derin from Woodhouse College gave a clear presentation of his reasoning:

At the first draw, when there are 6 balls in the bag, you want one of your 3 numbers to be picked, so the chance of that happening is $\frac{3}{6}$ (or $\frac{1}{2}$).

For your second ball to be chosen, when there are five balls left in the bag and you want one of your 2 remaining numbers to be picked, the chance of that happening is $\frac{2}{5}$.

For the last ball to be chosen, when there are four balls left in the bag and you want your one remaining number to be picked, the chance of that happening is $\frac{1}{4}$.

In order for all three of your balls to be picked sequentially, you must multiply the probabilities of each being chosen on their own, i.e:$\frac{3}{6}\times\frac{2}{5}\times\frac{1}{4}= \frac{1}{20}$ therefore you have a one-in-twenty (or 0.05) chance of winning.

Ayden from Melbourn Village College calculated the solution to the rest of the problem. He also noticed that the chances of picking two balls out of six and four balls out of six is the same.

If you had a 2 out of 6 ball lottery it would increase your chances of winning:

$\frac{2}{6}\times\frac{1}{5}= \frac{1}{15}$

Drawing 4 balls instead of 2 will keep the chance of winning the same:

$\frac{4}{6}\times\frac{3}{5}\times\frac{2}{4}\times\frac{1}{3}= \frac{1}{15}$

Having a 1-ball lottery will dramatically increase your chance of winning to $\frac{1}{6}$.

Having a 5 ball lottery will give you the same chance of winning:

$\frac{5}{6}\times\frac{4}{5}\times\frac{3}{4}\times\frac{2}{3}\times\frac{1}{2}= \frac{1}{6}$

To minimise the chance of winning a 10-ball lottery, the organisers should require participants to pick 5 balls.

Phil from Wilson's School provided the explanation:

If the Mathsland Lottery changes to a 10-ball lottery and wants to minimise the chance of winning, then they should ask participants to pick 5 balls correctly to win. This is because if you look at the six-ball lottery the chance of winning was its lowest when three balls were needed to be picked correctly. This is half of three, and on either side of 3 balls, 2 and 4, the chance of winning went
back up.

When 5 out of ten must be picked, the chance of winning is

$\frac{5}{10}\times\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}= \frac{24}{6048}=\frac{1}{252}$

which is smaller than when 4 balls have to be picked:

$\frac{4}{10}\times\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}=\frac{1}{210}$

and when 6 balls have to be picked:

$\frac{6}{10}\times\frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}\times\frac{2}{6}\times\frac{1}{5}=\frac{1}{210}$

As we get further away from picking 5, the probabilities keep increasing:

When 3 balls have to be picked:

$\frac{3}{10}\times\frac{2}{9}\times\frac{1}{8}=\frac{1}{120}$

and when 8 balls have to be picked:

$\frac{8}{10}\times\frac{7}{9}\times\frac{6}{8}\times\frac{5}{7}\times\frac{4}{6}\times\frac{3}{5}\times\frac{2}{4}\times\frac{1}{3}=\frac{1}{45}$

Therefore if the Mathsland lottery is organising a ten-ball lottery, participants should be ask to predict 5 balls, since this results in the lowest winning chances.

It is worth pointing out that picking 2 balls out of 6 balls is essentially the same as picking 4 balls out of 6 balls, since in both cases 2 balls need to be identified and separated. The two processes are equivalent, so the probabilities match. The same goes for picking 1 out of 6 and picking 5 out of 6, etc.

The chance of winning the UK National lottery, in which participants choose 6 numbers from 49, is:

$\frac{6}{49}\times\frac{5}{48}\times\frac{4}{47}\times\frac{3}{46}\times\frac{2}{45}\times\frac{1}{44}=\frac{1}{13,983,816}$ or approximately 1 in 14 million.

Well done to everyone!

## Teachers' Resources

**Why do this problem?**

This problem offers an engaging context in which to develop students' understanding of theoretical probability. They can calculate theoretical probabilities, perhaps by listing at first, but then by moving towards multiplying fractions based on conditional probabilities.

### Possible approach

You could introduce the problem by simulating a lottery using numbered balls or digit cards in a bag.

### Key questions

How often would you expect to win?

Why is the probability of winning the two from six lottery the same as the probability of winning the four from six lottery?

### Possible support

You may want to experiment with this lottery simulator before moving on to the theoretical probabilities

### Possible extension

Some students may wish to extend the 10-ball lottery question to the general case: what is the hardest lottery to win with $n$ balls? This could be a good opportunity to introduce them to factorial notation and the binomial coefficients.