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.