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Probability of winning the irish lottery

  • 22-10-2010 11:04am
    #1
    Closed Accounts Posts: 353 ✭✭


    Someone in my office is setting up a syndicate for the Irish lottery, she said the jackpot is 16 million euros.

    What's the probability of winning the lottery ?

    For an investment of 5 euros in tickets (the amount we are required to invest), what is the expected value of the return ?

    Is it more or less than 5 euros

    I know it depends on the probability of winning and the total number of tickets bought........but can someone do the maths for me !


    I'll join the syndicate if it makes mathematical sense.


«13

Comments

  • Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    The odds of winning the Irish lotto are 1 in 8,145,060. i.e 1 divided by 45 choose 6. But I know nothing about how syndicates work or anything like that as I've played the lotto once in my life so maybe another poster can help you


  • Closed Accounts Posts: 39,022 ✭✭✭✭Permabear


    This post has been deleted.


  • Registered Users, Registered Users 2 Posts: 68,317 ✭✭✭✭seamus


    The odd are 1 in 8,145,060.

    This means that if everyone in Ireland bought a single ticket tonight with different numbers on them, there is still roughly a 50% chance that no-one will win.

    In the syndicate, if there are nine of you paying a fiver each, then you can buy 30 tickets. This improves the chances of winning to a paltry 1 in 271,502.

    I'm not sure what you mean by "expected return". You should expect to get zero. The "possible return" if there are nine in the syndicate is 350,000%.

    Interestingly, the odds of winning "something" are about 1 in 40. In your syndicate of nine, this ups the odd significantly of winning "something" to about 75%. However this "something" is a small amount - usually about a fiver, divided by nine people, represents a return of 11% of your original sum. Depending on what way you want to look at it, your likely ROI is about -95%, that is, you stand to lose about 95% of your original investment, though you may lose all of it.


  • Closed Accounts Posts: 39,022 ✭✭✭✭Permabear


    This post has been deleted.


  • Registered Users, Registered Users 2 Posts: 2,793 ✭✭✭oeb


    This is based on the UK lotto, but an interesting little tool.

    http://lottery.reevo.com/


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  • Closed Accounts Posts: 6,943 ✭✭✭abouttobebanned


    Token post:

    But someone has to win!!!!


  • Registered Users, Registered Users 2 Posts: 68,317 ✭✭✭✭seamus


    Pfft, let's not complicate it :)

    I should add that syndicates aren't an entirely bad idea. In the example I give above, playing 3 lines on your own your chances of winning €16 million all to yourself is roughly 1 in 2.7 million. On the other hand, 10 people playing 30 lines has ten times better odds of winning, but the amount you win is only €1.6m each. Obviously here the smart choice is the same amount of investment, with a better chance of return, given the levels of money involved.


  • Registered Users, Registered Users 2 Posts: 10,807 ✭✭✭✭28064212


    Those are just the odds of winning the top prize. It's much, much more complicated than that, especially when you take in the Lotto Plus draws (€2 a line). Here are the results of last Wednesday's draws, with the number of winners and prize values:
    Winners|Numbers Matched|Prize
    Lotto
    0|Jackpot Winner|€14,764,690
    4|Match 5 + Bonus|€25,000
    86|Match 5||€1,761
    225|Match 4 + Bonus|€168
    4,433|Match 4|€53
    5,196|Match 3 + Bonus|€30
    71,881|Match 3|€5 scratchcard
    LottoPlus2
    0|Top Prize Winner|€350,000
    1|Match 5 + Bonus|€3,500
    93|Match 5|€350
    180|Match 4 + Bonus|€35
    3,772|Match 4|€15
    5,341|Match 3 + Bonus|€9
    62,453|Match 3|€3 scratchcard
    LottoPlus2
    1|Top Prize Winner|€250,000
    2|Match 5 + Bonus|€2,500
    75|Match 5|€250
    209|Match 4 + Bonus|€25
    3,716|Match 4|€10
    5,215|Match 3 + Bonus|€5
    60,388|Match 3|€1 scratchcard

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  • Registered Users, Registered Users 2 Posts: 399 ✭✭Dermot2468


    Interestingly the odds of dying (on average) any hour in Ireland are 1 in 1.1 million (Death rate 4 per hour). So therefore If you do the lottery more than ~ 8 hours before the draw you have a higher chance of being dead before the draw than winning.


  • Closed Accounts Posts: 353 ✭✭MungoMan


    Thanks everyone , 8 million to 1 against..........

    If you bought every combination of numbers, it would cost 12 million

    The jackpost is 16 million, therefore buying a ticket tonight could make mathematical sense

    The expected return (probably the wrong term) on a 1.50 investment is 2 euros (that would be true if you were the only person who bought a ticket)........it's actually slightly higher than 2 euros because of the match 5s and match 4s.

    But of course there is a risk that someone else might win too, and the jackpot might be divided among several people


    I think buying a ticket for a 16 million jackpot probably makes sense, but it depends on how many other tickets are competing for the jackpot.


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  • Closed Accounts Posts: 353 ✭✭MungoMan


    Dermot2468 wrote: »
    Interestingly the odds of dying (on average) any hour in Ireland are 1 in 1.1 million (Death rate 4 per hour). So therefore If you do the lottery more than ~ 8 hours before the draw you have a higher chance of being dead before the draw than winning.

    That's definitely true if you are aged 80, but if you are a 17 year old who doesn't drive or who isn't on a rock climbing holiday, I think the odds of winning the lottery are higher than being dead in the 8 hours before the lottery.


  • Closed Accounts Posts: 39,022 ✭✭✭✭Permabear


    This post has been deleted.


  • Closed Accounts Posts: 353 ✭✭MungoMan


    I think I have the maths

    12 million is the figure you'd want to win for overcoming odds of 8'000'000 to one if you buy 1 ticket. (for it to be a neutral bet)


    But if you do overcome the odds of 8'000'000 to 1, can you expect to win 12'000'000 ?

    The answer is no !

    Because there is a probability that someone else will also win....

    The probability of someone else winnng is

    number of tickets sold / 8'000'000


    Let's say there are 3'000'000 tickets sold, there is a 36% chance that if you win, you'd have to share it with someone else




    Therefore the prize for overcoming the 8'000'000 to 1 odds is reduced from 16'000'000 to say (16'000'000 minus 0.36*16'000'000) = 10.3 million


    If you overcome odds of 8 million to 1, you can expect to get 10.3 million in return, when you'd need to get 12 million to 1 for it to be a fair value bet...


    I'm not joining the syndicate ! I'm sticking to poker


  • Registered Users, Registered Users 2 Posts: 13,031 ✭✭✭✭bnt


    Plus, of course, the fact that no Lottery pays out all its winnings. Not even close: they're raising money for various causes in the process. From the National Lottery website:
    The National Lottery's mission is to operate a world class lottery raising funds for good causes on behalf of the Government and we are proud to have delivered on this again in 2009, raising €263.5 million.
    This brings the total amount raised since the establishment of the National Lottery to over €3.4 billion
    That's €3.4 billion that went to all kinds of things that would otherwise be funded by taxation, and didn't go to lottery winners. In other words: lotteries are a tax on those who don't know basic statistics. :p

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  • Registered Users, Registered Users 2 Posts: 36,169 ✭✭✭✭ED E


    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.


  • Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    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.

    The logistics would be horrendous - I can't remember if there's 8 or 10 boxes on a slip; you'd need to either have (roughly) 1,000,000 or 800,000 slips filled out (this could be done with a printer and software).
    The longest gap between draws is roughly 3 days 23 hours 40 minutes (iirc the terminals close at 7:45 so you would have approx from 8:05 on the Sat night to 7:45 on the Weds night). That's 344400 seconds. At a conservative estimate of 5 seconds per slip (to insert the slip into the reader, read, print the lotto ticket and eject the slip), that gives you 68880 print cycles (assuming you can find a 24-hour shop *and* the shopkeeper is willing to work for almost 4 days straight :)). To print off all 1m (800k) slips would need 15 (12) people working continuously for 4 days. There is no way that would go unnoticed by the either the general public or the NL themselves. As being somewhat knowledgeable in the field of IT, I've no doubt the NL would have algorithms to spot unusual buying patterns coming up to a draw.

    To top it all off, if the top prize was shared, you'd be buggered. The pools for the match 5 etc is proportional to the main prize fund, so you'd just force down the individual prize amount for those ones. The only way to make money would be to play would have been when there was a (relatively large) fixed prize for the lesser results (match 4 etc). When the Scruffy Murphy syndicate attempted this in the early 90's, it was around a bank holiday when there was £100 for the match 4 (iirc). Shortly after this attempt, the draw went bi-weekly :)


  • Moderators, Science, Health & Environment Moderators Posts: 1,851 Mod ✭✭✭✭Michael Collins


    Yakuza wrote: »
    When the Scruffy Murphy syndicate attempted this in the early 90's, it was around a bank holiday when there was £100 for the match 4 (iirc). Shortly after this attempt, the draw went bi-weekly :)

    Yeah, and they added several new numbers - thus vastly increasing the number of possible combinations and making any brute force attempts effectively impossible.

    They marketed this on TV as:

    "you know the way you couldn't pick numbers 37, 38 and 39 - well the nice people at the National Lottery have now added these for your enjoyment..."

    and the majority of the public went

    "Ah sure that's great!"


  • Registered Users, Registered Users 2 Posts: 4,586 ✭✭✭sock puppet


    seamus wrote: »
    The odd are 1 in 8,145,060.

    This means that if everyone in Ireland bought a single ticket tonight with different numbers on them, there is still roughly a 50% chance that no-one will win.

    That's a very unrealistic assumption though. In a real-life scenario would you not be practically guarenteed a winner if that number of people played?


  • Registered Users, Registered Users 2 Posts: 1,015 ✭✭✭rccaulfield


    This post has been deleted.

    They should have you as guest speaker for all the lotto winners bash we've had through the years!;)


  • Closed Accounts Posts: 834 ✭✭✭Reillyman


    That's a very unrealistic assumption though. In a real-life scenario would you not be practically guarenteed a winner if that number of people played?

    ???

    Population ~ 4,500,000

    [latex]\frac{1}{8145060} \times 4500000 = 55.25[/latex]%


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  • Moderators, Science, Health & Environment Moderators Posts: 1,851 Mod ✭✭✭✭Michael Collins


    Reillyman wrote: »
    ???

    Population ~ 4,500,000

    [latex]\frac{1}{8145060} \times 4500000 = 55.25[/latex]%

    Maybe sock puppet's refering to the fact that you have to buy at least two lines at a time?


  • Registered Users, Registered Users 2 Posts: 4,586 ✭✭✭sock puppet


    Reillyman wrote: »
    ???

    Population ~ 4,500,000

    [latex]\frac{1}{8145060} \times 4500000 = 55.25[/latex]%

    But that has everyone picking different numbers though which is why I asked the question. If it was a real-life scenario where it's possible for more than person to pick each combination would the probability of at least one person winning the jackpot not be very high?


    edit: I'm not sure whether I'm thinking about it right or not. Is the logic the same as that for the birthday problem or do I have it wrong?


  • Moderators, Science, Health & Environment Moderators Posts: 1,851 Mod ✭✭✭✭Michael Collins


    But that has everyone picking different numbers though which is why I asked the question. If it was a real-life scenario where it's possible for more than person to pick each combination would the probability of at least one person winning the jackpot not be very high?

    Nope, the probability of at least one person winning is maximized if we all pick different numbers. Since this doesn't happen in actuality, the probability of at least one person winning the jackpot is less than

    [latex] N_{players} \cdot \frac{1}{8145060} [/latex]

    This assumes that the numbers are random and there's no fiddling going on of course ;)


  • Registered Users, Registered Users 2 Posts: 2,122 ✭✭✭c montgomery


    But that has everyone picking different numbers though which is why I asked the question. If it was a real-life scenario where it's possible for more than person to pick each combination would the probability of at least one person winning the jackpot not be very high?


    edit: I'm not sure whether I'm thinking about it right or not. Is the logic the same as that for the birthday problem or do I have it wrong?

    The opposite would be true. If people pick the same numbers thats less individual combinations resulting in a lower chance of the lottery being won.

    In fact i know 2 people who when they do the lotto pick the numbers 1,2,3,4,5,6. They reckon they have as good a chance as any of coming up:)
    They dont know each other so i never mentioned it to them.


  • Closed Accounts Posts: 23,316 ✭✭✭✭amacachi


    The chances of winning the Lotto are just like everything else, 50/50, you either win it or you don't. :pac:


  • Registered Users, Registered Users 2 Posts: 2,122 ✭✭✭c montgomery


    amacachi wrote: »
    The chances of winning the Lotto are just like everything else, 50/50, you either win it or you don't. :pac:


    In that case ill bet you 5 euro every draw that you wont win. Ill even give ya odds of 100/1:p:p:p


  • Registered Users, Registered Users 2 Posts: 4,586 ✭✭✭sock puppet


    Nope, the probability of at least one person winning is maximized if we all pick different numbers. Since this doesn't happen in actuality, the probability of at least one person winning the jackpot is less than

    [latex] N_{players} \cdot \frac{1}{8145060} [/latex]

    This assumes that the numbers are random and there's no fiddling going on of course ;)

    Yeah I see how I went wrong now. I was thinking of the probability of any 2 players picking the same number. Pretty big mistake:pac:


  • Registered Users Posts: 332 ✭✭Paddy The Pirate


    i never thought there were such odds against you winning the lotto!! :eek:


  • Registered Users, Registered Users 2 Posts: 68,317 ✭✭✭✭seamus


    Yeah I see how I went wrong now. I was thinking of the probability of any 2 players picking the same number. Pretty big mistake:pac:
    Funnily enough, the nature of humans is such that the odds of two people picking the same six numbers are shockingly high.
    We tend to have an inbuilt penchant for patterns, so if you pick a common sequence like 1-6 or 2,4,6,8,10,12, or anything remotely similar you are guaranteed to only win a very tiny amount of the jackpot as a surprisingly large amount of people pick these sequences. Presumably because they think that nobody else is clever enough to do that.

    I'm almost sure I've read that this happened somewhere - the numbers 1 - 6 came out in the lottery draw and about 50 people turned up, quite disappointed with their share of the jackpot :D

    If you want to ensure that you have the best chance of being the sole winner of a jackpot, use a quickpick. The odds of your quickpick matching someone else's ticket are exactly the same as the odds of winning the lotto.


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  • Closed Accounts Posts: 39,022 ✭✭✭✭Permabear


    This post has been deleted.


  • Moderators, Science, Health & Environment Moderators Posts: 1,851 Mod ✭✭✭✭Michael Collins


    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...


  • Registered Users, Registered Users 2 Posts: 1,005 ✭✭✭Enkidu


    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.


  • Closed Accounts Posts: 23,316 ✭✭✭✭amacachi


    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.


  • Registered Users Posts: 236 ✭✭acurno


    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.


  • Moderators, Science, Health & Environment Moderators Posts: 1,851 Mod ✭✭✭✭Michael Collins


    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.


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  • Registered Users, Registered Users 2 Posts: 10,807 ✭✭✭✭28064212


    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

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  • Registered Users Posts: 236 ✭✭acurno


    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.


  • Registered Users, Registered Users 2 Posts: 7,476 ✭✭✭Brussels Sprout


    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


  • Posts: 0 [Deleted User]


    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.

    :o


  • Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    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.


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  • Posts: 0 [Deleted User]


    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.

    :confused:


  • Registered Users, Registered Users 2 Posts: 338 ✭✭ray giraffe


    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.


  • Closed Accounts Posts: 11,812 ✭✭✭✭evolving_doors


    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),,,


  • Registered Users Posts: 43 kouffaley


    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.


  • Closed Accounts Posts: 834 ✭✭✭Reillyman


    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]


  • Closed Accounts Posts: 4,436 ✭✭✭c_man


    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


  • Registered Users, Registered Users 2 Posts: 56,619 ✭✭✭✭walshb


    Permabear wrote: »
    This post had been deleted.

    What a f%$£^ng spoil sport:p


  • Registered Users, Registered Users 2 Posts: 4,586 ✭✭✭sock puppet


    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.


  • Registered Users, Registered Users 2 Posts: 1,583 ✭✭✭alan4cult


    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).


  • Closed Accounts Posts: 1 bopt


    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/


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