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)

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!

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

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 ... 😂

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

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.

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

9 LEE wrote:

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 🙂

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 😎

tournedos wrote:

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' ... 😐 😁