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Point taken! You've been really helpful, and I certainly appreciate it.
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?
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)?
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
I would really like to see the formula. Thank you for explaining.
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.
Yes, one person with two tickets not winning anything. Thanks for your help!
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!
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