# How to Calculate the Probability of Winning the National Lottery in the UK

*I am a former maths teacher and owner of Doingmaths. I love writing about maths, its applications and fun mathematical facts.*

## National Lottery Stands

## The National Lottery

The National Lottery has been running in the United Kingdom since November 1994, when Noel Edmonds presented the first draw live on the BBC and the original jackpot of £5 874 778 was shared by 7 winners.

Since then, the National Lottery draw has happened every weekend (and also every Wednesday since February 1997) creating numerous millionaires and donating many millions of pounds to charities through the Big Lottery Fund.

## How does the National Lottery Work?

A person playing the National Lottery chooses six numbers between 1 and 59 inclusive. During the draw, six numbered balls are drawn without replacement from a set of balls numbered 1-59. A bonus ball is then drawn after this.

Anybody who matches all six numbers (order of draw does not matter) wins the jackpot (shared with anybody else who matches the six numbers). There are also prizes in descending order of value for matching five numbers + the bonus ball, five numbers, four numbers or three numbers.

## Prize Value

Anybody who matches three balls wins a set £25. The other prizes are all calculated as a percentage of the prize fund and so alter depending upon how many tickets were sold that week.

Generally four balls wins roughly £100, five balls wins roughly £1000, five balls and a bonus ball wins roughly £50 000, while the jackpot can vary from between approximately £2 million to a record of approximately £66 million. (Note: these are the total jackpot amounts. They are usually shared between multiple winners).

## How to Work out the Probability of Winning the National Lottery

## Calculating the Probability of Winning the Jackpot

In order to calculate the probability of winning the jackpot, we need to know how many different combinations of six numbers it is possible to get from the 59 available.

To do this, let's think of the draw as it happens.

The first ball is drawn. There are 59 possible values this can have.

The second ball is drawn. As the first ball is not replaced, there are only 58 possible values for this one.

The third ball is drawn. There are now only 57 possible values.

This continues so that the fourth ball has 56 possible values, the fifth ball has 55 possible values and finally the sixth ball has 54 possible values.

This means that in total there are 59 x 58 x 57 x 56 x 55 x 54 = 32 441 381 2180 possible different ways that the numbers could come up.

However, this total does not take into account the fact that it does not matter what order the numbers are drawn in. If we have six numbers, they can be arranged in 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways, so in reality we need to divide our first figure by 720 to get a total of 45 057 474 different combinations of six numbers.

Obviously, only one of these combinations is the winning combination, so the probability of winning the jackpot is ^{1}/_{45 057 474}.

## What About the Other Prizes?

Calculating the probability of winning the other prizes is slightly trickier, but with a bit of thought, it is certainly possible. We have already worked out the first part by calculating the total number of possible combinations of numbers that can be drawn. To work out the probability of any smaller prize, we now need to work out how many ways they can occur too.

In order to do this we are going to use a mathematical function know as 'choose' (often written nCr or as two numbers vertically stacked within brackets). For ease of typing, I will use the nCr format which is the one generally used on scientific calculators).

nCr is calculated as follows: nCr = ^{n!}/_{r!(n-r)!} where the ! means factorial. (A number factorial equals the number itself multiplied by every positive whole number below it e.g. 5! = 5 x 4 x 3 x 2 x 1).

If you look back at what we did to work out our total of 45 057 474, you will see that we actually calculated 59C6. In short nCr tells us how many different combinations of r objects we can get from a total of n objects, where the order of choice does not matter.

For example, suppose we had the numbers 1, 2, 3 and 4. If we were to choose two of these numbers, we could choose 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4 or 3 and 4, giving us a total of 6 possible combinations. Using our earlier formula 4C2 = ^{4!}/_{2!(4-2!} = 6, the same answer.

## The probability of matching three balls

To find the probability of winning the smaller prizes, we need to split our problem up into two separate parts: the matching balls and the non-matching balls.

Firstly, let's look at the matching balls. We need 3 of our 6 numbers to match up. To work out how many ways this can happen we need to do 6C3 = 20. This means that there are 20 different combinations of 3 numbers out of a set of 6.

Now, let's look at the non-matching balls. We need 3 numbers out of the 53 numbers that have not been drawn, so there are 53C3 = 23 426 ways of doing this.

To find the number of possible combinations of 3 matching numbers and 3 non-matching numbers, we now multiply these two together to get 20 x 23 426 = 468 520.

Therefore, the probability of matching exactly 3 numbers is this last number over our total number of combinations of 6 numbers, so ^{468 520}/_{45 057 474} or approximately ^{1}/_{96}.

## The probability of matching four balls

To find the probability of matching exactly four numbers, we use the same idea.

This time we need 4 of our 6 numbers to match, so 6C4 = 15. We then need 2 further non-matching numbers out of the 53 numbers that have not been drawn, so 53C2 = 1378.

This gives us a probability of ^{15 x 1378}/_{45 057 474} = ^{20 670}/_{45 057 474 }or approximately ^{1}/_{2180}.

## The probability of matching five balls with or without the bonus ball

The probability of matching 5 numbers is a little trickier because of the use of the bonus ball, but to start with we will do the same thing.

There are 6C5 = 6 ways to match 5 numbers from 6 and there are 53C1 = 53 ways to get the final number from the 53 remaining numbers so there are 6 x 53 = 318 possible ways of matching exactly 5 numbers.

However, remember that the bonus ball is then drawn and matching our remaining number to this will increase the prize. There are 53 balls remaining when the bonus ball is drawn, hence there is a ^{1}/_{53} chance of our remaining number matching this.

This means that out of the 318 possibilities for matching 5 numbers, ^{1}/_{53} x 318 = 6 of them will also include the bonus ball, leaving the remaining 318 - 6 = 312 not matching the bonus ball.

Our probabilities are therefore:

Prob (exactly 5 balls and no bonus ball) = ^{312}/_{45 057 474} or approximately ^{1}/_{144 415}

Prob ( 5 balls and the bonus ball) = ^{6}/_{45 057 474} or ^{1}/_{7 509 579}.

## Probabilities summary

P(3 numbers) = ^{1}/_{96}

P(4 numbers) ≈ ^{1}/_{2180}

P(5 numbers) ≈ ^{1}/_{144 415}

P(5 numbers + bonus ball) ≈ ^{1}/_{7 509 579}

P(6 numbers) ≈ ^{1}/_{45 057 474}

## Questions & Answers

**Question:** A state lottery has 1.5 million tickets of which 300 are prize winners . What is the probability of getting a prize by buying just one ticket ?

**Answer:** The probability of winning a prize is 300/1.5 million, which simplifies to 1/5000 or 0.0002.

## Comments

**David (author)** from West Midlands, England on October 31, 2018:

No problem. Thank you for the comment.

**Larry Slawson** from North Carolina on October 22, 2018:

Understanding probability was never one of my strong suits haha. But this is helpful. Thank you for sharing.