tournedos:

Peter :

Still don't agree with this. If you had a 550 side dice as the random number generator, you will always have a 1 in 550 chance of your number coming up no matter what is rolled before. There is no possible link. It is likely if you roll the dice 550 times that it may come up but if you have rolled it 549 times and it has not come up, you still have a 1 in 550 chance only!

 

But the probability will NEVER get to be 1, now matter how many times you repeat it!

how ?

Assume we have one prize - 550 competitors, and no one wins till the last roll .... than you have 1:1 chance to win ?

Am I right ?

John

You are repeatedly asnswering a different question! The question you are answering is  which is what is the probability on a single draw (i.e. one event) of winning a prize. (Random benoulli trials as the textbooks call them).

After this event  has been repeated 120 times with a population of 550 people who all have an equal chance of winning  you will find that some people have won  no prizes, some people have won one and there will be a smaller number of people who have won two or more. How this breaks down can be modelled with a fairly high degree of accuracy so you would be able to predict the probability that any one individual would have won no prizes, one prize, or two or more prizes using the known properties of the binomial distribution, and thereby answer the question what is the likelihood af any individual winning one or more prizes.

According to Minitab (Statistical software) the chance of winning one prize is  0.175698, or 17.57% so I agree with Mika, but when you include the chances of winning more than one prize the probabilities are:

Probability Density Function

Binomial with n = 120 and p = 0.00181818

x  P( X = x )
0    0.803820
1    0.175698
2    0.019042
3    0.001364
4    0.000073
5    0.000003

so the chances of winning a prize in any one year are around 19.6%

Moral of the story: Never argue with a Statistician!

Simon

Puncher:

Is it really this difficult? -

If I throw a dice then the chance of my number coming up is 1/6.

If I throw it twice then chance of my number coming up in any of the throws is 1/6 + 1/6 = 2/6 = 1/3.

If I throw it six times then the theoretical probability of one of the six attempts being my number is 6/6 i.e 1.

The probabilities of multiple trials added together - it makes sense that, if you enter the draw 550 times then you should have won at least once.

Puncher

The logic you describe would work if you were, for example, turning over six cards numbered one to six.If you have alrady turned over 5 cards and one of them is yours, the last card you turn over will be yours.

With dice there is no reason a number cannot happen again, so in all probability after 6 rolls there will be at least one number appearing two or more times, and at least one number that has not apeared at all.

The chances of any ne number appearing are shown below:

Probability Density Function

Binomial with n = 6 and p = 0.166666

x  P( X = x )
0    0.334900
1    0.401878
2    0.200938
3    0.053583
4    0.008037
5    0.000643

so there is around a 2/3 chance of your chosen number appearing after 6 dice rolls.

Simon

wirralsimon:

 

...the chances of winning a prize in any one year are around 19.6%

Moral of the story: Never argue with a Statistician!

 

Simon

So basically, what one needs to do is sign up, then get another four friends and relatives to sign up, then you are guaranteed a prize.

Sorted!

Ilikehifi:

wirralsimon:

 

...the chances of winning a prize in any one year are around 19.6%

Moral of the story: Never argue with a Statistician!

 

Simon

 

 

So basically, what one needs to do is sign up, then get another four friends and relatives to sign up, then you are guaranteed a prize.

 

Sorted!

In which case there is still a 33% chance that you will win nothing at all, but on the plus side Beoworld gets more money to pay it's running costs!

Simon

If you toss a coin you have a 50/50 chance it will be heads. 

If you toss a coin ten times in a row and each times it falls tails, the eleventh time you toss it, it will still be 50/50 it will land heads.

The law of probability would suggest that the greater number of consecutive results should mean that the following toss will bring a different result. But there are still only two outcomes of a single coin toss (assuming we discount the possibility of it landing on its edge but lets not get into that!)

I conclude that you must have a 1 in however many eligible members at the time of the draw.

Opman

Thanks for the run-through, guys. Factor in the variables (multiple wins, etc), then the probability is reduced.

- seems I am as thick as I feared!

I actually dug out my maths books last night to have a bit of a "refresher"............... the sad thing is, I used to know stuff like this. I blame Wednesdays© - I must lose something useful everytime jandyt fills my head with another useless fact - I think the merits of sock weighing may have been responsible for me losing this particular corner of what's left of my maths brain

Anyhoo, thanks to saf and Mika for keeping us right, the chance of winning any prize within a year is indeed 19.6 percent, and as the draws roll by -

You are not thick at all, Puncher – and I can confess a sudden need for a couple of cans (“probably” at least two) yesterday in order to refresh what I’d learned at school long time ago…

At least we all came with something (an easy formula perhaps? ) Lee can now “probably” adapt to keep people’s chances of winning a price here on BeoWorld constantly attractive … as documented, they are indeed just that!

PS Mika should teach in his spare time, btw – he could make some extra money!

saf:

You are not thick at all, Puncher – and I can confess a sudden need for a couple of cans (“probably” at least two) yesterday in order to refresh what I’d learned at school long time ago…

.........that's where I went wrong! I wouldn't care but it's my normal solution to problem solving!

and, slightly more worringly, here is the probability of throwing a dice and getting at least one six, across 25 attempts - no wonder I'm crap at board games!

So..it is actually possible to win the top prize more than Once in a year?

What are the odds on a single individual doing that?

DavidOd:

So..it is actually possible to win the top prize more than Once in a year?

What are the odds on a single individual doing that?

Approximately a 0.02% chance.

Simon

These odds are surely greater than the weekends lotto draw.

What is a good optimal chance of winning?

With 550 prize candidates, at what point (number of potential winners) do the number of prizes have to be increased ( to say 20 prizes ) to maintain  fair winning odds?

Compared to most draws the odds are fantastic! I don't think you will get many draws where the top prize is only a 1 in 550 chance when it is worth several thousand pounds and the monthly stake is £2.50. And you get the chance to win 9 other prizes as well! I think we expected rather more people to sign up for this - so make hay whilst you can!

wirralsimon:

Approximately a 0.02% chance.

Simon

Hmmm - excellent odds, compared to what people otherwise gamble on.

Any one draw = 1 / number of gold members.

Simple solution = remove lots of gold members to increase your odds*......;-)

The chance of winning 1 draw is 1/550

The chance of winning a prize is 120 goes at 1/550.

Then argue if you think this makes you feel you have a better chance of winning anything!

Always wondered why the draw was held in reverse order? if you can win multiple prizes in each draw it doesn't matter what order its done it!

*Now where's my white cat?

Probably of winning in one draw =1/550

Probability of not winning in one draw = 1 - 1/550 = 549/550

Probability of not winning any prizes in 120 draws (a year) = (549/550)^120 = 0.8

Probability of winning more than zero prizes in a year = 1 - Probability of not winning any prizes in 120 draws (a year) = 1 - 0.8 = 0.2 or 20% => 1 in 5

On any particular draw, the probability of winning a prize is still 1 in 550; the draw machine (hopefully) has no memory of previous draws

I did a maths degree over 20 years ago so this may be flakey

Justin:

Probably of winning in one draw =1/550

Probability of not winning in one draw = 1 - 1/550 = 549/550

Probability of not winning any prizes in 120 draws (a year) = (549/550)^120 = 0.8

Probability of winning more than zero prizes in a year = 1 - Probability of not winning any prizes in 120 draws (a year) = 1 - 0.8 = 0.2 or 20% => 1 in 5

On any particular draw, the probability of winning a prize is still 1 in 550; the draw machine (hopefully) has no memory of previous draws

I did a maths degree over 20 years ago so this may be flakey

(549/550)^120 = 0.804 rather than 0.8000, so you have the same result (80.4% of winning, 19.6% chance of not winning a prize) as a few of us.

Simon