# Blog

## Archive

Show me a Random Blog Post**2017**

### Mar 2017

The End of Coins of Constant WidthDragon Curves II

### Feb 2017

The Importance of Estimation Error### Jan 2017

Is MEDUSA the New BODMAS?**2016**

**2015**

**2014**

**2013**

**2012**

## Tags

folding paper folding tube maps london underground platonic solids rhombicuboctahedron raspberry pi weather station programming python php inline code news royal baby probability game show probability christmas flexagons frobel coins reuleaux polygons countdown football world cup sport stickers tennis braiding craft wool emf camp people maths trigonometry logic propositional calculus twitter mathslogicbot oeis pac-man graph theory video games games chalkdust magazine menace machine learning javascript martin gardner reddit national lottery rugby puzzles advent game of life dragon curves fractals pythagoras geometry triangles european cup dates palindromes chalkdust christmas card bubble bobble asteroids final fantasy curvature binary arithmetic bodmas statistics error bars estimation accuracy misleading statistics pizza cutting**2015-10-08 04:38:58**

## How Much Will I Win on the New National Lottery?

This post also appeared on the Chalkdust Magazine blog. You can read the excellent second issue of Chalkdust here, including the £100 prize crossnumber which I set.

From today, the National Lottery's Lotto draw has 59 balls instead of 49. You may be thinking that this means there is now much less chance of winning. You would be right, except the prizes are also changing.

Camelot, who run the lottery, are saying that you are now "more likely to win a prize" and "more likely to become a millionaire". But what do these changes actually mean?

### The Changes

Until yesterday, Lotto had 49 balls. From today, there are 59 balls. Each ticket still has six numbers on it and six numbers, plus a bonus ball, are still chosen by the lottery machine. The old prizes were as follows:

Requirement | Estimated Prize |

Match all 6 normal balls | £2,000,000 |

Match 5 normal balls and the bonus ball | £50,000 |

Match 5 normal balls | £1,000 |

Match 4 normal balls | £100 |

Match 3 normal balls | £25 |

50 randomly picked tickets | £20,000 |

The prizes have changed to:

Requirement | Estimated Prize |

Match all 6 normal balls | £2,000,000 |

Match 5 normal balls and the bonus ball | £50,000 |

Match 5 normal balls | £1,000 |

Match 4 normal balls | £100 |

Match 3 normal balls | £25 |

Match 2 normal balls | Free lucky dip entry in next Lotto draw |

One randomly picked ticket | £1,000,000 |

20 other randomly picked tickets | £20,000 |

### Probability of Winning a Prize

The probability of winning each of these prizes can be calculated. For example, the probability of matching all 6 balls in the new lotto is $$\mathbb{P}(\mathrm{matching\ ball\ 1})\times \mathbb{P}(\mathrm{matching\ ball\ 2})\times...\times\mathbb{P}(\mathrm{matching\ ball\ 6})$$ $$=\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{2}{55}\times\frac{1}{54}$$ $$=\frac{1}{45057474},$$ and the probability of matching 4 balls in the new lotto is $$(\mathrm{number\ of\ different\ ways\ of\ picking\ four\ balls\ out\ of\ six})\times\mathbb{P}(\mathrm{matching\ ball\ 1})\times\\...\times\mathbb{P}(\mathrm{matching\ ball\ 4})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 5})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 6})$$ $$=15\times\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{53}{55}\times\frac{52}{54}$$ $$=\frac{3445}{7509579}.$$ In the second calculation, it is important to include the probabilities of not matching the other balls to prevent double counting the cases when more than 4 balls are matched.

Calculating a probability for every prize and then adding them up gives the probability of winning a prize. In the old draw, the probability of winning a prize was \(0.0186\). In the new draw, it is \(0.1083\). So Camelot are correct in claiming that you are now more likely to win a prize.

But not all prizes are equal: these probabilities do not take into account the values of the prizes. To analyse the actual winnings, we're going to have to look at the expected amount of money you will win. But first, let's look at Camelot's other claim: that under the new rules you are more likely to become a millionaire.

### Probability of Winning £1,000,000

In the old draw, the only way to win a million pounds was to match all six balls. The probability of this happening was \(0.00000007151\) or \(7.151\times 10^{-8}\).

In the new lottery, a million pounds can be won either by matching all six balls or by winning the millionaire raffle. This will lead to different probabilities of winning on Wednesdays and Saturdays due to different numbers of people buying tickets. Based on expected sales of 16.5 million tickets on Saturdays and 8.5 million tickets on Wednesdays, the chances of becoming a millionaire on a Wednesday or Saturday are \(0.0000001398\) (\(1.398\times 10^{-7}\)) and \(0.00000008280\) (\(8.280\times 10^{-8}\)) respectively.

These are both higher than the probability of winning a million in the old draw, so again Camelot are correct: you are now more likely to become a millionaire...

But the new chances of becoming a millionaire are actually even higher. The probabilities given above are the chances of winning a million in a given draw. But if two balls are matched, you win a lucky dip: you could win a million in the next draw without buying another ticket. We should include this in the probability calculated above, as you are still becoming a millionaire due to the original ticket you bought.

In order to count this, let \(A_W\) and \(A_S\) be the probabilities of winning a million in a given draw (as given above) on a Wednesday or a Saturday, let \(B_W\) and \(B_S\) be the probabilities of winning a million in this draw or due to future lucky dip tickets on a Wednesday or a Saturday (the values we want to find) and let \(p\) be the probability of matching two balls. We can write $$B_W=A_W+pB_S$$ and $$B_S=A_S+pB_W$$ since the probability of winning a million is the probability of winning in this draw (\(A\)) plus the probability of winning a lucky dip ticket and winning in the next draw (\(pB\)). Substituting and rearranging, we get $$B_W=\frac{A_W+pA_S}{1-p^2}$$ and $$B_W=\frac{A_S+pA_W}{1-p^2}.$$

Using this (and the values of \(A_S\) and \(A_W\) calculated earlier) gives us probabilities of \(0.0000001493\) (\(1.493\times 10^{-7}\)) and \(0.00000009736\) (\(9.736\times 10^{-8}\)) of becoming a millionaire on a Wednesday and a Saturday respectively. These are both significantly higher than the probability of becoming a millionaire in the old draw (\(7.151\times 10^{-8}\)).

Camelot's two claims—that you are more likely to win a prize and you are more likely to become a millionaire—are both correct. It sounds like the new lottery is a great deal, but so far we have not taken into account the size of the prizes you will win and have only shown that a very rare event will become slightly less rare. Probably the best way to measure how good a lottery is is by working out the amount of money you should expect to win, so let's now look at that.

### Expected Prize Money

To find the expected prize money, we must multiply the value of each prize by the probability of winning that prize and then add them up, or, in other words,

$$\sum_\mathrm{prizes}\mathrm{value\ of\ prize}\times\mathbb{P}(\mathrm{winning\ prize}).$$
Once this has been calculated, the chance of winning due to a free lucky dip entry must be taken into account as above.

In the old draw, after buying a ticket for £2, you could expect to win 78p or 83p on a Wednesday or Saturday respectively. In the new draw, the expected winnings have changed to 58p and 50p (Wednesday and Saturday respectively). Expressed in this way, it can be seen that although the headline changes look good, the overall value for money of the lottery has significantly decreased.

Looking on the bright side, this does mean that the lottery will make even more money that it can put towards charitable causes: the lottery remains an excellent way to donate your money to worthy charities!

### Similar Posts

How to Kick a Conversion | "Uncanny" Royal Coincidence | The End of Coins of Constant Width | Dragon Curves II |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2014-04-11 13:25:00**

## Countdown Probability, pt. 2

As well as letters games, the contestants on Countdown also take part in numbers games. Six numbers are chosen from the large numbers (25,50,75,100) and small numbers (1-10, two cards for each number) and a total between 101 and 999 (inclusive) is chosen by CECIL. The contestants then use the six numbers, with multiplication, addition, subtraction and division, to get as close to the target number as possible.

The best way to win the numbers game is to get the target exactly. This got me wondering: is there a combination of numbers which allows you to get every total between 101 and 999? And which combination of large and small numbers should be picked to give the highest chance of being able to get the target?

To work this out, I got my computer to go through every possible combination of numbers, trying every combination of operations. (I had to leave this running overnight as there are a lot of combinations!)

### Getting Every Total

There are 61 combinations of numbers which allow every total to be obtained. These include the following (click to see how each total can be made):

- 5 6 8 9 10 100
- 5 6 7 8 10 100
- 4 6 7 8 9 100
- 3 6 7 8 10 100
- 3 5 7 8 9 100
- 2 5 6 8 9 100
- 2 6 7 8 9 100
- 5 6 8 9 75 100
- 3 6 8 10 75 100
- 2 6 9 10 75 100

By contrast, the following combination allows no totals between 101 and 999 to be reached:

- 1 1 2 2 3 3

The number of attainable targets for each set of numbers can be found here.

### Probability of Being Able to Reach the Target

Some combinations of numbers are more likely than others. For example, 1 2 25 50 75 100 is four times as likely as 1 1 25 50 75 100, as (ignoring re-orderings) in the first combination, there are two choices for the 1 tile and 2 tile, but in the second combination there is only one choice for each 1 tile. Different ordering of tiles can be ignored as each combination with the same number of large tiles will have the same number of orderings.

By taking into account the relative probability of each combination, the following probabilities can be found:

Number of large numbers | Probability of being able to reach target |

0 | 0.964463439 |

1 | 0.983830962 |

2 | 0.993277819 |

3 | 0.985770510 |

4 | 0.859709475 |

So, in order to maximise the probability of being able to reach the target, two large numbers should be chosen.

However, as this will mean that your opponent will also be able to reach the target, a better strategy might be to pick no large numbers or four large numbers and get closer to the target than your opponent, especially if you have practised pulling off answers like this.

Edit: Numbers corrected.

Edit: The code used to calculate the numbers in this post can now be found here.

### Similar Posts

Countdown Probability | Pointless Probability | How Much Will I Win on the New National Lottery? | "Uncanny" Royal Coincidence |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2016-07-20**

Matthew

**2016-07-20**

Matthew

**2016-07-20**

I'm a fan of the game myself (but then I'm French, so to me it's the original, "Des chiffres et des lettres"), but for the numbers game, this is pretty much irrelevant to the language and country :)

Francis Galiegue

**Add a Comment**

**2014-04-06 19:32:00**

## Countdown Probability

On Countdown, contestants have to make words from nine letters. The contestants take turns to choose how many vowels and consonants they would like. This got me wondering which was the best combination to pick in order to get a nine letter word.

Assuming the letters in countdown are still distributed like this, the probability of getting combinations of letters can be calculated. As the probability throughout the game is dependent on which letters have been picked, I have worked out the probability of getting a nine letter word on the first letters game.

### The Probability of YODELLING

YODELLING has three vowels and six consonants. There are 6 (3!) ways in which the vowels could be ordered and 720 (6!) ways in which the consonants can be ordered, although each is repeated at there are two Ls, so there are 360 distinct ways to order the consonants. The probability of each of these is:

$$\frac{21\times 13\times 13\times 6\times 3\times 5\times 4\times 8\times 1}{67\times 66\times 65\times 74\times 73\times 72\times 71\times 70\times 69}$$
So the probability of getting YODELLING is:

$$\frac{6\times 360\times 21\times 13\times 13\times 6\times 3\times 5\times 4\times 8\times 1}{67\times 66\times 65\times 74\times 73\times 72\times 71\times 70\times 69} = 0.000000575874154$$
### The Probability of Any Nine Letter Word

I got my computer to find the probability of every nine letter word and found the following probabilities:

Consonants | Vowels | Probability of nine letter word |

0 | 9 | 0 |

1 | 8 | 0 |

2 | 7 | 0 |

3 | 6 | 0.000546 |

4 | 5 | 0.019724 |

5 | 4 | 0.076895 |

6 | 3 | 0.051417 |

7 | 2 | 0.005662 |

8 | 1 | 0.000033 |

9 | 0 | 0 |

So the best way to get a nine letter word in the first letters game is to pick five consonants and four vowels.

### Similar Posts

Countdown Probability, pt. 2 | Pointless Probability | How Much Will I Win on the New National Lottery? | "Uncanny" Royal Coincidence |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**Add a Comment**

**2013-12-15 18:49:00**

## Pointless Probability

Last week, I was watching Pointless and began wondering how likely it is that a show features four new teams.

On the show, teams are given two chances to get to the final—if they are knocked out before the final round on their first appearance, then they return the following episode. In all the following, I assumed that there was an equal chance of all teams winning.

If there are four new teams on a episode, then one of these will win and not return and the other three will return. Therefore the next episode will have one new team (with probability 1). If there are three new teams on an episode: one of the new teams could win, meaning two teams return and two new teams on the next episode (with probability 3/4); or the returning team could win, meaning that there would only one new team on the next episode. These probabilities, and those for other numbers of teams are shown in the table below:

N^{o} of new teams today | |||||

N
^{o}of new teams tomorrow | 1 | 2 | 3 | 4 | |

1 | 0 | 0 | \(\frac{1}{4}\) | 1 | |

2 | 0 | \(\frac{1}{2}\) | \(\frac{3}{4}\) | 0 | |

3 | \(\frac{3}{4}\) | \(\frac{1}{2}\) | 0 | 0 | |

4 | \(\frac{1}{4}\) | 0 | 0 | 0 |

Call the probability of an episode having one, two, three or four new teams \(P_1\), \(P_2\), \(P_3\) and \(P_4\) respectively. After a few episodes, the following must be satisfied:

$$P_1=\frac{1}{4}P_3+P_4$$
$$P_2=\frac{1}{2}P_2+\frac{3}{4}P_3$$
$$P_3=\frac{3}{4}P_3+\frac{1}{2}P_4$$
$$P_4=\frac{1}{4}P_1$$
And the total probability must be one:

$$P_1+P_2+P_3+P_4=1$$
These simultaneous equations can be solved to find that:

$$P_1=\frac{4}{35}$$
$$P_2=\frac{18}{35}$$
$$P_3=\frac{12}{35}$$
$$P_4=\frac{1}{35}$$
So the probability that all the teams on an episode of Pointless are new is one in 35, meaning that once in every 35 episodes we should expect to see all new teams.

Edit: This blog answered the same question in a slightly different way before I got here.

### Similar Posts

Countdown Probability, pt. 2 | Countdown Probability | How Much Will I Win on the New National Lottery? | "Uncanny" Royal Coincidence |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**Add a Comment**

**2013-07-24 15:18:00**

## "Uncanny" Royal Coincidence

A news story on the BBC Website caught my eye this morning. It reported the following "uncanny coincidence" between a Northern Irish baby and a Royal baby:

But both new mothers share the name Catherine, the same birthday - 9 January - and now their sons also share the same birth date.

I decided to work out just how uncanny this is.

The Office for National Statistics states that 729,674 babies are born every year in the UK. This works out at 1,999 babies born each day, assuming that births are uniformly distributed, so there will be approximately 1,998 babies who share Price Nameless's birthday.

So, what is the chance of the mother of one of these babies having the same birthday as Princess Kate? To work this out I used a method similar to that which is used in the birthday "paradox", which tells us that in a group of 23 people there is a more than 50% chance of two people sharing a birthday, but that's another story.

First, we look at one of our 1,998 mothers. The chance that she shares Princess Kate's birthday is 1/365 (ignoring leap days). The chance that she does

**not**share Princess Kate's Birthday is 364/365.Next we work out the probability that none of our 1,998 mothers shares Princess Kate's birthday. As our mothers' birthdays are independent we can multiply the probabilities together to do this (this is why we are looking at the probability of

**not**sharing a birthday instead of sharing a birthday). Our probability therefore is \(\left(\frac{364}{365}\right)^{1998} = 0.00416314317\).Back to the original question, we wanted to know the probability that one of our mothers shares Princess Kate's birthday. To calculate this we do take 0.00416314317 away from 1. This gives 0.99583685682 or 99.6%.

There is a 99.6% chance that there is a resident of the UK who shares the same birthday as Princess Kate and had a child on the same day.

Uncanny.

But let's be fair. The mother in our story is also called Kate. So what are the chances of

*that*? In fact, the same method can be followed, working with the probability of having neither the same birthday or name as Princess Kate.I think it is safe to assume that this would still be considered news-worthy if our non-princess was called Katie, Cate, Cathryn, Katie-Rose or any other name which is commonly shortened to Kate, so I included a number of variations and used this fantastic tool to find the probability of a mother being called Kate. The data only goes back to 1996, but as the name is dropping in popularity, we can assume that before 1996 at least 1.5% of babies were called Kate. Disregarding males, we can estimate that 3% of mothers are called Kate.

If anyone would like the details of the rest of the calculation, please comment on this post and I will include it here. For anyone who trusts me and isn't curious, I eventually found that the probability of none of our 1,998 mothers share the same name and birthday as Princess Kate is 0.84855028964. So the probability of another Kate having a child on the same day and sharing Princess Kate's birthday is 0.15144971035 or 15.1%. Just over one in seven.

So this is as uncanny as anything else which has a probability of one in seven, such as the Royal baby being born on a Monday (uncanny!).

### Similar Posts

How Much Will I Win on the New National Lottery? | The End of Coins of Constant Width | The Importance of Estimation Error | Euro 2016 Stickers |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**Add a Comment**

Add a Comment