Bachelor Thesis

ZÁPADOCESKÁ UNIVERZITA V PLZNI

FAKULTA APLIKOVANÝCH VED

KATEDRA MATEMATIKY

BAKALÁRSKÁ PRÁCE

Loterie a testování náhodnosti tažených císel

2011 MILAN MRÁZEK

UNIVERSITY OF WEST BOHEMIA IN PILSEN

FACULTY OF APPLIED SCIENCES

DEPARTMENT OF MATHEMATICS

BACHELOR THESIS

Lotteries and Testing the Randomness of theNumbers Drawn

2011 MILAN MRÁZEK

DECLARATION

I declare that I made the bachelor thesis titled Lotteries and Testing the Random-ness of the Numbers Drawn alone and that all materials I used are mentioned in thebibliography.

In Plzenauthor’s signature

i

ACKNOWLEDGEMENT

I would like to thank my supervisor, Ing. Jan Pospíšil Ph.D., for his suggestions, piecesof advice and patient guidance throughout the time of developing this thesis.

ii

Preface

The subject of the bachelor thesis are the lotteries and testing randomness of numbersdrawn. It features derivation of probability formulas regarding the Lotto games, whichare further used for analysis of particular Lotto games within the European Union.These games are analysed and compared with respect to the probabilities of winningcategories.

The next part of the thesis includes a study of the χ2 test for testing equidistribu-tion of the sets of balls drawn. Since numbers are drawn without replacement, theχ2 statistics is not the usual χ2 distribution with N − 1 degrees of freedom, where Nis the number of imaginary cells. In this case the χ2 statistics behaves asymptoticallyas a sum of independent weighted random χ2 variables. Because of this behaviour aspecial method of computation for p-values has to be used in order to decide whetherthe tested sets of balls are drawn with equal probability. This modified χ2 test is thenapplied to the available data obtained from the lottery companies within the EuropeanUnion and the results are presented.

Thesis also includes an analysis of the discrepancies found in the article by Genestet al. (2002) published in the Journal of the Royal Statistical Society.

Plzen, May, 2011.

iii

Contents

1 Introduction to Lottery 1

2 Theory 42.1 Defining the Lottery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Probability theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Calculations 73.1 Choosing k from N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Adding bonus numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.2.1 Drawing from separate set of balls . . . . . . . . . . . . . . . . . . 93.2.2 Drawing a bonus ball after the first k numbers were drawn . . . . 103.2.3 Drawing two bonus balls after the first k numbers were drawn . 11

4 Testing the Randomness of Numbers Drawn 124.1 Randomness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Pearson’s standard goodness-of-fit test . . . . . . . . . . . . . . . . . . . 134.3 Asymptotic null distribution for subsets of size c = 1, ..., k . . . . . . . . . 144.4 Other approaches to testing uniformity . . . . . . . . . . . . . . . . . . . 15

5 Computation of p-values 165.1 P-value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.2 Method of Imhof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Discrepancies in the χ2 and the Lottery Article 20

7 Probabilities 257.1 Austria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277.2 Belgium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297.3 Bulgaria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307.4 Denmark, Estonia, Finland, Lithuania and Sweden . . . . . . . . . . . . . 317.5 Czech Republic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327.6 France . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347.7 Germany and Luxembourg . . . . . . . . . . . . . . . . . . . . . . . . . . 357.8 Greece and Cyprus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

v

7.9 Hungary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397.10 Ireland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.11 Italy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437.12 Latvia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457.13 Malta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467.14 Poland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477.15 Portugal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487.16 Romania . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497.17 Slovakia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507.18 Slovenia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517.19 Spain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.20 United Kingdom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537.21 Austria, Belgium, France, Ireland, Luxembourg, Portugal, Spain and the

United Kingdom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

A Content of the CD 56

Bibliography 58

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Chapter 1

Introduction to Lottery

Lottery is perhaps the most widely known and by far the oldest game of chance. Sincethe very beginning with slips, woods or simple drawing of lots to nowadays most pop-ular form of lottery with randomly selected balls, the basic structure, technical proce-dure and the simplicity of this game of chance has remained preserved.

The English word lottery has roots in Dutch word loterij, which is derived from theDutch noun lot meaning f ate. But the roots of lottery itself can be traced back to thesecond millennium B.C. There is a reference to a game of chance known as ’the drawingof wood’ in the early Chinese collection of poems and songs. This game of chance incontext appeared to describe drawing of lots.

The first signs of lottery come from the Han Dynasty between 205 and 187 B.C.,where ancient Keno slips were discovered. It is believed that proceeds from these lot-teries helped to finance the government projects. Also the first known European occur-rence of lottery during the Roman Empire organized by the Roman Emperor AugustusCaesar was used as a way how to raise money, in this case the proceeds went for repairsto the city of Rome. The winners were given prizes in the form of valuable articles.

The first records of lotteries with prizes in the form of money date to 1443-1449 andthey come from the Low Countries, which are the historical lands around the low-lyingdelta of three rivers, the Rhine, Scheldt and Meuse. The Dutch were the first to shift thelottery prizes to solely monetary prizes and also they were the first to base the prizeson the actual odds.

Thanks to the popularity of lotteries, they were often used as a ’painless’ form oftaxation. Official lottery in England was designed during the 16th century to raisemoney for public reparations. Followed by France in the 17th century, where the lotter-ies became one of the main resources for religious congregations in the 18th century. Incolonial America between 1744 and 1776 there were sanctioned over two hundred lot-teries and they played a huge role in the financing of both private and public ventures.

The lotteries still remain very popular these days. The most common are the stateand national number lotteries, offered and run by states whose regulations allow thistype of game of chance. The popularity of lottery is partly due to its transparency andalso simplicity. There is no opponent, no dealer, no strategy that can affect the course

1

of the game. All the components are clearly visible: the urn with the balls, the shufflingdevice, the numbers on the player’s ticket. The players only choose the numbers, buyticket and wait for the numbers to be drawn, which happens usually once or twice aweek. It is very common that the drawings are broadcasted on national TV. Besidesthe transparency or simplicity, the most important element that helps to the public’sfascination with the lottery games are the amounts of winnings.

Lotteries usually offer the highest amount of winnings among the legal gamblinggames available, which every day attracts the players that buy the tickets and dreamabout their numbers matching the winning combination. The possibility of winningis mathematically very improbable. The high prizes, especially the highest winningcategories are of course compensated with the low winning probabilities. These proba-bilities vary nationally or internationally due to different set of rules or game matrices.Each lottery matrix can be described and numerical probabilities for each matrix can befound. The probabilities are basically the entire ensemble of the lottery game since noreal strategy how to win the lottery exists. Some say that the lottery is a tax on peoplewho are bad at mathematics. The following story says something different.

In 1992 a group of 28 members organized by 43-year-old businessman Stefan Klin-cewicz tried to buy all the possible combinations and thus guarantee a jackpot win,which reached £1.7 million. At initial cost of £0.50 for one combination, covering allpossible combinations in 6/36 game matrix would cost only £973,896. So the plan wasset. The Irish National Lottery noticed an unusual high amount of sold lottery ticketsand tried to scupper this plan by limiting the number of tickets any machine could sell,and by turning off terminals, which Klincewicz’s team of ticket purchasers was usingheavily. Despite all the company’s efforts, Klincewicz’s team had the winning numberson the night. Unfortunately two other winning tickets were sold too, so the groupcould claim only one-third of the jackpot, or £568,682. But many smaller match-5 andmatch-4 prizes brought its total winnings to approximately £1,166,000. To avoid similarschemes, the National Lottery changed later that year the game matrix to a 6/39 inorder to raise the jackpot odds.1

One of the subjects of the mathematical interest connected with the lottery is theso called lottery problem. There are several articles published dealing with the lotteryproblem like developing Monte Carlo algorithm seeking the smallest possible num-ber of tickets to guarantee at least one winning ticket with m correct matches for anyt-subset for lottery (N, t, m). This particular approach can be found in paper by Braver-man and Gueron [1]. Another article on the lottery problem written by Füredi, Székelyand Zubor [2] contains proof that 100 tickets are needed to guarantee 2 correct matchesin the Hungarian Lottery. Results of this work were further used by Bougard in thearticle The lotto numbers L(n, 3, p, 2) [3]. The problem what is the minimum numberof tickets so that there is at least one ticket with particular matching combination is in-vestigated also in the article by Jans and Degraeve [4] or in the article A Lotto SystemsProblem [5] written by Russel and Griffiths.

Several strategies for Lotto games are examined such as the numbers that should

1source: www.independent.co.uk

2

be ’due’ in the article by Heinze [6] or generally the ’proven’ strategies for Lotto gamesin the book [7] by Heinze and Riedwyl. Mathematical models for various playing sys-tems are described in the book The Mathematics of Lottery: Odds, Combinations, Systemsby Barboianu [8]. Modelling the probability distribution of prize winnings is anothertopic described in article by Baker and McHale [9], which delivers a spin-off result, thatlottery players may increase the expected value of their tickets by choosing numberswhich are less popular with other lottery players.

Another researched subject connecting mathematics and lottery is testing the ran-domness of the numbers drawn. Some methods how to test the randomness of the ballsdrawn are described in the article by Haigh [10] or Johnson and Klotz [11]. Article writ-ten by Genest, Lockhart and Stephens [12] shows one way how to test the randomnessof the numbers drawn using χ2 properly as opposed to the usual approach to this testusing χ2 that can be found for example in the book by Woolfson [13]. Another approachalso cited in the article [12] is described in the article written by Joe [14].

The aim of this paper is to study and implement the test of randomness introducedby Genest, Lockhart and Stephens in [12] for testing real data obtained from the lot-tery companies within the European Union. Each lottery game is analysed and com-pared with respect to the probabilities for its winning categories. In order to analyse theprobabilities we established a probability space on which to work in Chapter 2. Chap-ter 3 describes how to derive formulas for calculating probabilities for various types ofgames. Next Chapter 4 discusses why to test the randomness of numbers and why touse the method introduced in [12] instead of the usual approach. Chapter 5 shows howto calculate p-values, which differs due the alternative approach to the testing as de-scribed in [12]. Chapter 6 contains the analysis of the discrepancies found in the article[12]. The results of the tests for the data obtained from the lottery companies with com-mentary to the p-values can be found together with the analysis of winning categoriesin Chapter 7.

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Chapter 2

Theory

2.1 Defining the Lottery

The most popular form of lottery is that which uses balls with numbers inscribed onthem and the rules for giving prizes are based on the quantity of correct numbers pre-dicted by the player that are randomly drawn. Let’s define the following parameters:

N – the total number of lottery numbers, i.e. numbers that can be drawn are 1, .., Nk – the number of balls drawn out of urn without replacement

The whole process of the number lottery game can be described as follows:Player buys a ticket before the draw by marking k predicted numbers on a printedmatrix of N numbers on an entry form. The form is scanned electronically and a ticketis printed out and given to the player as a record. Then on the established date andtime the draw is performed and the k winning numbers are determined. Both lotterycompany and player check the winning numbers with the numbers on the bought/soldtickets. If there was a ticket sold that matches some of the winning prize categories, theplayer is awarded a prize according to the category. Another option how to pick thenumbers these days is to use lottery number generator, most lottery companies providethis service. In this case the numbers are pseudorandomly generated by a computer.

Each lottery has its own awarding system and numerical parameters. However, wecan already distinguish between the various games by referring to them as Lotto k/N.For example we may now refer to a game where 6 winning numbers ranging from 1 to49 are drawn without a replacement as „Lotto 6/49”. k/N represents a certain lotterymatrix, the most common within the EU is 6/49, but there are also 5/35, 5/50, 5/90,6/42, 6/45, 6/48, 6/90, 7/39. See Chapter 7 for EU Lotto games descriptions.

4

2.2 Probability theory

What we are interested in the lottery game as in every game of chance is some descrip-tion of possible outcomes. In probability theory we call them events and in our lotterycase events are the occurrences of certain numbers or groups of numbers. Machine thatperforms the drawing generates the outcomes: combinations of k different numbersout of N numbers. We can think of these combinations as the sample space of our ex-periment, which is drawing k numbers from N numbers without replacement. Samplespace is the set of all possible outcomes. All of these events are equally possible to bedrawn which is a necessary condition for our probability model. Let’s denote the sam-ple space Ω. Such set has Ck

N elements which is all combinations of k numbers takenout of N.

Game matrix No. of elements

6/90 6226146305/90 439492687/39 153809376/49 139838166/48 122715126/42 52457865/50 21187605/35 324632

Table 2.1: Number of elements for lottery matrices within EU national lotteries in de-creasing order

We consider the field of events F as being the set of parts of the sample space, sothis set is also finite. The field of events is suitable for a function P given by the classi-cal definition of probability on a finite field of events with equally possible elementaryevents. The probability P of event E is a number expressing the chance that event Ewill occur, in other words it is a ratio between the number of outcomes favourable forE to occur and the number of equally possible outcomes.

On a finite field of events P is a function P : F → R and satisfies these axioms:

1. P(E) ≥ 0, ∀E ∈ F2. P(Ω) = 13. P(E1 ∪ E2) = P(E1) + P(E2), for any E1, E2 ∈ F that E1 ∩ E2 = ∅

With P being probability function we have built a probability space (Ω,F , P) thatensures basic probability model on which to work.

5

Taking for example matrix 5/35, Ω = (1, 2, 3, 4, 5), (1, 2, 3, 4, 6), ..., (31, 32, 33, 34, 35),that is C5

35 elements. We can build similar models for working with number of numbersdrawn up to k for predicting events such as drawing various subsets of numbers. Hav-ing the matrix 5/35 and four numbers already drawn the sample space of the probabil-ity model for the last number in this particular case would contain 31 elements (35− 4) .

6

Chapter 3

Calculations

3.1 Choosing k from N

Let us start with what most people are interested in – what are the chances of winningthe lottery? Considering that the condition for winning the highest prize is predictingcorrectly all the k drawn numbers from N numbers in the urn (we play Lotto k/N),let this be an event Ek and so probability of winning the lottery is now P(Ek). We candemonstrate the chance of winning in the following way:

Starting with an unmarked matrix on an entry form, there are N numbers we can

choose to mark as the first one and so there is probability1N

of predicting the numbercorrectly. As soon as we pick the first number there is N− 1 numbers left, which means

there is probability1

N − 1of predicting the second one correctly. Keeping in mind that

the drawn balls are not returned back, we can see that there are N(N − 1) ways howto choose the first two numbers. Therefore the probability of predicting the first two

numbers correctly would be1

N(N − 1). We can continue this way ending up picking

the last k-th number which we will mark correctly with probability1

N − k + 1. And this

way we get:

P(E) =1

N(N − 1)...(N − k + 1),

which can be also written as:

P(E) =1

N!/(N − k)!

However, this P(E) is not the probability of winning the lottery, in fact it is even

7

smaller number than our desired probability, because we are taking into account theorder of the numbers, which is not significant during the draw. It does not matter ifwe pick up and mark on the entry form the last drawn number as our first, it will stillcount as correctly predicted. Therefore we have to divide the denominator by k!, whichis the number of possible orders of k numbers in which they could be drawn. Thus theprobability of winning the lottery denoted as P(Ek) is:

P(Ek) = 1/

N!(N − k)!k!

What we have now in the denominator is the numberN!

(N − k)(k!), which is the

number of all possible combinations of k numbers drawn from N numbers. This num-ber can be also written as Ck

N or more generally as:

(Nk

)=

N!(N − k)!k!

C stands for combinations and CkN is the number of ways of picking k unordered

outcomes from N possibilities. It is also known as choice number and read "N choosek" or as a binomial coefficient or combinatorial number.

Now move on to the next possibilities. When drawing k numbers out of N thereis only way how to predict them correctly - pick exactly the one unique combination,but for subsets of k there is more than one combination of k numbers that can matchthe subset and therefore there is a higher probability. We have already established Ck

N ,which is the number of possible combinations for a group of k numbers taken out of N.As written above for predicting k numbers out of k correctly there is of course only oneunique combination:

(kk

)=

k!(k− k)!k!

=k!

0!k!= 1

But there is more than one combination for Ckn if n < k. Thus for predicting correctly

n balls of the k balls drawn there is(

kn

)ways how to do that. Moreover there are still

k − n losing balls which are drawn from N − k numbers and these can be chosen in(N − kk− n

)ways. Therefore there are in total

(kn

)(N − kk− n

)ways that gives the result of

picking correctly n balls out of the draw containing k numbers.

8

We can now write a formula how to calculate a probability of predicting n numbersmatching the k balls drawn:

P(En) =

(kn

)(N − kk− n

)/N!

(N − k)!k!.

The number in the denominator is again CkN , which are the all combinations possi-

ble. If we put n = k we get exactly the P(Ek)

P(En=k) =

(kk

)(N − kk− k

)/N!

(N − k)!k!=

1N!/(N − k)!k!

.

3.2 Adding bonus numbers

Many lotteries draw an additional bonus number, a bonus ball. There are two kindsof these numbers. Either the bonus balls are drawn from a separate urn from the mainlottery or they are drawn from the same urn after the main k numbers were drawn.

3.2.1 Drawing from separate set of balls

Let B be the number of bonus numbers and l be the number of bonus numbers drawnout of B and m number of correctly predicted drawn bonus numbers. Let Dm be theevent of predicting correctly m of l drawn bonus numbers. For calculating the proba-bility we would use the same scheme as for the main lottery thus:

P(Dm) =

(lm

)(B− ll −m

)/ (Bl

).

For a lottery game of matrix N/k and B/l matrix for bonus numbers we can calcu-late the probability of matching n numbers of the main lottery and m bonus numbersthis way:

P(Am,n) = P(En)P(Dm)

Now with the above formula we can fully analyse the probabilities of winning ina European lottery called Euro Millions. This lottery uses main game matrix 5/50 andtwo additional bonus numbers are drawn from separate board containing 10 numbers.The results can be viewed in the following Table 3.1.

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Table 3.1: Probability analysis of winning categories for Euro Millions lottery

Category Main Bonus Approximate WinningNumber Numbers Numbers Odds Probability

1. 5 2 1:76275360 0.00000001312. 5 1 1:5448240 0.00000018353. 5 0 1:3632160 0.00000027534. 4 2 1:339001 0.00000294985. 4 1 1:24214 0.00004129776. 4 0 1:16143 0.00006194667. 3 2 1:7705 0.00012979298. 3 1 1:550 0.00181710069. 2 2 1:538 0.001860364910. 3 0 1:367 0.002725650911. 1 2 1:102 0.009766915612. 2 1 1:38 0.0260451081

When we sum all the probabilities in the above Table 3.1, we obtain the probabilityof winning anything when buying one lottery ticket. The probability is 4.2%.

3.2.2 Drawing a bonus ball after the first k numbers were drawn

The other case is when a bonus ball is drawn from the same urn after the main k num-bers were drawn. Games based on this scheme are more common than the previouscase. Let’s establish formula for computing probability of predicting correctly n + bnumbers, n being the number of correctly predicted numbers from the k numbers and bis either 0 that is not predicting the bonus number correctly or 1 predicting the numbercorrectly. Let this be an event Ln+b.

Of course the probability of matching n numbers + the bonus number will be lowerthan when matching only n numbers, but also we can not forget that when calculatingprobability of winning the category, where n numbers must be matched and wherecategory n+1 also exists, we must omit the combinations including the bonus numberfor the category n, which is in fact n + 0.

Number of combinations matching n numbers out of k is(

kn

)(N − kk− n

). Now we

consider the case where after the k winning numbers there is the bonus number drawnout of the remaining N − k numbers. In case we want the number of combinations

matching the bonus ball also we have to multiply(

kn

)(N − kk− n

)by

k− nN − k

, which is the

ratio of combinations(

kn

)(N − kk− n

)that contains the bonus number and we get:

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P(Ln+1) =

(kn

)(N − kk− n

)k− nN − k(

Nk

) .

In the second case the ratio of combinations(

kn

)(N − kk− n

)that are not matching the

bonus number is:

(N − k)− (k− n)N − k

,

which can be also derived from:

1− k− nN − k

.

therefore the probability of event Ln+0 is:

P(Ln+0) =

(kn

)(N − kk− n

)(N − k)− (k− n)

N − k(Nk

) .

3.2.3 Drawing two bonus balls after the first k numbers were drawn

Viking Lotto draws two bonus numbers after the main k numbers were drawn. Thistwo balls play a role in determining the winner of the second highest category, which ismatching 5 numbers of six drawn and matching one of the two bonus balls, see page 31for the list of winning categories and the probability analysis of the Viking Lotto. Sincethere is only one ball left for matching one of the bonus balls we can use the formula

P(Ln+1) =

(kn

)(N − kk− n

)k− nN − k(

Nk

) .

We only have to double the probability of the chance of matching the bonus ball,since there are two of them in the rest of the balls. Thus the probability of winning thesecond category of the Viking Lotto can be calculated in following way:

P(L5+1) =

(65

)(421

)2(6− 5)48− 6(

486

) = 0.0000009779.

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Chapter 4

Testing the Randomness ofNumbers Drawn

4.1 Randomness

We have already described the way the lottery game works in Chapter 2. The k numbersare selected from N numbers at random from a rotating drum that ejects them individ-ually without any human influence. Drawing this way should guarantee the k numbersto be produced without any bias. Taking as an example data for Latvian Latloto game,when we look at the Picture 4.1 the first thing we notice is that there is a large range offrequencies. When we have a closer look at the Table 4.1 we may observe that the rangeis from 162 for number 4 to 216 for number 5. This might lead us to consider whetherthe selection was flawed in some way. To check whether all the numbers forming win-ning combination come up with equal probability we will use the Pearson’s standardgoodness-of-fit test [12].

Table 4.1: Observed frequency of occurrence of balls 1-35 in the five-number winningcombination of the first n=1320 draws of Latvian 5/35 Lotto spanning January 4th,1997, and December 29th, 2010

(1) 208 (8) 182 (15) 196 (22) 204 (29) 174(2) 189 (9) 181 (16) 187 (23) 183 (30) 206(3) 184 (10) 192 (17) 175 (24) 202 (31) 194(4) 162 (11) 195 (18) 185 (25) 178 (32) 193(5) 216 (12) 192 (19) 181 (26) 177 (33) 169(6) 167 (13) 195 (20) 203 (27) 192 (34) 183(7) 212 (14) 179 (21) 183 (28) 197 (35) 184

12

1 6 11 16 21 26 31 350

50

100

150

200

250

Figure 4.1: Observed frequency of occurrence of balls 1-35 in the five-number winningcombination of the first n=1320 draws of Latvian 5/35 Lotto spanning January 4th,1997, and December 29th, 2010. The blue line is at level 188.6, which is the mean.

4.2 Pearson’s standard goodness-of-fit test

For testing one number at a time, a classic approach is to determine the observed fre-quency Oi with which the numbers i = 1, . . . , N occurred among the k winning num-bers in n lottery draws, and then to attempt to compare these observed counts with theexpected counts Ei, which we can express as

Ei =nkN

.

Then we would be ready to use the traditional Pearson statistics

χ2 =N

∑i=1

(Oi − Ei)2

Ei.

In most cases the resulting statistics would be compared with with χ2 with N − 1degrees of freedom, denoted by χ2

N−1 under the null hypothesis of equiprobability. Butin our lottery case the statistics is not the usual χ2 – distribution with N − 1 degreesof freedom, because the observations, or winning numbers are not drawn with replace-ment. After the number was once selected among the k winning numbers, it is not goingback to the drum and thus can not be chosen again in the same draw; the variability ofthe standard statistics is thereby reduced.

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When testing the null hypothesis of equiprobability of subsets of winning numbersof size c = 1, . . . , k, the statistics χ2 behaves asymptotically as a sum of c independentweighted χ2 random variables. There are two ways how to approach this either tryto adapt Pearson’s χ2 statistics in such a way that its limiting distribution remains asimple χ2-distribution, which explained Joe [14], or to use the equation above and findthe weights in its asymptotic distribution according to Genest, Richard and Stephens[12], which we will use in this paper.

4.3 Asymptotic null distribution for subsets of size c = 1, ..., k

We already established the formula for calculating the expected counts for c = 1. Inthe same way we are able to test if all subsets of size c = 1, ..., k are drawn with equalprobability in n lottery draws among the k numbers chosen from the set of N balls. LetPc denote the collection of such subsets, we expand the statistics according to Genest,Richard and Stephens [12] and we may write

χ2 = ∑s∈Pc

(Os − Es)2

Es,

where Os denotes the observed count for the subset s ∈ Pc and

Es = ec ≡ n(

N − ck− c

)/ (Nk

)stands for the expected count for the same subset. The expected count for a subset

of size c may be also written as

ec = n(

kc

)/ (Nc

).

The idea is that in every draw we have c subsets among k drawn numbers and wedivide these by all possible combinations of size c taken from N numbers. We of coursemultiply this expected count n times since with more than one draw the expected countfor each subset will be n times higher. The equation above taken from Genest, Richard

and Stephens [12] is somewhat more general and works with all combinations(

Nk

),

thus for every subset of size c we have(

N − ck− c

)combinations from

(Nk

).

14

It is proved in Appendix A of Genest, Richard and Stephens [12] that the asymptoticdistribution of χ2 defined above is a linear combination of c independent χ2 randomvariables, i.e.

c

∑l=1

wlχ2vl

,

where

wl =

(k− lk− c

)(N − c− l

k− c

)/ (N − ck− c

)and

vl =

(Nl

)−(

Nl − 1

)=

(Nl

)N − 2l + 1N − l + 1

.

When k = c, we have w1 = ... = wc = 1 and in this case χ2 is asymptotically dis-

tributed as a χ2 random variable with(

Nc

)− 1 degrees of freedom. This is in fact

a drawing with replacement, because after each k-winning combination is drawn theballs are going back to the drum and they are ready for the next draw which includes

all N numbers again, thus there are(

Nc

)combinations to be drawn.

4.4 Other approaches to testing uniformity

A very popular way for testing uniformity of frequencies for the N Lotto balls is demon-strated for example by Michael M.Woolfson in the book Everyday probability and statistics[13]. The test is done on the first n = 1130 draws of the UK lottery, where N = 49 andk = 6. In fact it is the very same χ2 goodness-of-fit test as introduced in chapter 4. Thetest uses the classic formula for the Pearson Statistics

χ2 =N

∑i=1

(Oi − Ei)2

Ei,

where

Ei =nkN

.

But after the test statistics is obtained, it is compared with the χ2 table giving prob-abilities for N − 1 degrees of freedom. However as previously stated, this approachshould not be used since the numbers are drawn without replacement and because ofthat the Pearson’s statistics does not follow a simple χ2-distribution that can be foundin tables.

15

Chapter 5

Computation of p-values

5.1 P-value

One way how to decide in statistical significance testing is on the account of p-value.P-value is the probability of obtaining a test statistics that is at least as extreme as theone that was observed, assuming that the null hypothesis is true. We often reject thenull hypothesis if the p-value is less than 0.01 or 0.05, these two numbers are the mostcommon values of significance level of the test. The significance level is representedby Greek letter α. Significance level of the test determines the probability of error ofthe first kind, or type I error. This is the error of rejecting a null hypothesis when it isactually true.In our lottery case for c = k, we can compute the p-value as follows

p-value = P(χ2(N

c )−1> x),

wherex = χ2,

which is the test statistics obtained by Pearson’s formula.(

Nc

)− 1 is in this case the

number of degrees of freedom. This is the probability of getting more extreme statisticsthan the one that was observed. We can use Matlab to obtain this number by simplecommand

P_VALUE=1-chi2cdf(x,degrees_of_freedom).

This is for the case where the statistics follows a simple χ2 distribution, thus fordrawing with replacement. But for c < k, where χ2 behaves asymptotically as a sum

16

of c independent weighted χ2 random variables we can not use this method. To obtainthese p-values we will use method of Imhof [15] instead.

5.2 Method of Imhof

χ2 as defined by the Pearson’s classic formula can be also written according to [12] as

χ2 =

(Nk

)/ (N − ck− c

)Y′nYn,

where

Yn = (O− E)/√

n

is a random vector of length(

Nc

)with O a E being the vectors of Os and Es, where

s ∈ Pc, which is the collection of all subsets. Prime symbol indicates the transposeoperation.

As stated in [12] the null distribution of Yn is normal and has mean 0 and covariancematrix Σ. With the number n of draws approaching +∞, standard results imply that χ2

converges in distribution to

(Nk

)/ (N − ck− c

) (Nc )

∑l=1

λlZ2l ,

where Zl ; l = 1, ..., (Nc ), are mutually independent standard normal variables and λl

are eigenvalues of Σ. As pointed out in [12] λls take only c possible distinct non-zerovalues κl for l = 1, .., c with multiplicity vl . Consequently the asymptotic distributionof χ2 is of the form

c

∑l=1

wlχ2vl

,

with the weights wl being wl = κl/ec.

17

A formula how to calculate the probability

P(m

∑r=1

λrχ2hr> x)

is given in the article by Imhof [15] and can be found as (2.1) in the article togetherwith the proof. The formula is

P[ m

∑r=1

λrχ22vr

> x]

=p

∑k=1

1(vk − 1)!

[∂vk−1

∂λvk−1 Fk(λ, x)]

λ=λk ,

where

Fk(λ, x) = λn−1 exp−x/(2λ)m

∏l 6=k

(λk − λr)−vr .

As we can see hr = 2vr(r = 1, ..., m), n = ∑m1 vr and p is such that

λ1 > λ2 > ... > λp > 0 > λp+1 > ... > λm.

The formula is very convenient to use when all vk are small, but with large vk asin our lottery case it becomes very unstable due to the corresponding derivatives ofFk(λ, x) and large factorials. Also it requires the degrees of freedom to be even. We willtherefore use a numerically more convenient formula (3.2) from the next section of thearticle [15] instead:

P(Q > x) =12+

1π

∫ ∞

0

sinθ(u)uρ(u)

du, (5.1)

where

θ(u) =12

m

∑1[hr tan−1(λru) + δ2

r λru)(1 + λ2r u2)−1]− 1

2xu,

ρ(u) =m

∏1(1 + λ2

r u2)14 hr exp

12

m

∑1(δrλru)2

/(1 + λ2

r u2)

and

Q =m

∑r=1

λrχ2hr ;δ2

r.

18

δ2r is the non-centrality parameter. In our case

Q =c

∑l=1

wlχ2vl

,

the non-centrality parameter δ2r = 0, which simplifies the θ(u) and ρ(u) functions

in (5.1), thus we have

P(c

∑l=1

wlχ2vl> x) =

12+

1π

∫ ∞

0

sinθ(u)uρ(u)

du,

where

θ(u) =12

c

∑1[vl tan−1(wlu)]−

12

xu,

ρ(u) =c

∏1(1 + w2

l u2)14 vl .

The function uρ(u) increases monotonically towards +∞, therefore the integrationin formula (5.1) can be carried only over a small finite range 0 ≤ u ≤ U. In our casechoosing U = 1 was sufficient enough.

19

Chapter 6

Discrepancies in the χ2 and theLottery Article

During the work with the article [12] we found some discrepancies regarding the tables1 and 2 in the article [12] on pages 252 and 253. The test of randomness is demonstratedthere on the data for the Canada’s Lotto 6/49. Nevertheless, the test statistics for c = 5in Table 1 and for c = 6 in Table 2 do not correspond to the p-values stated in thetables. The p-values are calculated by the method of Imhof and the formula used is5.1. However if we put x = 1906878, which is the original value of the test statistics forc = 5 from the Table 1 in the article into the formula we obtain p-value equal to onehalf. The function

f (u) =sinθ(u)uρ(u)

,

where

θ(u) =12

c

∑1[vl tan−1(wlu)]−

12

xu,

ρ(u) =c

∏1(1 + w2

l u2)14 vl

can be seen in Figure 6.1. Similarly the same function for c = 6 from the Table 2,where x = 13983809 is portrayed in Figure 6.2 .

20

0 1 2 3 4 5 6 7 8

x 10−3

−0.35

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

u

f(u)

Figure 6.1: A graph of f (u) = sinθ(u)uρ(u) for c = 5, k = 6, x = 1906878

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 10−3

−0.35

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

u

f(u)


21

Note that both integrals of these functions are really small numbers. Moreover ac-cording to the formula

P(Q > x) =12+

1π

∫ ∞

0

sinθ(u)uρ(u)

du,

they get divided by π, which leads to the result 0.5 in both cases. Having the datafor the first n = 1798 draws of the Canada’s Lotto 6/49 available, we were able to carryout the same tests as in [12]. The Tables 6.1b and 6.2b show the results. The Tables 6.1aand 6.2a show the original values given by the authors of the article [12].

Table 6.1: Test of equidistribution for subsets of c = 1, ..., 6 balls for Canada’s Lotto6/49 using the first 1798 draws spanning June 12th, 1982, and April 14th, 2001

(a) Original table

c ec χ2 p-value

1 220.1633 54.34 0.1042 22.9337 1190.95 0.3003 1.9518 18416.4 0.4764 0.1273 211899.2 0.4795 0.0056 1906878 0.5346 0.0001 13982018 0.633

(b) Corrected table

c ec χ2 p-value

1 220.1633 54.34 0.1042 22.9337 1190.95 0.2993 1.9518 18416.4 0.4764 0.1273 211899.2 0.4795 0.0057 1906702 0.5346 0.0001 13982018 0.633

Table 6.2: Test of equidistribution for subsets of c = 1, ..., 7 balls for Canada’s Lotto6/49 using the same data of 1798 draws spanning June 12th, 1982, and April 14th, 2001

(a) Original table

c ec χ2 p-value

1 256.85714 57.64 0.0442 32.10714 1218.06 0.1643 3.41565 18487.51 0.3574 0.29701 212471.8 0.2385 0.01980 1906599 0.5446 0.00090 13983809 0.8537 0.00002 85898786 0.555

(b) Corrected table

c ec χ2 p-value

1 256.85714 57.64 0.0442 32.10714 1218.06 0.1643 3.41565 18487.51 0.3574 0.29701 212471.8 0.2385 0.01980 1906599 0.5446 0.00090 13977896 0.8537 0.00002 85898786 0.555

Surprisingly, the p-values were actually correct, but the test statistics given in thearticle were wrong. As shown previously the p-values computed based on the teststatistics given in the article were not the same as in the tables. On the other handfor the real test statistics obtained from the data for the first n = 1798 draws the p-values were matching those in the article. The following Figures 6.3 and 6.4 show the

22

integrand for the correct test statistics.

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

10

20

30

40

50

60

70

80

90

u

f(u)


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 10−3

0

500

1000

1500

2000

2500

3000

u

f(u)


23

We can also observe two rounding errors in the Table 6.1a. First is for the expectedfrequency e5. According to

ec = n(

N − ck− c

)/ (Nk

),

where c is number of balls in the subsets, n is number of draws, N is number of ballsin the lottery and k is number of drawn balls out of N. For c = 5, k = 6,N = 49,n = 1798we get:

e5 = 1798(

49− 56− 5

)/ (496

)= 79112/13983816 = 0.005657397093898,

thus

e5=0.0057.

Another different value is the p-value for subset of two balls in Table 6.1a. The resultusing the method of Imhof was

p-value=0.299390046140098,

which after correct rounding is 0.299.

24

Chapter 7

Probabilities

Tables in this section show the approximate odds as well as the probabilities of winningfor categories of particular lottery games within the European Union. The analysis isabsent only for Netherlands, where the eldest running lottery called Staadsloterij1exists,but unfortunately there were no data available in the electronic form.

Another part of the report shows the results of testing for equidistribution for theavailable data. One way how to interpret small p-values is that such an event is rare toappear. Actually there are some p-values that are less than 5% significance level, see forexample Belgium Lotto for c = 1 in Table 7.4a on on page 29 or Greek Lotto for c = 3looking at the whole history in Table 7.18 on page 38, which is significant at the 10%level. These would lead to rejection of the null hypothesis of equidistribution. But at the5% significance level we can reject very few of the tested hypothesis of equidistribution.We can observe very low p-values for the Italian Gioco Lotto in Table 7.28 on page 44,which can be possibly explained by the long history during which balls and machineswere changed several times. But taking modern lotteries for example Czech Sportka,see Table 7.10 on page 33 or German Lotto 6aus45 in Table 7.15 on page 36 the p-valuesdo not provide us any serious ground for suspecting a lack of uniformity. Worth a noticeis the comparison of p-values for the draws including the bonus numbers with thosewithout the bonus number. There is a tendency for lower p-values to occur when takinga bonus number as part of the draw. This test however does not take into account theorder of the numbers which is important when determining the prizes. Another remarkwhich is also mentioned in the article [12] is that for large subsets, for example takingthe classic matrix 6/49 and c = 6, p-value may be for example of value 0.821 as in Table7.12a on page 34, but in fact such statistics is the lowest possible for n = 4858 draws.Having 4858 cells with one count and the rest of (49

6 ) cells having zero count, the realp-values in such cases are 1. Although this is not the case for all the lotteries. Duringthe history of German Lotto there appeared two draws that had the same outcome, thatmeans one cell with 2 counts, n− 1 cells with one count and the rest were zeros. Thisfact can be observed in the Tables 7.15 and 7.14a on pages 36 and 35, p-values for c = 6are actually lower. Similar thing can be also observed in Latvia’s Lotto where there were

1www.staatsloterij.nl

25

www.staatsloterij.nl

three different outcomes that reappeared during the 1320 draws. It may be seen thenas shown in the article [12] that the asymptotic distribution of χ2 slowly deteriorates asc increases, but there is no reason to doubt its reliability for c = 1, 2, 3, 4 regarding theclassic matrices.

26

7.1 Austria

Name: Lotto "6 aus 45"Since: 1986Run by: Österreichische Lotterien Ges.m.b.H.2

Type of lottery game: 6/45No. of Bonus numbers: 1Separate board game: No

Table 7.1: Probability analysis of winning categories for Lotto "6 aus 45"

Category Numbers Approximate odds Probability

1. Match 6 of 6 1:8145060 0.00000012282. Match 5 of 6 + Bonus 1:1357510 0.00000073663. Match 5 of 6 1:35724 0.00002799244. Match 4 of 6 + Bonus 1:14290 0.00006998115. Match 4 of 6 1:772 0.00129464986. Match 3 of 6 + Bonus 1:579 0.00172619977. Match 3 of 6 1:48 0.02071439628. Match Bonus 1:8 0.1315364159

Probability of not winning anything is 84.46%. Probability of not matching any ballis 33.89% (40.06% not including the bonus number).

Table 7.2: Test of equidistribution for subsets of c balls for Austria’s Lotto 6/45 usingn = 1962 draws spanning September 7th, 1986, and December 29th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 261.600000 41.33 0.3652 29.727273 985.98 0.4083 2.765328 14141.02 0.5494 0.197523 148665.35 0.6875 0.009635 1219120.09 0.9446 0.000241 8143098.00 0.686

(b) c = 1, ..., 7

c ec χ2 p-value

1 305.200000 42.42 0.2762 41.618182 993.07 0.3313 4.839323 14243.24 0.3564 0.460888 149417.72 0.2725 0.033724 1221655.93 0.5176 0.001686 8143187.16 0.6667 0.000043 45377658.02 0.582

2www.lotterien.at

27

www.lotterien.at

7.2 Belgium

Name: LottoSince: 1978Run by: Loterie Nationale de Belgique3

Type of lottery game: 6/42No. of Bonus numbers: 1Separate board game: No

Table 7.3: Probability analysis of winning categories for Belgium Lotto


1. Match 6 of 6 1:5245786 0.00000019062. Match 5 of 6 + Bonus 1:874298 0.00000114383. Match 5 of 6 1:24980 0.00004003214. Match 4 of 6 1:555 0.00180144605. Match 3 of 6 1:37 0.0272218501


Table 7.4: Test of equidistribution for subsets of c balls for Belgium Lotto 6/42 usingn = 2617 draws spanning February 4th, 1978, and December 29th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 373.857143 50.32 0.0472 45.592334 897.30 0.1623 4.559233 11684.36 0.1284 0.350710 112217.61 0.2965 0.018458 850351.92 0.5886 0.000499 5243169.00 0.790

(b) c = 1, ..., 7

c ec χ2 p-value

1 436.166667 60.74 0.0022 63.829268 974.41 0.0143 7.978659 11640.27 0.1954 0.818324 111934.42 0.4755 0.064605 848679.43 0.8876 0.003492 5243503.03 0.7427 0.000097 26975711.00 0.639

3www.loterie-nationale.be

28

www.loterie-nationale.be

7.3 Bulgaria

Name: LotoRun by: Bulgarian Sports Totalizator4

Since: 1958Type of lottery game: 6/49No. of Bonus numbers: 0Separate board game: No

Table 7.5: Probability analysis of winning categories for Bulgarian Loto


1. Match 6 of 6 1:13983816 0.00000007152. Match 5 of 6 1:54201 0.00001844993. Match 4 of 6 1:1032 0.00096861974. Match 3 of 6 1:57 0.0176504039

Probability of not winning anything is 98.14%. Probability of not matching any ballis 43.60%.

Table 7.6: Test of equidistribution for subsets of c = 1, ..., 6 balls for Bulgarian Lotto6/49 using n = 7989 draws spanning 1958 and 2010

c ec χ2 p-value

1 978.244898 53.49 0.1202 101.900510 1229.92 0.1263 8.672384 18846.55 0.0354 0.565590 213379.73 0.0235 0.025137 1909313.31 0.1196 0.000571 13993330.84 0.036

4www.toto.bg

29

www.toto.bg

7.4 Denmark, Estonia, Finland, Lithuania and Sweden

Name: Viking LottoRun by: Danske Spil5, Eesti Loto6, Veikkaus7,Perlas8and Svenska Spel9


V IKI N ŲG

Table 7.7: Probability analysis of winning categories for Viking Lotto



Probability of not winning anything is 98.02%. Probability of not matching any ballis 30.82% (42.75% not including the bonus numbers).

9www.danskespil.dk9www.eestiloto.ee9www.veikkaus.fi9www.perlas.lt9www.svenskaspel.se

30

www.danskespil.dk

www.eestiloto.ee

www.veikkaus.fi

www.perlas.lt

www.svenskaspel.se

7.5 Czech Republic

Name: SportkaRun by: Sazka, a.s.10


Table 7.8: Probability analysis of winning categories for Czech Sportka




Table 7.9: Test of equidistribution for subsets of c balls for Czech Sportka 6/49 usingn = 5190 draws spanning January 1st, 1977, and December 29th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 635.510204 35.36 0.8052 66.198980 1069.13 0.9443 5.633956 18099.08 0.8974 0.367432 211656.71 0.6055 0.016330 1909301.24 0.1206 0.000371 13984014.75 0.485

(b) c = 1, ..., 7

c ec χ2 p-value

1 741.428571 38.45 0.6402 92.678571 1110.14 0.7503 9.859422 18078.53 0.8704 0.857341 210320.02 0.9585 0.057156 1903954.47 0.8916 0.002598 13989056.38 0.1767 0.000060 85895394.07 0.654

10www.sazka.cz

31

www.sazka.cz

Table 7.10: Test of equidistribution for subsets of c = 1, ..., 6 balls for Czech Sportka6/49 using n = 6822 draws spanning April 21st, 1957, and December 29th, 2010

c ec χ2 p-value

1 835.346939 41.81 0.5272 87.015306 1124.21 0.7253 7.405558 18237.07 0.7534 0.482971 211556.23 0.6555 0.021465 1905457.46 0.7556 0.000488 13981093.62 0.697

32

7.6 France

Name: LotoRun by: La Française des Jeux11

Since: 1976Type of lottery game: 5/4912

No. of Bonus numbers: 0Separate board game: ChanceType of Separate board game: 1/10

Table 7.11: Probability analysis of winning categories for French Loto


1. Match 5 of 5 + Chance 1:19068840 0.00000005242. Match 5 of 5 1:2118760 0.00000047203. Match 4 of 5 1:8668 0.00011537154. Match 3 of 5 1:202 0.00496097305. Match 2 of 5 1:14 0.06945362176. Match Chance 1:11 0.0925469509


Table 7.12: Test of equidistribution for subsets of c balls for French Loto 6/49 usingn = 4858 draws spanning May 19th, 1976, and October 4th, 2008

(a) c = 1, ..., 6

c ec χ2 p-value

1 594.857143 36.04 0.7802 61.964286 1124.40 0.7233 5.273556 18274.44 0.7024 0.343928 210387.31 0.9745 0.015286 1902595.88 0.9816 0.000347 13978958.00 0.821

(b) c = 1, ..., 7

c ec χ2 p-value

1 694.000000 33.20 0.8512 86.750000 1091.88 0.8363 9.228723 18188.01 0.7644 0.802498 210384.80 0.9515 0.053500 1904903.67 0.7966 0.002432 13985997.02 0.3497 0.000057 85895726.05 0.644

12www.fdj.fr12since 2008

33

www.fdj.fr

7.7 Germany and Luxembourg

Name: Lotto 6 aus 49Run by: Deutsche Lotto- und Totoblock13

Since: 1955Type of lottery game: 6/49No. of Bonus numbers: 1Separate board game: SuperzahlType of Separate board game: 1/10

Table 7.13: Probability analysis of winning categories for Lotto 6 aus 49


1. Match 6 of 6 + Superzahl 1:139838160 0.00000000722. Match 6 of 6 1:15537573 0.00000006443. Match 5 of 6 + Bonus 1:2330636 0.00000042914. Match 5 of 6 1:55491 0.00001802085. Match 4 of 6 + Bonus 1:22197 0.00004505216. Match 4 of 6 1:1083 0.00092356767. Match 3 of 6 + Bonus 1:812 0.00123142358. Match 3 of 6 1:61 0.0164189803


Table 7.14: Test of equidistribution for subsets of c balls for German Lotto 6/49 using n= 4885 draws spanning June 17th, 1956, and December 29th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 598.163265 37.39 0.7262 62.308673 1128.14 0.7013 5.302866 18290.49 0.6784 0.345839 212727.96 0.1275 0.015371 1907110.86 0.4556 0.000349 13984656.20 0.437

(b) c = 1, ..., 7

c ec χ2 p-value

1 697.857143 35.90 0.7522 87.232143 1099.98 0.8003 9.280015 18077.60 0.8704 0.806958 211477.44 0.6585 0.053797 1904564.46 0.8356 0.002445 13979064.92 0.8007 0.000057 85895698.95 0.645

13www.lotto.de

34

www.lotto.de

Table 7.15: Test of equidistribution for subsets of c = 1, ..., 6 balls for German Lotto6/49 using n = 4921 draws spanning October 9th, 1955, and December 29th, 2010

c ec χ2 p-value

1 602.571429 39.54 0.6322 62.767857 1143.57 0.6033 5.341945 18309.39 0.6494 0.348388 212707.71 0.1325 0.015484 1906807.95 0.5136 0.000352 13984578.33 0.443

The following draw (15, 25, 27, 30, 42, 48) appeared twice during the history of Ger-man Lotto. First on Saturday, December 20th, 1986, for the second time on Wednesday,June 21st , 1995.

35

7.8 Greece and Cyprus

Name: LottoRun by: OPAP14


Table 7.16: Probability analysis of winning categories for Greek Lotto




Table 7.17: Test of equidistribution for subsets of c balls for Greek Lotto 6/49 usingn = 104 draws spanning January 3rd, 2010, and December 29th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 12.734694 50.69 0.1852 1.326531 767.71 1.0003 0.112896 18328.12 0.6204 0.007363 211945.82 0.4555 0.000327 1906260.00 0.6186 0.000007 13983712.00 0.508

(b) c = 1, ..., 7

c ec χ2 p-value

1 14.857143 40.79 0.5302 1.857143 786.98 1.0003 0.197568 18225.85 0.7194 0.017180 211495.63 0.6515 0.001145 1904700.00 0.8206 0.000052 13983088.00 0.5517 0.000001 85900479.80 0.503

14www.opap.gr

36

www.opap.gr

Table 7.18: Test of equidistribution for subsets of c = 1, ..., 6 balls for Greek Lotto 6/49using n = 1844 draws spanning December 5th, 1990, and December 29th, 2010

c ec χ2 p-value

1 225.795918 45.10 0.3812 23.520408 1209.57 0.2053 2.001737 18787.66 0.0584 0.130548 212604.01 0.1645 0.005802 1905471.62 0.7526 0.000132 13981972.00 0.636

37

7.9 Hungary

Name: ÖtöslottóRun by: Szerencsejáték Rt15


Table 7.19: Probability analysis of winning categories for Hungarian Ötöslottó



Probability of not winning anything is 97.67%. Probability of not matching any ballis 74.63%.

Table 7.20: Test of equidistribution for subsets of c = 1, ..., 5 balls for Hungarian Lotto5/90 using n = 2808 draws spanning March 7th, 1976, and December 25th, 2010

c ec χ2 p-value

1 156.000000 97.51 0.1622 7.011236 3951.38 0.6703 0.239019 117715.69 0.3174 0.005495 2554617.53 0.5975 0.000064 43946459.98 0.618

15www.szerencsejatek.hu

38

www.szerencsejatek.hu

Name: HatoslottóRun by: Szerencsejáték Rt16


Table 7.21: Probability analysis of winning categories for Hungarian Hatoslottó




Table 7.22: Test of equidistribution for subsets of c balls for Hungarian Lotto 6/45 usingn = 794 draws spanning 1988, and July 22th, 2007

(a) c = 1, ..., 6

c ec χ2 p-value

1 105.866667 32.16 0.7892 12.030303 921.26 0.8343 1.119098 14145.97 0.5404 0.079936 148894.51 0.5525 0.003899 1223662.87 0.1256 0.000097 8144266.00 0.578

(b) c = 1, ..., 7

c ec χ2 p-value

1 123.511111 35.36 0.6032 16.842424 915.04 0.8143 1.958421 13841.49 0.8984 0.186516 148023.03 0.8925 0.013648 1220472.39 0.7456 0.000682 8151225.73 0.0777 0.000017 45378825.99 0.533

16www.szerencsejatek.hu

39

www.szerencsejatek.hu

Table 7.23: Test of equidistribution for subsets of c balls for Hungarian Lotto 6/45 usingn = 973 draws spanning 1988, and December 26th, 2010

c ec χ2 p-value

1 129.733333 33.97 0.7132 14.742424 939.15 0.7353 1.371388 14138.07 0.5554 0.097956 148773.76 0.6255 0.004778 1223036.42 0.2196 0.000119 8144087.00 0.595

40

7.10 Ireland

Name: LottoRun by: An Post National Lottery Company17


Table 7.24: Probability analysis of winning categories for Irish Lotto

1. Match 6 of 6 1:8145060 0.00000012282. Match 5 of 6 + Bonus 1:1357510 0.00000073663. Match 5 of 6 1:35724 0.00002799244. Match 4 of 6 + Bonus 1:14290 0.00006998115. Match 4 of 6 1:772 0.00129464986. Match 3 of 6 + Bonus 1:579 0.00172619977. Match 3 of 6 1:48 0.0207143962


17www.lotto.ie

41

www.lotto.ie

7.11 Italy

Name: SuperEnalottoRun by: Sisal Sport Italia S.p.A.18


Table 7.25: Probability analysis of winning categories for Italian SuperEnalotto




Table 7.26: Test of equidistribution for subsets of c balls for Italian SuperEnalotto 6/90using n = 902 draws spanning January 3rd, 2005, and December 30th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 60.133333 86.61 0.3992 3.378277 3482.67 1.0003 0.153558 116749.41 0.9004 0.005295 2550724.98 0.9655 0.000123 43960097.44 0.1316 0.000001 622613722.56 0.510

(b) c = 1, ..., 7

c ec χ2 p-value

1 70.155556 93.68 0.1912 4.729588 3731.12 0.9923 0.268727 116759.20 0.8714 0.012355 2550167.43 0.9675 0.000431 43934966.39 0.9156 0.000010 622608312.38 0.5697 0.00000019 7471374658.00 0.503

18www.superenalotto.com191.207274340255491e-007

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www.superenalotto.com

Name: Gioco del LottoRun by: Lottomatica S.p.A.20


Table 7.27: Probability analysis of winning categories for Italian Gioco del Lotto

Game type Numbers Approximate odds Probability

Cinquina Match 5 of 5 with 5 numbers 1:43949268 0.0000000228Quaterna Match 4 of 5 with 4 numbers 1:511038 0.0000019568Terno Match 3 of 5 with 3 numbers 1:11748 0.0000851209Ambo Match 2 of 5 with 2 numbers 1:401 0.0024968789Ambata Match 1 of 5 with 1 numbers 1:18 0.0555555556

Gioco del Lotto is different from the classic lotto games and offers different gamesaccording to the number of numbers bet.

Table 7.28: Test of equidistribution for subsets of c = 1, ..., 5 balls for Italian Lotto 5/90using n = 48422 draws spanning January 7th, 1939, and December 30th, 2010

c ec χ2 p-value

1 2690.111111 122.55 0.0042 120.903870 4225.90 0.0103 4.121723 117983.03 0.1614 0.094752 2556683.72 0.2585 0.001102 43975271.64 0.003

20www.lottomaticaitalia.it

43

www.lottomaticaitalia.it

7.12 Latvia

Name: LatlotoRun by: Latvijas Loto21

Since: 1997Type of lottery game: 5/35No. of Bonus numbers: 0Separate board game: PapildskaitlisType of Separate board game: 1/10

Table 7.29: Probability analysis of winning categories for Latvian Latloto


1. Match 5 of 5 + Papildskaitlis 1:3246320 0.00000030802. Match 5 of 5 1:360702 0.0000027724

Probability of not winning anything is 99.9997 %. Probability of not matching anyball is 36.58% (43.90% not including the bonus number).

Table 7.30: Test of equidistribution for subsets of c = 1, ..., 5 balls for Latvian Latloto5/35 using n = 1320 draws spanning January 4th, 1997, and December 29th, 2010

c ec χ2 p-value

1 188.571429 29.16 0.5142 22.184874 555.41 0.7623 2.016807 6383.43 0.8724 0.126050 52043.20 0.8185 0.004066 324787.60 0.423

During the history of Latvian Latlotto these 3 draws appeared twice:(1,2,3,21,26), on August 17th, 2005 and May 20th 2009.(1,21,23,30,34), on August 31st, 2005 and October 27th, 2010.(10,22,23,31,35), on October 18th, 2000 and January 1st, 2002.

21www.latloto.lv

44

www.latloto.lv

7.13 Malta

Name: LottoRun by: Maltco Lotteries22


Table 7.31: Probability analysis of winning categories for Malta’s Lotto

Game type Numbers Approximate odds Probability

Quaterno I Match 4 of 5 with 4 numbers 1:511038 0.0000019568Quaterno II Match 3 of 5 with 4 numbers 1:2937 0.0003404835Quaterno III Match 2 of 5 with 4 numbers 1:67 0.0149812734

Terno Match 3 of 5 with 3 numbers 1:11748 0.0000851209

Ambo Match 2 of 5 with 2 numbers 1:401 0.0024968789

Prima Match 1 of 5 with 1 numbers 1:18 0.0555555556

Malta’s lottery offers a bit different scheme of game, where players determine howmuch numbers they would like to bet: one number in Prima game, two in Ambo game,three in Terno game or four in Quaterno. But for each game numbers must match thosefive drawn.

22www.maltco.com.mt

45

www.maltco.com.mt

7.14 Poland

Name: LottoRun by: Totalizator Sportowy23


Table 7.32: Probability analysis of winning categories for Polish Lotto




Table 7.33: Test of equidistribution for subsets of c = 1, ..., 6 balls for Polish Lotto 6/49using n = 4946 draws spanning January 27th, 1957, and December 30th, 2010

c ec χ2 p-value

1 605.632653 59.00 0.0442 63.086735 1267.32 0.0433 5.369084 18704.17 0.1084 0.350158 212526.58 0.1905 0.015563 1907280.17 0.4226 0.000354 13978870.00 0.825

23www.lotto.pl

46

www.lotto.pl

7.15 Portugal

Name: TotolotoRun by: Jogos Santa Casa24

Since: 1985Type of lottery game: 5/49No. of Bonus numbers: 0Separate board game: SorteType of Separate board game: 1/13

Table 7.34: Probability analysis of winning categories for Totoloto


1. Match 5 of 5 + Sorte 1:24789492 0.00000004032. Match 5 of 5 1:2065791 0.00000048413. Match 4 of 5 1:8668 0.00011537154. Match 3 of 5 1:202 0.00496097305. Match 2 of 5 1:14 0.06945362176. Match Sorte 1:14 0.0711899623


24www.jogossantacasa.pt

47

www.jogossantacasa.pt

7.16 Romania

Name: LotoRun by: Compania Nationala Loteria Romana25


Table 7.35: Probability analysis of winning categories for Romanian Loto



Probability of not winning anything is 98.14 %. Probability of not matching any ballis 43.60%.

Table 7.36: Test of equidistribution for subsets of c = 1, ...6 balls for Romanian Loto6/49 using n = 960 draws spanning August 8th, 1993, and December 31st, 2010

c ec χ2 p-value

1 117.551020 47.30 0.2942 12.244898 1140.02 0.6273 1.042119 18285.16 0.6864 0.067964 211483.36 0.6905 0.003021 1907083.01 0.4606 0.000069 13982856.00 0.572

25www.loto.ro

48

www.loto.ro

7.17 Slovakia

Name: LotoRun by: TIPOS, a.s.26


Table 7.37: Probability analysis of winning categories for Slovak Loto




Table 7.38: Test of equidistribution for subsets of c = 1, ..., k balls for Slovak Loto 6/49using n = 2862 draws spanning September 11th, 1994, and December 29th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 350.448980 33.79 0.8572 36.505102 1144.66 0.5963 3.106817 18494.10 0.3524 0.202619 212742.59 0.1235 0.009005 1909256.12 0.1246 0.000205 13980954.00 0.706

(b) c = 1, ..., 7

c ec χ2 p-value

1 408.857143 36.91 0.7092 51.107143 1137.82 0.5913 5.436930 18329.96 0.5804 0.472777 211643.28 0.5875 0.031518 1908016.00 0.3136 0.001433 13991702.16 0.0817 0.000033 85897722.09 0.586

26www.tipos.sk

49

www.tipos.sk

7.18 Slovenia

Name: LotoRun by: Loterija Slovenije27


Table 7.39: Probability analysis of winning categories for Slovenian Loto


1. Match 7 of 7 1:15380937 0.00000006502. Match 6 of 7 + Bonus 1:2197277 0.00000045513. Match 6 of 7 1:70880 0.00001410844. Match 5 of 7 1:1477 0.00067720195. Match 4 of 7 1:89 0.01128669866. Match 3 of 7 + Bonus 1:98 0.0102285706


Table 7.40: Test of equidistribution for subsets of c = 1, ..., k balls for Slovenian Loto7/39 using n = 1264 draws spanning October 13th, 1991, and December 29th, 2010

(a) c = 1, ..., 7

c ec χ2 p-value

1 226.871795 33.04 0.4142 35.821862 779.59 0.1323 4.840792 9353.29 0.1174 0.537866 82735.91 0.1885 0.046103 577193.96 0.1426 0.002712 3261887.31 0.6037 0.000082 15379673.00 0.590

(b) c = 1, ..., 8

c ec χ2 p-value

1 259.282051 33.54 0.3362 47.762483 768.18 0.1663 7.745268 9456.48 0.0664 1.075732 83737.68 0.0175 0.122941 579838.33 0.0076 0.010848 3268161.28 0.0507 0.000657 15386035.58 0.2028 0.000021 61522483.90 0.545

27www.loterija.si

50

www.loterija.si

7.19 Spain

Name: Lotto 6/49Run by: Loteria de Catalunya28


Table 7.41: Probability analysis of winning categories for Spanish Lotto 6/49




28www.loteriadecatalunya.cat

51

www.loteriadecatalunya.cat

7.20 United Kingdom

Name: Lotto29

Run by: Camelot Group PlcSince: 1994Type of lottery game: 6/49No. of Bonus numbers: 1Separate board game: No

Table 7.42: Probability analysis of winning categories for English Lotto




Table 7.43: Test of equidistribution for subsets of c = 1, ..., k balls for UK’s Lotto 6/49using n = 1567 draws spanning November 19th, 1994, and December 29th, 2010

(a) c = 1, ..., 6

c ec χ2 p-value

1 191.877551 52.22 0.1472 19.987245 1172.89 0.4093 1.701042 18161.42 0.8424 0.110938 210942.26 0.8875 0.004931 1904377.77 0.8886 0.000112 13982249.00 0.616

(b) c = 1, ..., 7

c ec χ2 p-value

1 223.857143 54.30 0.0842 27.982143 1202.93 0.2233 2.976824 18402.54 0.4754 0.258854 212089.01 0.3885 0.017257 1907007.17 0.4756 0.000784 13983045.79 0.5547 0.000018 85899016.86 0.548

29www.national-lottery.co.uk

52

www.national-lottery.co.uk

7.21 Austria, Belgium, France, Ireland, Luxembourg, Portugal,Spain and the United Kingdom

Name: Euro Millions30

Since: 2004Type of lottery game: 5/50No. of Bonus numbers: 0Separate board game: Lucky StarType of Separate board game: 2/10

Table 7.44: Probability analysis of winning categories for Euro Millions


1. Match 5 of 5 + 2 Lucky Star 1:76275360 0.00000001312. Match 5 of 5 + 1 Lucky Star 1:5448240 0.00000018353. Match 5 of 5 1:3632160 0.00000027534. Match 4 of 5 + 2 Lucky Star 1:339002 0.00000294985. Match 4 of 5 + 1 Lucky Star 1:24214 0.00004129776. Match 4 of 5 1:16143 0.00006194667. Match 3 of 5 + 2 Lucky Star 1:7705 0.00012979298. Match 3 of 5 + 1 Lucky Star 1:550 0.00181710069. Match 2 of 5 + 2 Lucky Star 1:538 0.0018603649

10. Match 3 of 5 1:367 0.002725650911. Match 1 of 5 + 2 Lucky Star 1:102 0.009766915612. Match 2 of 5 + 1 Lucky Star 1:38 0.0260451081

Probability of not winning anything is 95.75%. Probability of not matching any ballis 35.88% (57.66% not including the bonus numbers).

30www.euro-millions.com

53

www.euro-millions.com

Table 7.45: Test of equidistribution for subsets of c = 1, ..., 5 balls for Euro Millions 5/50using n = 326 draws spanning February 7th, 1994, and December 31st, 2010

c ec χ2 p-value

1 32.600000 42.82 0.5702 2.661224 942.20 1.0003 0.166327 19334.11 0.8764 0.007078 229517.73 0.8645 0.000154 2118434.00 0.563

54

Appendix A

Content of the CD

CD contains file with the thesis in folder THESIS, obtained data from the lottery com-panies in folder DATA. In folder TEST can be found a test of randomness configuratedfor the Austrian lottery using the Method of Imhof for computing p-values. Directorystructure of the CD:

• DATA• TEST• THESIS

55

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[9] Baker, R. D. and McHale, I. G. Modelling the probability distribution of prizewinnings in the uk national lottery: consequences of conscious selection. Journalof the Royal Statistical Society: Series A (Statistics in Society), 2009, vol. 172, no. 4,pp. 813–834.

[10] Haigh, J. The statistics of the national lottery. Journal of the Royal Statistical Society:Series A (Statistics in Society), 1997, vol. 160, no. 2, pp. 187–206.

[11] Johnson, R. and Klotz, J. Estimating hot numbers and testing uniformity for thelottery. Journal of the American Statistical Association, 1993, vol. 88, no. 422, pp. 662–668.

[12] Genest, C., Lockhart, R. A., and Stephens, M. A. χ2 and the lottery. Journal of theRoyal Statistical Society: Series D (The Statistician), 2002, vol. 51, no. 2, pp. 243–257.

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[13] Woolfson, M. M. Everyday Probability and Statistics. 1st edition. Imperial CollegePress, 2008. ISBN 1-84816-032-1.

[14] Joe, H. Tests of uniformity for sets of lotto numbers. Statistics & Probability Letters,1993, vol. 16, no. 3, pp. 181–188.

[15] Imhof, J. P. Computing the distribution of quadratic forms in normal variables.Biometrika, 1961, vol. 48, no. 3-4, pp. 419–426.

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Bachelor Thesis

Documents

Transcript of Bachelor Thesis