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This is the first Archived Forum which was active between 17th April 2007 and 1st March February 2012

 

Latest post 06-29-2009 9:42 AM by wirralsimon. 68 replies.
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  • 06-27-2009 7:18 AM

    • 9 LEE
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    Any mathematicians out there?

    Hi all,

    I remember there being some confusion about this, so i'd like a correct answer please.

    A member has 120 chances a year to win a prize on BeoWorld.  Each prize randomly picks out a number, then starts again. Remember when we had a double-winner on one draw? That proved beyond doubt that you do indeed have ten chances of winning every month. That's that sorted then.

    We have around 550 Gold Members. So i'd like to know the approximate odds of winning a prize over the 12 Month Gold Membership period.

    Is it 1 in 4.58, or am i as bad at maths as i was at school?

    Lee

     

    BeoWorld - Everything Bang & Olufsen

  • 06-27-2009 7:43 AM In reply to

    Re: Any mathematicians out there?

    I may be equally bad at statistics (lies, damn lies and bl**dy statistics)

    The way I look at it is in two ways:-

    The overall odds of winning at least one prize is 120 prizes / (550 members x 12 draws) which is 1.8% or 1 in 55

    The odds of not winning one prize in a year is 120 prizes / ((550 members - 1 unlucky member ) x 12 draws) which is 1 in 54.9

     

    10% (thats 10 in every 100)

  • 06-27-2009 7:58 AM In reply to

    Re: Any mathematicians out there?

    Sorry Gents - you're both wrong!

    Each prize draw is a completely random event, and therefore you cannot add up individual probabilities to give you a grand total of probabilities over a given time period. So the chances of winning are always the same - i.e. whatever the chances are for each prize draw (1 in 550 according to Lee's figures). The 1 in 4.58 thing is just wishful thinking!

    President, Beomaster 8000 Appreciation Society

  • 06-27-2009 8:01 AM In reply to

    Re: Any mathematicians out there?

    hi im a founder member never won so i think its zero lol

    but on the plus side got thousands of people i can talk to if i want to

    so what the hell.

    Lee keep up the good work

    malcolm

  • 06-27-2009 8:01 AM In reply to

    Re: Any mathematicians out there?

    First I'm a really bad mathematician !

    But one thing I know, my chances are 0:0 to win first prize ....

     

     

     

     

     

     

     

    I'm not gold yet ... Laughing

     

    when your Black Label begin to taste like juice just take shot or two of Absinthe and after that quench with some vodka, if you still feel juice like take beer with grappa !

  • 06-27-2009 8:38 AM In reply to

    • 9 LEE
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    Re: Any mathematicians out there?

    I don't follow the maths.

    120 individual prizes for 550 members.  If we had a one Prize Draw with 120 prizes up for grabs, and 550 'entrants' - what would the odds be? That's the same odds but spread over a year, surely?

    Lee

    BeoWorld - Everything Bang & Olufsen

  • 06-27-2009 8:40 AM In reply to

    Re: Any mathematicians out there?

    John is completely correct that you have a 1 in 550 chance of winning each prize. If one could only win one prize in each year, then the chances would increase for those left but this is not the case. However because the draw is completely random, it is likely that the prizes will distribute themselves relatively evenly and this does seem to have been the case. It would be interesting to see just how many winners there have been since the inception of the draw. And how many have won more than one.

  • 06-27-2009 9:02 AM In reply to

    • 9 LEE
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    Re: Any mathematicians out there?

    Okay - so the chances of winning a prize over the year are 1 in 550. Even though you have 120 '1 in 550' chances.

    Don't the law of averages and probabilities come in and give a helping hand to the odds?

    Lee

    BeoWorld - Everything Bang & Olufsen

  • 06-27-2009 9:06 AM In reply to

    Re: Any mathematicians out there?

    9 LEE:

    120 individual prizes for 550 members.  If we had a one Prize Draw with 120 prizes up for grabs, and 550 'entrants' - what would the odds be? That's the same odds but spread over a year, surely?

    Yes, but it just makes it harder to count - keep it as 120 draws with one prize each, please Smile

    The way I see it is, since the draws are completely independent random events,

    • the probability to win a prize in a single draw (of which we have 10 each month) P1 = 1 / 550
    • the probability to not win a prize in a single draw (ditto) P2 = 1 - (1 / 550)

    so the probability to win one and just one prize in 120 draws is then

    P = (P1 * P2^119) * 120 (because we need one win and 119 non-wins, and the win can be any of the 120)

    ~= 17.57%

    Of course, the real probability for winning at least one prize is larger, because you are not excluded once you have already won. Unless we want to get into binomials, it's easiest to calculate the probability of not winning even once and complement that, so the answer to the original question is

    P = 1 - (P2 ^ (12*10)) ~= 0.196 = 19.6%

    Likewise, the chance of me not winning a single prize in all my past 17 draws is then P2^(17*10) ~= 73%, which probably explains it all Stick out tongue

     

    -mika

  • 06-27-2009 9:46 AM In reply to

    • saf
    • Top 150 Contributor
    • Joined on 04-17-2007
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    Re: Any mathematicians out there?

    tournedos:
    Of course, the real probability for winning at least one prize is larger, because you are not excluded once you have already won.

    But let's perhaps wish this probability will actually decrease - if not, it would mean that the no. of gold members is not increasing (enough)!

    Of course, Lee could always attempt to employ a combination of some more advanced calculus with actual data on gold membership as input, while increasing the no. of draws, eg - to keep chances of winning (a) 'constant' ... Cool Big Smile

     

  • 06-27-2009 10:16 AM In reply to

    • 9 LEE
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    Re: Any mathematicians out there?

    Stop getting all tecchie on me.. just gimme a number. Geeked

    Whistle

    Lee

     

    BeoWorld - Everything Bang & Olufsen

  • 06-27-2009 10:44 AM In reply to

    Re: Any mathematicians out there?

    I always find probabilities paradoxical.

    But in this case, since no member is eliminated after winning, the odds of winning remain 1/550 for each draw (if there are no additional Gold members); and the odds of winning a prize after a year of draws, are 120/550, if we have only unique winners. The distribution of actual winners will vary, of course, with some (JandyT) getting more than others!

    If the odds aggregated each month, we'd end up with more winners than there are prizes, and that's paradoxical, but probably quite fitting for this forum!

    120/550 is excellent odds.

     

     

  • 06-27-2009 10:52 AM In reply to

    Re: Any mathematicians out there?

    19.6% ~= 1 to 5.1 Big Smile

    -mika

  • 06-27-2009 10:57 AM In reply to

    Re: Any mathematicians out there?

    tournedos:

    9 LEE:

    120 individual prizes for 550 members.  If we had a one Prize Draw with 120 prizes up for grabs, and 550 'entrants' - what would the odds be? That's the same odds but spread over a year, surely?

    Yes, but it just makes it harder to count - keep it as 120 draws with one prize each, please Smile

    The way I see it is, since the draws are completely independent random events,

    • the probability to win a prize in a single draw (of which we have 10 each month) P1 = 1 / 550
    • the probability to not win a prize in a single draw (ditto) P2 = 1 - (1 / 550)

    so the probability to win one and just one prize in 120 draws is then

    P = (P1 * P2^119) * 120 (because we need one win and 119 non-wins, and the win can be any of the 120)

    ~= 17.57%

    Of course, the real probability for winning at least one prize is larger, because you are not excluded once you have already won. Unless we want to get into binomials, it's easiest to calculate the probability of not winning even once and complement that, so the answer to the original question is

    P = 1 - (P2 ^ (12*10)) ~= 0.196 = 19.6%

    Likewise, the chance of me not winning a single prize in all my past 17 draws is then P2^(17*10) ~= 73%, which probably explains it all Stick out tongue

     

    HA !

    I know you are mathematician ! Cool

     

    when your Black Label begin to taste like juice just take shot or two of Absinthe and after that quench with some vodka, if you still feel juice like take beer with grappa !

  • 06-27-2009 11:02 AM In reply to

    Re: Any mathematicians out there?

    I'm not exactly Mr Maths, but I think it could be argued to be misleading to quote the figures across the year. As each draw is a random event that has no bearing on those that pro/preceded it, then the chances of winning are straightforward in that they're 1/550. As no one is knocked out after each one of the ten monthly draws has taken place, then that's got to be correct.

    Let me put it another way: if there were 550 draws each year (550/12 = 46) or 46 draws per month, that would give the probability of winning a prize as 550/550 = 1 - and that cannot be the case for any randomised draw that takes place. I'm happy to stand corrected. Anyone here a real mathematician???

    President, Beomaster 8000 Appreciation Society

  • 06-27-2009 11:16 AM In reply to

    Re: Any mathematicians out there?

    You were right the first time - each prize is unrelated so each prize has a 1 in 550 chance of being one by any one member. It does not matter how many times you do it, you will always have a 1 in 550 chance. The chances of every one winning one prize is no greater than one person winning every prize. The likely event is that there will be a mix of multiple, single and non winners. If you did an infinite number of draws, the winning numbers would tend to even out, probably! Laughing The only way to increase your chances is to form a syndicate! If you have more entries, you have more chances of winning - but you will only win a share of the prize!

  • 06-27-2009 11:20 AM In reply to

    Re: Any mathematicians out there?

    9 LEE:

    I don't follow the maths.

    120 individual prizes for 550 members.  If we had a one Prize Draw with 120 prizes up for grabs, and 550 'entrants' - what would the odds be? That's the same odds but spread over a year, surely?

    Lee

    The problem you are having is that you are increasing the top figure but not the bottom! For one prize, you have one chance in 550. If you add your chance here to the chance in the next prize, you have two chances but so does everyone else - hence two chances in 1100 =1 in 550!

     

  • 06-27-2009 11:24 AM In reply to

    Re: Any mathematicians out there?

    Yes. I know my suggestion seems way too simplistic, but you can't aggregate the odds relative to the number of draws, without also aggregating the number of "drawees." And that would be 550x12=6600. Which means you're back where you started, really.

    And this is why it comes down to the very prosaic 120/550. (And for each double or triple a member strikes during the year, the odds are reduced, of course, but it won't change them significantly, unless JandyT goes on a streak!)

  • 06-27-2009 11:26 AM In reply to

    • saf
    • Top 150 Contributor
    • Joined on 04-17-2007
    • Posts 458
    • Founder

    Re: Any mathematicians out there?

    What about - to check the result - a simple analogy with a local raffle draw?

    10 prize draws, 12 times a year, 550 members, all being random, makes it like a raffle draw affair where:

    there are 66 000 tickets ... 10x12x550

    120 per member

    120 prizes

    So chances of not winning anything in each prize draw are:

    (66 000-120)/66000=0,9982

    and there are 120 of draws, so probability of not winning anything in each of them is: 0,9982^120=0,8038

    That means that probability of winning at least 1 prize is: 1-0,8038=0,1962 ... 19,62pct.

    Hello tournedos Big Smile

  • 06-27-2009 11:30 AM In reply to

    Re: Any mathematicians out there?

    I'd get thrown out of the casino if I tried those complicated scribbles.

    120x100/550=21.8%

  • 06-27-2009 11:33 AM In reply to

    Re: Any mathematicians out there?

    I'm sorry to repeat myself, but it really is as simple as this:

    me:

    Of course, the real probability for winning at least one prize is larger, because you are not excluded once you have already won. Unless we want to get into binomials, it's easiest to calculate the probability of not winning even once and complement that, so the answer to the original question is

    P = 1 - (P2 ^ (12*10)) ~= 0.196 = 19.6%

    I.e the probability that you will never win, is the probability of (did not win in draw 1) AND (did not win in draw 2) AND (did not win in draw 3) AND....

    ...and the probability that was asked for is 1 - this.

    If you want to argue further, please go read the books first Stick out tongue

    EDIT: high five, saf Big Smile

    -mika

  • 06-27-2009 11:33 AM In reply to

    • Puncher
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    Re: Any mathematicians out there?

    Am I being thick here?Unsure

    Yes the chance of winning any prize is 1 in 550 but I have 120 individual discrete chances of winning a prize (at the same odds) over the course of the year. Surely the chance of winning a single prize over the course of the year must be 120/550= Lee's 1 in 4.58!

    If you did 550 trials you would have a theoretical probability of 1 (obviously experimental probability will vary).

    Generally speaking, you aren't learning much if your lips are moving.

  • 06-27-2009 11:49 AM In reply to

    Re: Any mathematicians out there?

    But just because you did not win once does not mean that you have any more chance of winning the next time. You do not have 120 chances out of 550! You have 120chances/ 550*120 . Of course every time there is a prize, you have a chance  1:550. But it can never be greater than that (except for the fact that Lee and I put the prizes back if we win!) Laughing

    The raffle ticket analogy is not accurate as as well as the winning ticket being removed, so are the 549 losing tickets for each prize. all you are doing is using big numbers to confuse the issue! Smile

  • 06-27-2009 11:49 AM In reply to

    Re: Any mathematicians out there?

    tournedos:

    I'm sorry to repeat myself, but it really is as simple as this:

    me:

    Of course, the real probability for winning at least one prize is larger, because you are not excluded once you have already won. Unless we want to get into binomials, it's easiest to calculate the probability of not winning even once and complement that, so the answer to the original question is

    P = 1 - (P2 ^ (12*10)) ~= 0.196 = 19.6%

    I.e the probability that you will never win, is the probability of (did not win in draw 1) AND (did not win in draw 2) AND (did not win in draw 3) AND....

    ...and the probability that was asked for is 1 - this.

    If you want to argue further, please go read the books first Stick out tongue

    EDIT: high five, saf Big Smile

    I want to argue with you Mika because I think you're wrong! When I studied probability for Maths A Level back in 1996 I recall that for discrete events the probability is always the same no matter how many times the discrete event is repeated, as event A has no bearing on event B and event C. You seem to be saying that this is the case, and I don't see how that is correct. I might be mistaken here as it was a long time ago, but I can't understand how you can aggregate the probabilities across the draws when each one is a stand alone event that does not have any bearing on those that went before or after it?

    What you're saying is akin to me stating that I have greater odds of winning the lottery here in the UK if I buy one ticket per week over 52 weeks of the year than I would if I bought 52 tickets for one week's draw! It just doesn't make sense!!!

    President, Beomaster 8000 Appreciation Society

  • 06-27-2009 11:57 AM In reply to

    • 9 LEE
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    Re: Any mathematicians out there?

    Is a 19.6% chance of winning something like 1 in 4.58 anyway?

    saf, Tournedos - can you calculate the odds of me understanding what the heck you are talking about Laughing

    Lee

    BeoWorld - Everything Bang & Olufsen

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