Winning probability in raffle: There are 10,000 tickets available for 1,000 prizes...

Ihave1question

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There are 10,000 tickets available for 1,000 prizes. Each prize has its own drawing (1,000 total drawings). All 10,000 tickets are entered into each drawing. What is the probability wins a prize over the course of all drawings?

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My answer is 1/10,000. My friend believes 1/10. Just a friendly debate
 
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There are 10,000 tickets available for 1,000 prizes. Each prize has its own drawing (1,000 total drawings). All 10,000 tickets are entered into each drawing. What is the probability wins a prize over the course of all drawings?
The question as written makes no sense because there is no subject. What seems likely to be the subject is something like a "person who buys one ticket" is the subject. Also the question is ambiguous. Do you mean "at least one prize" or "exactly one prize." Please give the question exactly as written in your book.

Also tell us what you think would be the most plausible way to think about the problem.

EDIT: In response to your edit, please clarify your question
 
A person buys 1 out of 10,000 raffle tickets

A person buys 1 ticket out of 10,000 raffle tickets. There are 1,000 prize drawings. All tickets are entered into each prize drawing. What is the probability that the person who has 1 ticket wins at least 1 prize
 
Is a person allowed to win in more than one drawing? That is, if a person wins in one drawing, are they then barred from the subsequent drawings? If not, is you question "What is the probability a person will win at least one drawing?"

Assuming that a person is allowed to win in more than one drawing and that the question is "What is the probability that a person will win at least one drawing", then this is 1 minus the probability of not[ winning any drawing. The probability of winning any one drawing is 1/10000 so the probability of not winning a given drawing is 9999/10000. The probability of not winning any of the 1000 drawings is (9999/10000)^{1000}. The probability of winning at least one drawing is 1- (9999/10000)^{1000}.
 
Is a person allowed to win in more than one drawing? That is, if a person wins in one drawing, are they then barred from the subsequent drawings? If not, is you question "What is the probability a person will win at least one drawing?"

Assuming that a person is allowed to win in more than one drawing and that the question is "What is the probability that a person will win at least one drawing", then this is 1 minus the probability of not[ winning any drawing. The probability of winning any one drawing is 1/10000 so the probability of not winning a given drawing is 9999/10000. The probability of not winning any of the 1000 drawings is (9999/10000)^{1000}. The probability of winning at least one drawing is 1- (9999/10000)^{1000}.

Each of the 1,000 prize drawings would have 10,000 possible winners. If I own 1 ticket, my odds are 1/10,000 for each individual drawing. However, the equation above suggests I have a 10% chance to win at least one prize over the course of 1,000 drawings. That would also mean that all 10,000 ticket holders have a 10% chance to win a raffle prize?

I'm no math expert, so I'm not doubting you. I
 
A person buys 1 ticket out of 10,000 raffle tickets. There are 1,000 prize drawings. All tickets are entered into each prize drawing. What is the probability that the person who has 1 ticket wins at least 1 prize
The answer to that question can be answered two ways. The long way is add up the probabilities that the person wins exactly once, exactly twice, exactly thrice, etc.

The short way is to calculate the probability of not winning any prize and subtract that result from

\(\displaystyle 1 - \left \{ \dbinom{1000}{1000} * \left ( \dfrac{9999}{10000} \right )^{1000} * \left ( \dfrac{1}{10000} \right ) ^0 \right \} \approx 1 - 0.9048 = 0.0952 = 9.52\%\).

EDIT: Both ways give the same answer obviously. And Halls simplified the relevant computation, but his method gives the same answer as well.
 
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The answer to that question can be answered two ways. The long way is add up the probabilities that the person wins exactly once, exactly twice, exactly thrice, etc.

The short way is to calculate the probability of not winning any prize and subtract that result from

\(\displaystyle 1 - \left \{ \dbinom{1000}{1000} * \left ( \dfrac{9999}{10000} \right )^{1000} * \left ( \dfrac{1}{10000} \right ) ^0 \right \} \approx 1 - 0.9048 = 0.0952 = 9.52\%\).

EDIT: Both ways give the same answer obviously. And Halls simplified the relevant computation, but his method gives the same answer as well.

Thanks for the response JeffM. This would mean that every ticket has a 9.52% chance of winning at least 1 prize over the course of 1,000 drawings, even though the odds of winnig on any individual drawing would be 1/10,000?
 
Thanks for the response JeffM. This would mean that every ticket has a 9.52% chance of winning at least 1 prize over the course of 1,000 drawings, even though the odds of winnig on any individual drawing would be 1/10,000?
Probability requires very careful use of language.

Intuitively, you feel that any specific ticket, say ticket #227, has a 0.01% chance of winning on any specific draw, say draw #913. That is correct if the draws truly are random. And then you also feel that with 1000 draws the combined chance of winning on any draw is
1000 * 0.01% = 10%. That is wrong. Why?

The probability of a ticket having exactly one win is the sum of the probabilities of winning only on the first draw, of winning only on the second draw, of winning only on the third draw, etc. Each case involves 1 win and 999 losses.

\(\displaystyle \dfrac{1}{10000} * \left ( \dfrac{9,999}{10,000} \right )^{999} \approx 0.00009.\)

But there are a thousand such cases so the probability of exactly one win is a little less than 1%.

But, to calculate the probability of at least one win, we also have to consider the cases of exactly two wins, of exactly three wins, and on up to the unbelievably unlikely but theoretically possible case of 1000 wins.

When we add those all up, we get about 9.52% (though Halls and I used a shortcut method).

So you can say that the probability of any one ticket winning at least once is about 9.52%.

The probability that every ticket wins at least once is the same as the probability that every ticket wins exactly once. If one ticket wins more than once, then at least one ticket cannot win at all. The probability that every ticket wins once is virtually zero.

So to be careful, you should say that any one ticket has a 9.52% chance of winning at least once, but that the chance of every one of the ten thousand tickets winning at least once is essentially zero.

EDIT: Basic probability theory does not involve hard computations, but it does involve formulating questions and answers with great care.
 
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