Probability of winning the irish lottery

Michael Collins · 18-11-2010 6:29pm

donegalfella wrote: »

This post has been deleted.

Exactly. With 7 being the most popular number. So if you want to reduce the odds that you'll share your prize, it might be a good idea to leave out 7. There's a least popular number too but I can't say it for obvious reasons...

Enkidu · 19-11-2010 11:29am

seamus wrote: »

Funnily enough, the nature of humans is such that the odds of two people picking the same six numbers are shockingly high.

This is also why humans are terrible random number generators. Commonly if asked to give a sequence of random numbers we tend to choose numbers that are not round numbers, to seem more random. Of course a uniformly distributed sequence of random numbers, given enough time, will have as many 30s as 37s.

Not only that, but we tend to purposely "over jump" to seem more random, like "3, 81, 21, ...", our average distance between two numbers is too large. Also we tend to alternate between high and low, again to seem more random, however this means that the correlation between the nth number and the (n+2)th is too high.

Of course even computers fail to give random sequences if you watch them for long enough. The only real random number generators are atomic decays, e.t.c.

EDIT: The actual reason this is related to choosing a limited number of options in the Lotto is because we have a terrible natural affinity for statistics of any kind. Maths education studies show that statistics takes longer to have a concrete conceptualisation than other areas of mathematics.

amacachi · 22-11-2010 1:51am

donegalfella wrote: »

This post has been deleted.

Also when extra numbers are added people generally stick to their old numbers so when a couple of 40+ numbers come out it's less likely there's a winner unless it's a QuickPick. I'd love to see some statistics since they increased the range to 45, it seems the higher numbers come out more than average.

acurno · 26-11-2010 10:24pm

I've actually had a running debate with my Dad about the picking of numbers only from 1-31 taking into account the birthdays. He stopped doing his numbers that he did for 20 years after listening to some Mathematical punter on Ray Darcy a couple of years ago. If those numbers ever come up......

Anyway, I'm still failing to grasp the logic of it, I'm not from a mathematical background so I can claim ignorance.

Surely the probability of picking six numbers to a max of 31, are the same as picking six numbers up to 42. There's as much a chance, as has been said, of 1,2,3,4,5,6 coming out as there are any other six numbers, irregardless of whether there's 36 or 42 numbers in the draw. My Dad argues that if you leave out the numbers from 31-42, that you're automatically excluding the probability of numbers 31-42 coming up. I say that the fact that you always pick 6 numbers automatically excludes you from the probability of the remaining 36 numbers coming up.

If I don't make any sense above I'll give an example...

Line 1: 3, 7, 9, 17, 21, 25
Line 2: 2, 9, 17, 21, 25, 36

Line 1 excludes numbers above 25, Line 2 doesn't. Is my Dad right in saying Line 2 has a greater probability and statistical chance of winning than Line 1?

I say no. They have the same.

Michael Collins · 26-11-2010 11:18pm

acurno wrote: »

I've actually had a running debate with my Dad about the picking of numbers only from 1-31 taking into account the birthdays. He stopped doing his numbers that he did for 20 years after listening to some Mathematical punter on Ray Darcy a couple of years ago. If those numbers ever come up......

Anyway, I'm still failing to grasp the logic of it, I'm not from a mathematical background so I can claim ignorance.

Surely the probability of picking six numbers to a max of 31, are the same as picking six numbers up to 42. There's as much a chance, as has been said, of 1,2,3,4,5,6 coming out as there are any other six numbers, irregardless of whether there's 36 or 42 numbers in the draw. My Dad argues that if you leave out the numbers from 31-42, that you're automatically excluding the probability of numbers 31-42 coming up. I say that the fact that you always pick 6 numbers automatically excludes you from the probability of the remaining 36 numbers coming up.

If I don't make any sense above I'll give an example...

Line 1: 3, 7, 9, 17, 21, 25
Line 2: 2, 9, 17, 21, 25, 36

Line 1 excludes numbers above 25, Line 2 doesn't. Is my Dad right in saying Line 2 has a greater probability and statistical chance of winning than Line 1?

I say no. They have the same.

Assuming each number is equally likely (which may or may not be valid, you'd need to see the stats here to be sure there isn't some bias in the machine and/or the organisers of the lotto), then any given set of six, specific numbers are equally probable. So then your Line 1 and Line 2 are equally likely.

I think the point that was made is that if you exclude the higher numbers and just pick numbers between 1 - 31, for example, then you are more likely to share the jackpot, should your six numbers match the ones drawn - because you're part of the large group of people who use birthdates to pick their numbers, and hence it is more likely others will have those same numbers as yourself.

There's no way I know of to increase your odds of winning apart from buying more lines of (different) numbers.

27-11-2010 2:04am

acurno wrote: »

I've actually had a running debate with my Dad about the picking of numbers only from 1-31 taking into account the birthdays. He stopped doing his numbers that he did for 20 years after listening to some Mathematical punter on Ray Darcy a couple of years ago. If those numbers ever come up......

Anyway, I'm still failing to grasp the logic of it, I'm not from a mathematical background so I can claim ignorance.

Surely the probability of picking six numbers to a max of 31, are the same as picking six numbers up to 42. There's as much a chance, as has been said, of 1,2,3,4,5,6 coming out as there are any other six numbers, irregardless of whether there's 36 or 42 numbers in the draw. My Dad argues that if you leave out the numbers from 31-42, that you're automatically excluding the probability of numbers 31-42 coming up. I say that the fact that you always pick 6 numbers automatically excludes you from the probability of the remaining 36 numbers coming up.

If I don't make any sense above I'll give an example...

Line 1: 3, 7, 9, 17, 21, 25
Line 2: 2, 9, 17, 21, 25, 36

Line 1 excludes numbers above 25, Line 2 doesn't. Is my Dad right in saying Line 2 has a greater probability and statistical chance of winning than Line 1?

I say no. They have the same.

You are right, they have the same probablility of winning.

Generally, the simplest way of explaining such a situation to a lay-person is to reduce it to its simplest terms. Compare your Dad's choice to 1,2,3,4,5,6. According to your Dad, that automatically excludes the probablility of 7-45, so you are guaranteed to win. Obviously that's not true

The only effect it can have is if you do pick the right numbers based on dates etc, you are more likely to end up sharing a jackpot

acurno · 27-11-2010 8:41pm

And that was my point exactly with him. The probabilities of sharing the jackpot are higher, but the probabilities of winning it are the same.

Thanks for backing up my logic, the brother was equally unconvinced by my argument.

Brussels Sprout · 28-11-2010 9:34pm

Lethal_Bullet wrote: »

Remember with the buy every combination thing, it cant actually be done. AFAIK people have tried in the past but the NL reserve the right to shut terminals at their will.

It was done successfully in the early days of the lotto in Ireland back when it was a weekly draw with only 36 numbers.

In a 6/36 lottery, the odds of matching all six numbers and winning the jackpot are 1 in 1,947,792. At Lotto's initial cost of £0.50 per line, all possible combinations could be purchased for £973,896. This left Lotto vulnerable to a brute force attack, which happened when the jackpot reached £1.7 million for the May 1992 bank holiday drawing. A 28-member Dublin-based syndicate, organized and headed by Polish-Irish businessman Stefan Klincewicz, had spent six months preparing by marking combinations on almost a quarter of a million paper playslips. In the days before the drawing they tried to buy up all possible combinations and thus win all possible prizes, including the jackpot.
The National Lottery tried to foil the plan by limiting the number of tickets any single machine could sell, and by turning off the terminals Klincewicz's syndicate was known to be using heavily. Despite its efforts, the syndicate did manage to buy over 1.6 million combinations, spending an estimated £820,000 on tickets. It had the winning numbers on the night—but two other winning tickets were sold, too, so the syndicate could claim only one-third of the jackpot, or £568,682. Match-5 and match-4 prizes brought the syndicate's total winnings to approximately £1,166,000, representing a profit of approximately £310,000 before expenses.
Klincewicz later appeared on the television talk show Kenny Live and capitalized on his short-lived notoriety with a self-published lottery-system book entitled Win the Lotto.

http://en.wikipedia.org/wiki/National_Lottery_(Ireland)#Lotto_6.2F36:_1988.E2.80.9392

18-09-2011 4:33pm

Has there been a decent discussion on this forum about probability and Lotto numbers? If so, can anyone point me in its direction?

I'm trying to get my head around some questions about the probability of numbers (or groups of numbers) coming up in any given draw, and it's giving me a bit of a headache.

LeixlipRed · 18-09-2011 4:37pm

New Lotto Superthread made, all questions about lotto to go in here from now on. OP if you just scroll back you can see the previous posts.

18-09-2011 5:48pm

LeixlipRed wrote: »

New Lotto Superthread made, all questions about lotto to go in here from now on. OP if you just scroll back you can see the previous posts.

Thanks for that.

Having read this thread, I don't know if my questions are relevant, because they aren't about winning the Lotto.

What I'm more interested in is how the probability of six numbers coming up is calculated, and also "reverse engineering" that calculation.

In effect, what I'm trying to work out is how many numbers would you need to pick to have (exactly or approximately) a 50-50 chance of naming all six numbers drawn. Also, how many numbers would you need to pick to have a 50-50 chance of naming none of the six numbers drawn?

A mate of mine (we're talking about the UK lottery in this case) says it's half the numbers in either case. You can't have 24.5 numbers, so therefore his contention is that the answer is 24 or 25. I don't agree with him. It just seems wrong, but I don't have the statistical wherewithal to say why.

ray giraffe · 20-09-2011 12:51am

Deleted User wrote: »

how many numbers would you need to pick to have (exactly or approximately) a 50-50 chance of naming all six numbers drawn. Also, how many numbers would you need to pick to have a 50-50 chance of naming none of the six numbers drawn?

In the English lotto you pick 6 numbers from 49.

If you pick 5 numbers then there is a 50.5% chance that you have none of the winning numbers.

If you pick 44 numbers you have a 50.5% chance of having all the winning numbers.

(Notice that 44 = 49 - 5. Try to figure out what's going on!)

In the Irish lotto you pick 6 numbers from 45.

If you pick 5 numbers then there is a 47.1% chance that you have none of the winning numbers.

If you pick 40 numbers you have a 47.1% chance of having all the winning numbers.

(Notice that 40 = 45 - 5)

Explanation for 50.5%:

If you pick 1 number there is a (43/49) chance that it is not a winning number.

If you pick 2 numbers there is a (43/49) chance the first is not winning. If the first is not a winning number, there is a (42/48) chance the second is also not winning. So the probability both are not winning is (43/49)*(42/48).

For 5 numbers,

Answer = P1*P2*P3*P4*P5
=(43/49)*(42/48)*(41/47)*(40/46)*(39/45)
=0.505
=50.5%

where
P1 = Probability 1st number is not winning,
P2= Probability 2nd number is not winning if 1st is not winning,
P3= Probability 3rd number is not winning if 1st and 2nd are not winning,
P4= Probability 4th number is not winning if 1st, 2nd and 3rd are not winning,
P5= Probability 5th number is not winning if 1st, 2nd, 3rd and 4th are not winning.

Related:
Find out about the Birthday Problem.

evolving_doors · 23-09-2011 11:01pm

Just a few warnings with syndicates to the OP (who's probably made their decision by now!),,

Once you're in yer in for life (especially if it's the same numbers every draw),, i thought about leaving my syndicate but couldn't bare thinking if the numbers came up.. I got them to reduce the amount "invested " each month though...

Also sort out what happens if someone is absent at work (are they still in or out),,,

kouffaley · 18-10-2011 10:53am

Hi guys sory i didnt know where to post this, so i hope i will get someone to help me,i am new in programming and really i dont understand nothing on it for the moment.:

Develop an application that simulates a lottery draw. The lottery will randomly draw 4 distinct numbers from the numbers 1 - 10.
The player must choose 4 distinct numbers for each set of numbers (i.e., a line of numbers.)
The player starts out with a credit of M €. The value for M is an input to the program, and you charge 50 cents per each line of numbers.
For each play, the player can choose 1 to 4 lines of numbers. If the player enters 0 as the number of lines to choose, then the program stops playing.
The player must have enough credit to pay for the number of lines chosen
At the end of the game, the program displays the amount of € left and how much the player has won or lost in €s.

Reillyman · 19-10-2011 2:58pm

kouffaley wrote: »

Hi guys sory i didnt know where to post this, so i hope i will get someone to help me,i am new in programming and really i dont understand nothing on it for the moment.:

Develop an application that simulates a lottery draw. The lottery will randomly draw 4 distinct numbers from the numbers 1 - 10.
The player must choose 4 distinct numbers for each set of numbers (i.e., a line of numbers.)
The player starts out with a credit of M €. The value for M is an input to the program, and you charge 50 cents per each line of numbers.
For each play, the player can choose 1 to 4 lines of numbers. If the player enters 0 as the number of lines to choose, then the program stops playing.
The player must have enough credit to pay for the number of lines chosen
At the end of the game, the program displays the amount of € left and how much the player has won or lost in €s.

Your lacking a good bit of information mate, how do you win a prize? How much do you win? The general formula for calculating lottery problems is the hypogeometric distribution formula. In this case the chance of getting all 4 correct would be

[latex]\frac{(^4\mathrm{C}_4) \times (^6\mathrm{C}_0)}{^(10) \mathrm{C}_4} [/latex]

c_man · 19-10-2011 3:09pm

kouffaley wrote: »

Hi guys sory i didnt know where to post this, so i hope i will get someone to help me,i am new in programming and really i dont understand nothing on it for the moment.:

Develop an application that simulates a lottery draw. The lottery will randomly draw 4 distinct numbers from the numbers 1 - 10.
The player must choose 4 distinct numbers for each set of numbers (i.e., a line of numbers.)
The player starts out with a credit of M €. The value for M is an input to the program, and you charge 50 cents per each line of numbers.
For each play, the player can choose 1 to 4 lines of numbers. If the player enters 0 as the number of lines to choose, then the program stops playing.
The player must have enough credit to pay for the number of lines chosen
At the end of the game, the program displays the amount of € left and how much the player has won or lost in €s.

Seems like homework... Break it down to start with. You have experience with random number generators*?

* yeah I know they're not truly random, damn Maths forum! :P

walshb · 23-03-2012 5:28pm

Permabear wrote: »

This post had been deleted.

What a f%$£^ng spoil sport:p

sock puppet · 28-03-2012 6:40pm

A lottery in the US has broken the record for the largest lottery jackpot. Draw is in two days and it has a nominal value of almost half a billion dollars with the odds of winning the jackpot at roughly 175,000,000:1.

alan4cult · 30-03-2012 12:50pm

The US lotto is paid out as an annuity, i.e. $38,500 over 26 years for a $1,000,000 jackpot.

If you want to take the cash lump sum, it's much less (although you'll still be loaded).

bopt · 03-04-2012 1:32pm

alan4cult wrote: »

The US lotto is paid out as an annuity, i.e. $38,500 over 26 years for a $1,000,000 jackpot.

If you want to take the cash lump sum, it's much less (although you'll still be loaded).

yes, in New York if you win 1million for example you can take 900k ish and thats it

or you can take 27k a year for 37 years or somethig like that, that way you will recieve 100% of your money and the country will make profit from investment/

shanered · 03-04-2012 2:03pm

False advertising if ya ask me, if its yours, its yours!

MathsManiac · 06-04-2012 11:17pm

alan4cult wrote: »

The US lotto is paid out as an annuity, i.e. $38,500 over 26 years for a $1,000,000 jackpot.

If you want to take the cash lump sum, it's much less (although you'll still be loaded).

There was a question about this on a Leaving Cert Maths paper last year (working out the present value of all the future payments and all that). It's question 6 here:
http://www.examinations.ie/archive/exampapers/2011/LC003ALP130EV.pdf

There was someone from Tullamore (or somewhere in the midlands) who won the New York Lotto about a year ago. The advertised Jackpot prize was $21.5 million, and she ended up witha a cheque for $7.9 million. I remember them talking on the radio trying to figure out how come she only got this much and people were ringing in with all sorts of (incorrect) reasons - I think it was on the Ray D'Arcy show but I'm not sure. Lo and behold, the next Leaving Cert Maths paper explains it all!

Michael Collins · 07-04-2012 12:48am

Ah, I like it! No harm in students learning more about personal finance. I'm assuming that present-value etc is covered in the new Project Maths text books? (I'm assuming there are new textbooks!)

MathsManiac · 07-04-2012 6:50pm

Not sure about the textbooks, but the syllabus says that the students should be able to:
"use present value when solving problems involving loan repayments and investments"

and also indicates that they should be able to handle applications of geometric series, including financial applications:
"solve problems involving finite and infinite geometric series including applications such as recurring decimals and financial applications, e.g. deriving the formula for a mortgage repayment."

GarIT · 09-04-2012 6:23pm

Has anyone worked out the expected value of the lotto including all the smaller wins?

ArmCandyBaby · 11-04-2012 7:09pm

Some poor maths made it into the letters page of the IT yesterday on this subject. I hope the Project Maths students that she refers to in the first paragraph would be able spot the mistake made here in the second. Clearly the IT couldn't.

However, these odds can also be described another way: if you used the same six numbers in the lottery for the next 8,145,060 times you played, or twice a week for the next 78,318 years, then, statistically, you should win the jackpot once.

http://www.irishtimes.com/newspaper/letters/2012/0410/1224314566042.html

Brussels Sprout · 11-04-2012 7:56pm

ArmCandyBaby wrote: »

Some poor maths made it into the letters page of the IT yesterday on this subject. I hope the Project Maths students that she refers to in the first paragraph would be able spot the mistake made here in the second. Clearly the IT couldn't.

http://www.irishtimes.com/newspaper/letters/2012/0410/1224314566042.html

Alright, I'll bite. What's the error?

(I really hope it's not something to do with splitting the jackpot or the number of weeks per year being slightly greater than 52 as that'd be pretty pedantic)

locum-motion · 11-04-2012 8:58pm

Something that annoys me:
Ads for the National Lottery's 'Millionaire Raffle' are currently stating something like

"You're six times more likely to win a million than be attacked by a bear"

I think that this 'bears' (pun intended) some closer scrutiny.

There were 180,000 tickets for the Easter Millionaire Raffle, and one Millionaire winner drawn out. So, your chance of winning is 1 in 180,000. (or 179,999 to 1 against)

If this is "six times more likely" than being attacked by a bear, then your chances of being attacked by a bear must be 1 in 1,080,000. (or 1,079,999 to 1 against)

But they hold these raffles about 4 times a year, so to truly compare like with like, then we should say that - in any given year - you have a 4 in 1,080,000 chance of being attacked by a bear. That's the same as 1 in 270,000. (or 269,999 to 1 against)

All of these raffles are held in the Republic of Ireland. The population of the Republic of Ireland is 4,588,252.

Therefore, if the claim made in the ad is true, then 17 people have been attacked by bears in Ireland in the last year.

Isn't it amazing how none of those 17 attacks made it into the newspapers?

ArmCandyBaby · 11-04-2012 11:17pm

Brussels Sprout wrote: »

Alright, I'll bite. What's the error?

(I really hope it's not something to do with splitting the jackpot or the number of weeks per year being slightly greater than 52 as that'd be pretty pedantic)

You can't just add probabilities for events that are not mutually exclusive. It's like saying that if you have a 50% chance of a coin landing on heads, then in your next 2 throws, "statistically", you'll land on heads once. The equation you use is:

$330f93fe8ccc450d0a84f92189ebca2e.png$

and for the coin example the probability is 0.5 + 0.5 - (0.5)(0.5) = 0.75. "Statistically" isn't just some fudge word to insert when the probability you've calculated equals to (or is greater than!) 1. Statistical significance is usually where the likelihood of an event being random chance is less than a certain predefined probability eg. 0.05 or 0.01. More.

If you played the lotto "for the next 8,145,060 times" you're probability of winning at least once = 1 - (1 - (1/8145060))^8145060, an arbitrary 63.212%. To have a greater than 95% chance of winning the jackpot, you need to play it ln(0.05)/ln(1 - (1/8145060)) = 24,400,418 times.

Also, as an aside, it doesn't matter if you play the same numbers or not.

Brussels Sprout · 11-04-2012 11:37pm

ArmCandyBaby wrote: »

...and for the coin example the probability is 0.5 + 0.5 - (0.5)(0.5) = 0.75.

I take it you mean the probability of getting at least 1 head there yeah, as opposed to getting exactly 1 head?

Probability of winning the irish lottery

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