Question about raffle odds

ltps · 2010-11-29 09:25:34

Hello, everyone. I have a question about raffle odds. Here are the facts of the problem:

1. There are 2000 tickets in the raffle.
2. 500 of those people have two tickets each. In other words, 1000 people have one ticket each and 500 people have two, making a total of 2000 tickets.
3. There are 100 prizes.
4. People cannot win more than one prize.

I would like to calculate the probability of one person who has two tickets not winning a prize, from the first of the 100 draws to the last, assuming that this person does not win a prize.

My way of the math, I think that this would be the probability for the first draw:

(1998/2000) ^ 100, or 90.5 percent

In English: 1998 tickets that belong to other people divided by a total of 2000 tickets, raised to the power of the draw.

This means that the probability of getting a prize on the first draw, for a person who has two tickets, is about 9.5 percent.

On the second draw, the odds of not getting a price increase to about 90.6 percent:

(1997/1999) ^ 99

And so on, until, on the final draw, the odds would be 99.9 percent against.

Do I have the right approach here?

Thank you very much for your help!

bobbym · 2010-11-29 10:46:54

Hi ltps;

Some clarification is necessary. Are you interested in one particular person with 2 tickets not winning or any one of the 500 of them not winning?

ltps · 2010-12-03 02:49:30

Yes, one person with two tickets not winning anything. Thanks for your help!

bobbym · 2010-12-03 06:34:56

Hi ltps;

You have 2 tickets there are 100 draws the chance you do not win a prize is:

ltps · 2010-12-03 09:57:13

Thank you. It would be very helpful to me to know how you arrived at the number and whether the odds would be the same for the first draw as for the 100th draw.

bobbym · 2010-12-03 10:04:28

Hi;

The odds are not the same. This is a sampling without replacement. The odds change on each draw. Essentially this is an urn problem without replacement.

To do this I have used the Hypergeometric distribution which was invented for these types of problems. Would you like to see the formula you plug into... It is a little complicated.

ltps · 2010-12-03 11:11:36

I would really like to see the formula. Thank you for explaining.

bobbym · 2010-12-03 11:39:41

Hi;

winners = the two tickets I hold.
I need = 0, so I do not get a prize.
total = 2000 tickets
trials = 100

ltps · 2010-12-03 13:25:06

Cool. Thank you very much. I also found HYPGEOMDIST in Excel for this:

=HYPGEOMDIST(successes_in_sample,sample_size,number_of_successes,population)

=HYPGEOMDIST(0,2,100,2000) ≈0.902476238

That's on the first draw. On the last draw, P(x=0) ≈ 0.998948

bobbym · 2010-12-03 14:33:28

Hi;

ltps wrote:

That's on the first draw. On the last draw, P(x=0) ≈ 0.998948

You are thinking about it in the wrong way and reading that chart wrong. .90247 is the probability for all 100 draws, not just for the 100 th draw. That means there is a 90% chance that after 100 draws you have not won.

Also here is yet another way to do this problem that treats it like a birthday problem.

So we have two ways to solve it.

ltps · 2010-12-07 10:49:21

Thank you. That's very helpful. So the probability on each draw is much more basic, just the number of tickets you have in the urn divided by the total number of tickets left in the "urn" (since we don't replace winning tickets)?

bobbym · 2010-12-07 10:54:47

Hi itps;

P = S / ( S + F ) which reads Probability equals successes over successes plus failures. So the probabilty on the first draw of not getting a winning ticket is 1998 / 2000.

On the next draw since one ticket was picked ( not one of yours ) you have 1999 tickets left, 2 are yours, the chance of not getting one of yours is 1997 / 1999.

You multiply them to get the probability that in 2 draws you do not win.

For 3 draws:

For 100 draws this would be hard to write in the above way. So we use the math notation for a product.

That is where this comes from:

ltps · 2010-12-08 02:45:02

Right ... I think I'm getting it. So, even though the probability on the last draw itself is 1899/1901 ≈ 0.99895, the probability on that draw of not having already won a prize is the product of all prior probabilities on each draw ≈ 0.90247. And we might also say that even the least lucky person is not without almost 10% probability of getting a prize?

bobbym · 2010-12-08 08:06:34

Hi ltps;

What you have said on the surface sounds simple and intuitive. If you are just starting to grasp probability concepts then I know that you will want to be able to reason about them, to have a feel for them. I urge you to try to reason about them using mathematics only.

Had you asked any number of other questions when you first posted I would be still be working on this problem. What I am trying to say is that slightly rewording a probability problem can turn a doable one into a nightmare.

ltps · 2010-12-09 12:47:21

Point taken! You've been really helpful, and I certainly appreciate it.

bobbym · 2010-12-09 12:48:53

Your welcome. Do you like probability?

Math Is Fun Forum

#1 2010-11-29 09:25:34

Question about raffle odds

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Re: Question about raffle odds

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#4 2010-12-03 06:34:56

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#5 2010-12-03 09:57:13

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#6 2010-12-03 10:04:28

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#7 2010-12-03 11:11:36

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