Premium Bonds, Windfall Bond, etc.

LetMeJoin

It's been a long time since I did Statistics GSCE, but I don't think the 1 in 60 claim of winning the Windfall Bond is correct. 21 prizes in 15,000, is roughly 1 in 714 chance of a prize (with one bond), agreed. But they say that over 12 months this reduces the odds/ probability to 1 in 60, don't agree. This would only work with 12 entries into the same draw. Each month is a separate/ fresh draw, so it's still 1 in 714 , regardless of how many months in a row you do the draw. With one bond, the probability is 1 in 714 months i.e. 1 win in nearly 60 years. Could be Draw 1 or Draw 714, depending upon your luck. 
Can you confirm whether I'm right?

eskbanker

LetMeJoin said:

It's been a long time since I did Statistics GSCE, but I don't think the 1 in 60 claim of winning the Windfall Bond is correct. 21 prizes in 15,000, is roughly 1 in 714 chance of a prize (with one bond), agreed. But they say that over 12 months this reduces the odds/ probability to 1 in 60, don't agree. This would only work with 12 entries into the same draw. Each month is a separate/ fresh draw, so it's still 1 in 714 , regardless of how many months in a row you do the draw. With one bond, the probability is 1 in 714 months i.e. 1 win in nearly 60 years. Could be Draw 1 or Draw 714, depending upon your luck. 
Can you confirm whether I'm right?

No - if you have a 1 in 714 chance of winning in one draw then you have (very close to) a 12 in 714, i.e. 1 in 60, chance of winning over 12 draws.

masonic

LetMeJoin said:

It's been a long time since I did Statistics GSCE, but I don't think the 1 in 60 claim of winning the Windfall Bond is correct. 21 prizes in 15,000, is roughly 1 in 714 chance of a prize (with one bond), agreed. But they say that over 12 months this reduces the odds/ probability to 1 in 60, don't agree. This would only work with 12 entries into the same draw. Each month is a separate/ fresh draw, so it's still 1 in 714 , regardless of how many months in a row you do the draw. With one bond, the probability is 1 in 714 months i.e. 1 win in nearly 60 years. Could be Draw 1 or Draw 714, depending upon your luck. 
Can you confirm whether I'm right?

I don't think so. The numbers are derived from 21 prizes per month from 15,000 tickets. The only difference between numbers drawn in a single month and numbers drawn from one month to the next is that once drawn, a ticket is not included for the rest of that month's draw (so the odds of each selection within one month's draw are slightly better than 1/15,000 per ticket on average). It's easiest to assume replacement for all selections, since this is what happens from one month to the next and will slightly underestimate the actual odds. So then we can expand to 252 prizes per year with replacement of each winning ticket (so it could win any number of times). In that scenario, the probability of a ticket not winning a prize is (14999/15000)^252 = 98.334%, so the probability of it winning a prize is 1.666% or 1 in 60. Since winning tickets are excluded from some subsequent draws, non-winning tickets have a very slightly greater chance of winning their first prize, but the 1 in 60 figure will probably be only increased an infinitesimal amount if calculated precisely using the non-replacement rules.

LetMeJoin

I see how they get to 1 in 60. It just doesn't seem correct probability wise. November, 1 in 714 chance of winning. December, 1 in 714 chance etc. In a year, there are 12x21=252 prizes, with 12x15000=180000 possible winners. This is still 1 in 714. Is it just me???
Premium bonds do make sense, 1 in 34,500 odds, so with 50,000, you should win nearly 1.5 prizes per month, or 17-18 per year. This is the case so far, for me. Admittedly, just £25's.

masonic

LetMeJoin said:

I see how they get to 1 in 60. It just doesn't seem correct probability wise. November, 1 in 714 chance of winning. December, 1 in 714 chance etc. In a year, there are 12x21=252 prizes, with 12x15000=180000 possible winners. This is still 1 in 714. Is it just me???

It's just you.

Toss a fair coin once. The probability of not getting a 'heads' is 50%. We can all agree on that.

Toss a coin once per month for 1 year. By your logic, the probability of getting a 'heads' is (1 x 12) / (2 x 12) = 50%. Do you really believe half the people who do this experiment will get 12 'tails' in a row?

The correct way to calculate is as I set out above: probability of not getting a heads is (1/2)^12 = 0.0244%, so probability of getting at least one heads is 1-0.000244 = 99.98%

Which looks more plausible?

LetMeJoin said:

Premium bonds do make sense, 1 in 34,500 odds, so with 50,000, you should win nearly 1.5 prizes per month, or 17-18 per year. This is the case so far, for me. Admittedly, just £25's.

No, that's not how statistics works. You only have a 76.5% chance of winning at least one prize in any given month if you have a £50k holding (i.e. a 23.5% chance of winning nothing).

grumiofoundation

LetMeJoin said:

I see how they get to 1 in 60. It just doesn't seem correct probability wise. November, 1 in 714 chance of winning. December, 1 in 714 chance etc. In a year, there are 12x21=252 prizes, with 12x15000=180000 possible winners. This is still 1 in 714. Is it just me???
Premium bonds do make sense, 1 in 34,500 odds, so with 50,000, you should win nearly 1.5 prizes per month, or 17-18 per year. This is the case so far, for me. Admittedly, just £25's. 

I think this^ is your issue.

There aren’t 180k bonds, the 15,000 in month 1 is the same as in month 2 and so on

Month one probability of given bond winning is 21/15000 = 0.0014 = 1/714 

Month two probability of given bond winning is the same and so on 

so probability of winning over the year is 0.0014 x 12 = 0.0168  which js as said before 1/60

masonic

grumiofoundation said:

LetMeJoin said:

I see how they get to 1 in 60. It just doesn't seem correct probability wise. November, 1 in 714 chance of winning. December, 1 in 714 chance etc. In a year, there are 12x21=252 prizes, with 12x15000=180000 possible winners. This is still 1 in 714. Is it just me???
Premium bonds do make sense, 1 in 34,500 odds, so with 50,000, you should win nearly 1.5 prizes per month, or 17-18 per year. This is the case so far, for me. Admittedly, just £25's. 

I think this^ is your issue.

There aren’t 180k bonds, the 15,000 in month 1 is the same as in month 2 and so on

Month one probability of given bond winning is 21/15000 = 0.0014 = 1/714 

Month two probability of given bond winning is the same and so on 

so probability of winning over the year is 0.0014 x 12 = 0.0168  which js as said before 1/60

LetMeJoin is correct insofar as when the total number of tickets drawn approaches the total number of tickets in the population, this crude approximation of multiplying odds per draw by number of draws breaks down, which is why you have to compute the probability of the win not happening and then invert it. For example, probability of a win in 60 years is not 21/15000x12x60 = 100%, it is 1-(14999/15000)^(21x12x60) = 63.5%

grumiofoundation

masonic said:

grumiofoundation said:

LetMeJoin said:

I see how they get to 1 in 60. It just doesn't seem correct probability wise. November, 1 in 714 chance of winning. December, 1 in 714 chance etc. In a year, there are 12x21=252 prizes, with 12x15000=180000 possible winners. This is still 1 in 714. Is it just me???
Premium bonds do make sense, 1 in 34,500 odds, so with 50,000, you should win nearly 1.5 prizes per month, or 17-18 per year. This is the case so far, for me. Admittedly, just £25's. 

I think this^ is your issue.

There aren’t 180k bonds, the 15,000 in month 1 is the same as in month 2 and so on

Month one probability of given bond winning is 21/15000 = 0.0014 = 1/714 

Month two probability of given bond winning is the same and so on 

so probability of winning over the year is 0.0014 x 12 = 0.0168  which js as said before 1/60

LetMeJoin is correct insofar as when the total number of tickets drawn approaches the total number of tickets in the population, this crude approximation of multiplying odds per draw by number of draws breaks down, which is why you have to compute the probability of the win not happening and then invert it. For example, probability of a win in 60 years is not 21/15000x12x60 = 100%, it is 1-(14999/15000)^(21x12x60) = 63.5%

You are right is a crude approximation (should have clarified).

I think the 180k possible winners is the confusion - as there aren't really 180,000 possible winners (bonds) there are 15,000 possible bonds each having 12 chances to win. 

masonic

grumiofoundation said:

masonic said:

grumiofoundation said:

LetMeJoin said:

I see how they get to 1 in 60. It just doesn't seem correct probability wise. November, 1 in 714 chance of winning. December, 1 in 714 chance etc. In a year, there are 12x21=252 prizes, with 12x15000=180000 possible winners. This is still 1 in 714. Is it just me???
Premium bonds do make sense, 1 in 34,500 odds, so with 50,000, you should win nearly 1.5 prizes per month, or 17-18 per year. This is the case so far, for me. Admittedly, just £25's. 

I think this^ is your issue.

There aren’t 180k bonds, the 15,000 in month 1 is the same as in month 2 and so on

Month one probability of given bond winning is 21/15000 = 0.0014 = 1/714 

Month two probability of given bond winning is the same and so on 

so probability of winning over the year is 0.0014 x 12 = 0.0168  which js as said before 1/60

LetMeJoin is correct insofar as when the total number of tickets drawn approaches the total number of tickets in the population, this crude approximation of multiplying odds per draw by number of draws breaks down, which is why you have to compute the probability of the win not happening and then invert it. For example, probability of a win in 60 years is not 21/15000x12x60 = 100%, it is 1-(14999/15000)^(21x12x60) = 63.5%

You are right is a crude approximation (should have clarified).

I think the 180k possible winners is the confusion - as there aren't really 180,000 possible winners (bonds) there are 15,000 possible bonds each having 12 chances to win.

If calculating accurately, the denominator would be multiplied up for each draw (in fact each ticket drawn if calculating precisely) such that it represents the total number of possible outcomes, but the numerator must also be multiplied by the number of combinations there are of drawing tickets that includes one of yours. So both numerator and denominator become very big numbers. You don't do anything with the denominator if using the approximation, as you say.