mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

How much will I win on the new National Lottery?

 2015-10-08 
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:
RequirementEstimated 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:
RequirementEstimated 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 ballsFree 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!
                        
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "factor" in the box below (case sensitive):

Archive

Show me a random blog post
 2024 

Feb 2024

Zines, pt. 2

Jan 2024

Christmas (2023) is over
 2023 
▼ show ▼
 2022 
▼ show ▼
 2021 
▼ show ▼
 2020 
▼ show ▼
 2019 
▼ show ▼
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

programming sport logs golden ratio propositional calculus realhats mean people maths hyperbolic surfaces video games runge's phenomenon convergence talking maths in public dinosaurs guest posts stirling numbers pascal's triangle cambridge a gamut of games triangles royal institution pi approximation day light countdown approximation nine men's morris databet preconditioning bempp ternary tmip finite element method asteroids bubble bobble pythagoras martin gardner arithmetic news advent calendar simultaneous equations folding paper anscombe's quartet draughts harriss spiral dragon curves phd javascript manchester science festival folding tube maps game show probability manchester rhombicuboctahedron dates newcastle datasaurus dozen pi platonic solids speed gerry anderson graph theory binary chalkdust magazine misleading statistics christmas card wave scattering games national lottery crochet squares the aperiodical data machine learning noughts and crosses final fantasy edinburgh bodmas crossnumber london stickers youtube mathsteroids data visualisation error bars go electromagnetic field fractals recursion hexapawn fonts inline code graphs quadrilaterals london underground geogebra sound computational complexity menace cross stitch game of life golden spiral sobolev spaces fence posts determinants boundary element methods royal baby map projections football braiding inverse matrices craft probability books errors numbers world cup live stream reuleaux polygons plastic ratio gaussian elimination pizza cutting signorini conditions oeis ucl php puzzles matrix multiplication logic european cup turtles 24 hour maths reddit pac-man estimation coins exponential growth dataset interpolation sorting matrix of cofactors accuracy numerical analysis wool zines captain scarlet curvature weak imposition python hannah fry hats raspberry pi mathsjam geometry logo polynomials christmas matt parker matrices chess gather town standard deviation correlation flexagons tennis trigonometry palindromes weather station matrix of minors radio 4 rugby chebyshev latex frobel mathslogicbot finite group big internet math-off statistics

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2024