E: 30/9/05 Win Ipod With Walkers Crisps (MERGED)

slowmo_3

Hi,
just wanted to point out that if occuring events are independent of each other, the probability of events happening are calculated by multiplying them together.

Eg. Heads occuring in 1 toss of coin is 1/2 = 0.5, in 2 tosses with heads occuring in both tosses is 1/2 x 1/2 = 1/4 = 0.25. In 3 tosses with heads on all 3 tosses is 0.5x0.5x0.5=0.125

However, in this case, you do not multiply the probability together, as that will get the probabilty of getting the Ipod on every single draw that you entered. In order to calculate the chances of winning in this case, we just need an extra step of finding the probability of NOTwinning on every single draw that you entered, and subtract that from 1.0.
Eg. with 1 entry out of 241 entries, then that prob of not winning in that draw is 240/241. With 10 draws of the same prob, then the chances of not winning a single draw is (240/241) through the power of 10 = 0.95927. So, the chances of winning 1 or more from the 10 draws is 1-0.95927 = 0.04073 = 4.073%
So it should be that there is a 75% chance of at least 1 Head occuring in 2 tosses of the coin, not 50% chance as mentioned by bs7. In 2 tosses, there are 4 possibilities = Heads & Heads (HH), Heads & Tails (HT), TH, TT. And it can be clearly seen that 3/4 has at least a Head in the result... :-) The 50% chance that bs7 mentioned is correct if he means one and only one Head in that 2 tosses.

I will leave you to work out the rest of the probabilities, as I really should be sleeping now... :-P

gizmoleeds

You people seem to be missing an invaluable point here:

If there are 250 entries in a typical early morning draw, then if you added 250 entries you would have a 50% chance of winning. Codes from the npn website cost 10p as a text to enter - cost £25. Do this twice (£50) and statistically you have probably won.

I know entering 250 times in one draw is not really possible, but you can see my point, spread out the odds should be similar.

NickC_4

bs7 wrote:

But you can't add the probabilities of winning if you spread them since they are all totally separate and completely independent, i.e. a 1/100 chance in one 5 minute period, followed by another in the next period does not equal 2/100. It's simply two 1/100 chances.

I knew this would provoke a bit of discussion.

Sorry bs7, but of course you can add the probabilities. If you enter twice you have twice the chance of winning - or halving your odds. And when I went to school 2 x 1/100, (or 1/100 + 1/100) was the same as 2/100.

Slowmo is also quite correct - my numbers are of course only averages. If you toss a coin twice you would expect to get 1 head on average. You would not be at all suprised if you got two or none. The more times you toss the coin, the nearer to 50% you should get.

Best of luck whatever you do.

Nick.

gizmoleeds

NickC wrote:

I knew this would provoke a bit of discussion.

Sorry bs7, but of course you can add the probabilities. If you enter twice you have twice the chance of winning - or halving your odds. And when I went to school 2 x 1/100, (or 1/100 + 1/100) was the same as 2/100.

Slowmo is also quite correct - my numbers are of course only averages. If you toss a coin twice you would expect to get 1 head on average. You would not be at all suprised if you got two or none. The more times you toss the coin, the nearer to 50% you should get.

Best of luck whatever you do.

Nick.

The odds of winning twice would be 1/100 x 1/100 = 1/10,000
The odds of just the 1st winning would be 1/100 x 99/100 = 99/10,000
The odds of just the 2nd winning would be 99/100 x 1/100 = 99/10,000
The odds of not winning would be 99/100 x 99/100 = 9,801/10,000

Therefore the odds of winning are 199/10,000 or 0.0199

fluffieone

lol, Why argue about odds, someone with one code could win, but yes you have a better chance when there are less entries. Have you lot tried Martins quiz, the first to post wins, used to get a bottle of champayne but don,t know if you still do.

bs7

NickC wrote:

Sorry bs7, but of course you can add the probabilities. If you enter twice you have twice the chance of winning - or halving your odds. And when I went to school 2 x 1/100, (or 1/100 + 1/100) was the same as 2/100.

Slowmo is also quite correct - my numbers are of course only averages.

I'll admit i messed up - but slowmo just explained it to you and you've missed the point - i can't remember much statistics but i remember that independent events are not treated the same way as dependent ones. And gizmoleeds has just backed it up with figures:

you say 1/100 entried twice is 2/100 chance or 2%
but in reality it's as gizmoleeds states: "199/10,000 or 0.0199" or 1.99%

you can say that it's an approximation but my original point is true - the odds are still going to be better when you enter at the same time (2/100) which would give you a 2% chance.

Now you are still maintaining your chance "on average" is exactly the same - it's not. "On average" it's going to be close with such small figures but it's not the same and that was my original point.

Amba_Gambla

waoh!!!! I wasn't expecting a GCSE maths lesson!!!!

Marvo434

don't know if it will help at all, but I seem to remeber something about the square of the hipopotamus being equal to the sum of PI R squared in a circular triangle.

NickC_4

bs7 wrote:

I'll admit i messed up - but slowmo just explained it to you and you've missed the point - i can't remember much statistics but i remember that independent events are not treated the same way as dependent ones. And gizmoleeds has just backed it up with figures:

you say 1/100 entried twice is 2/100 chance or 2%
but in reality it's as gizmoleeds states: "199/10,000 or 0.0199" or 1.99%

you can say that it's an approximation but my original point is true - the odds are still going to be better when you enter at the same time (2/100) which would give you a 2% chance.

Now you are still maintaining your chance "on average" is exactly the same - it's not. "On average" it's going to be close with such small figures but it's not the same and that was my original point.

OK, I wasn't entirely paying attention. I understand what gizmoleeds explained now. What I have also realised is that by putting all your entries into one draw, you are reducing the probability of any one entry winning.

i.e. Your 2 entries are now therefore 2/101 which is 0.0198 or 1.98% - still less than the 1.99%. So I stand by my spread it around theory, even if the rest of it was a little off and with the amount of other entries involved it makes little difference!

Thanks all, Nick.

jadring

Has anyone won one of these yet?

I've tried many times, on the website and on the mobile but not won