Peter :

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!

Actually, in this case (in the calculation) the tickets are "returned back"! I agree though that if they were removed from the pool (in the case of BeoWorld draws), than the analogy/answer would not be correct, Peter. 

The result is correct ...

It's about an application of the binomial distribution - to make it sound even more weird . The binomial distribution gives the probability of obtaining a number X of "successes" out of an experiment that is repeated N times, with the probability of success held constant across the experiments...

I'd love for Tournedos and saf to take us through their reasoning in a detailed manner, but I do not see why not winning should have any relevance.

@Lee.

120x100/550=21.82%

21.82x4.58=99.93

I'm sure that if you gave us the third decimal, we'd get very close -- not proving anything if the reasoning is wrong, but I don't think it is. The result is skewed if some members waive their prizes, or if others have repeat wins, but unless there's a very anomalous behavior/distribution, it shouldn't matter much for the overall odds.

soundproof:

I'd love for Tournedos and saf to take us through their reasoning in a detailed manner, but I do not see why not winning should have any relevance.

The easiest way to calculate the probability of winning "at least one prize" is to calculate the probability of winning
no prizes at all, and subtract that value from 1 - binominals ...

But I'll try to explain it better ...

Great, because I want to learn.

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:

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.

I found for you (and Soundproof):

The probability of an event can be expressed as a binomial probability if its outcomes can be broken down into two probabilities p and q, where p and q are complementary (i.e. p + q = 1) For example, tossing a coin can be either heads or tails, each which have a (theoretical) probability of 0.5. Rolling a four on a six-sided die can be expressed as the probability (1/6) of getting a 4 or the probability (5/6) of rolling something else.

If an event has a probability, p, of happening, then the probability of it happening twice is p², and in general pⁿ for n successive trials.

And here is the difference when you compare to your calculation(s) ...

Puncher, it isn't that easy

If you throw a dice 6 times doesn't guarantee you your desired number.

That's the reason lotteries make money....

My statistic teacher at the university told me exactly what J0hn mentioned: Each chance is absolutely random; It doesn't guarantee a win after 550 draws

So if there's a number: chances to win is 10 times (Prizes within one draw is assumed not to been given to the same member)

10 x (1/550) = 1.81%

EACH draw, no accumulation....

    Gunther

saf:

Puncher:

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.

I found for you (and Soundproof):

The probability of an event can be expressed as a binomial probability if its outcomes can be broken down into two probabilities p and q, where p and q are complementary (i.e. p + q = 1) For example, tossing a coin can be either heads or tails, each which have a (theoretical) probability of 0.5. Rolling a four on a six-sided die can be expressed as the probability (1/6) of getting a 4 or the probability (5/6) of rolling something else.

If an event has a probability, p, of happening, then the probability of it happening twice is p², and in general pⁿ for n successive trials.

And here is the difference when you compare to your calculation(s) ...

OK - in 3 attempts with a dice, what is the probability that at leat one of the numbers rolled would be a 5?

DELETE

The probability of getting exactly k successes in n trials is

... use your figures!

EDIT: Oh, a little help:

.

eg 3!=3x2x1

It's getting late here

I remember playing "Ludo" (?, we call it "Mensch ärgere Dich nicht")) where you have to roll a "6" to get out in the game and you have 3 tries each round I was still sitting waiting after 2 rounds...

So much for the theory; I quit looking at saf's formulas 

Although the "n over k" thing reminds me more of the theoretical chances to win in a "6 out of 49" lucky numbers lottery...

    Gunther

soundproof:

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

That's why I , sadly enough, would never attempt to play poker ...

Puncher:

saf:

Puncher:

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.

I found for you (and Soundproof):

The probability of an event can be expressed as a binomial probability if its outcomes can be broken down into two probabilities p and q, where p and q are complementary (i.e. p + q = 1) For example, tossing a coin can be either heads or tails, each which have a (theoretical) probability of 0.5. Rolling a four on a six-sided die can be expressed as the probability (1/6) of getting a 4 or the probability (5/6) of rolling something else.

If an event has a probability, p, of happening, then the probability of it happening twice is p², and in general pⁿ for n successive trials.

And here is the difference when you compare to your calculation(s) ...

 

OK - in 3 attempts with a dice, what is the probability that at leat one of the numbers rolled would be a 5?

This is easiest to calculate with saf's binomial equations - but since your problem is simple, let us do it the hard way. In words, it means obviously that you get a 5 with first roll AND not the others, OR only on 2nd roll, OR only on 3rd roll, OR 1st AND 2nd, OR... you get the idea.

This can now be formulated as

P = (1/6 * 5/6 * 5/6) + (5/6 * 1/6 * 5/6) + (5/6 * 5/6 * 1/6) + (1/6 * 1/6 * 5/6) + (1/6 * 5/6 * 1/6) + (5/6 * 1/6 * 1/6) + (1/6 * 1/6 * 1/6)

   = 25/216 + 25/216 + 25/216 + 5/216 + 5/216 + 5/216 + 1/216

   = 91/216 ~= 0.4213

...or we can take the easy way out, and think of the complementary problem, i.e not getting a single fiver. That is

P = (5/6 * 5/6 * 5/6)

   = (5/6)^3

   = 125/216 ~= 0.5787

...and since we wanted to know the complement (getting at least one fiver), we subtract that from 1:

P = 1 - 125/216 = 91/216 ~= 0.4213

How's that for a coincidence?

saf:

The probability of getting exactly k successes in n trials is

... use your figures!

 

EDIT: Oh, a little help:

.

eg 3!=3x2x1

It's getting late here

UH .........

I see, this is overture connected to BL3 - master link to Avant .....

Easy ....

9 LEE:

 

 

Thank you, Mika!

Must understand this use of binomials.

Let's see. What if we make the components a little more manageable.

There's one prize. There are two contestants. It's a blind draw, just as the monthly draw here.

In the world I inhabit, there's the niggling thought that there's a 1/2 chance of either contestant winning. But in this new world I seem to be entering ... I send the ball over to you Tournedos!

Go binomial on us!

soundproof:

In the world I inhabit, there's the niggling thought that there's a 1/2 chance of either contestant winning. But in this new world I seem to be entering ... I send the ball over to you Tournedos!

If I understand that correctly, you want to get confirmed/shown that 1/2 chance of winning each month would mean app. 99,98pct probability of winning at least once a year?

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.

No No No Mr P.

The chances of your number coming up are 1 in 5 - you have to take the number you have chosen out of the odds which sort of blows this theory out of the water I'm afraid......

Chris.

As a professional poker player I am well versed in mathematical probabilities!

The answer you are looking for is : it's always 50:50.

You either win a prize, or you don't.

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!

Ilikehifi:

As a professional poker player I am well versed in mathematical probabilities!

The answer you are looking for is : it's always 50:50.

You either win a prize, or you don't.

I like this !

Peter :

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!

Every time when you rolled and you don't win you need to exclude one who get prize, in order to have better chance, if you don't exclude, you will always have 1 in 550 no meter how much you roll ....

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!

To clarify this, I am absolutely NOT claiming that the different draws/rolls would be somehow connected - quite the opposite!

But as soon as there are more than one draw, there are several possible ways to combine wins/non-wins, and every one of these combinations has its own probability. All of these probabilities have to be combined (=OR) to get the probability of the entire problem.

As soon as you roll your 550 side dice more than once, you have a greater than 1/550 probability of getting AT LEAST one win. But the probability will NEVER get to be 1, now matter how many times you repeat it!