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Is there a way to determine the probability of winning a particular sweepstakes with only the following information:

-the estimated odds of winning are 1 in 13,000

-contestants must be owners (property holders) within the corporation running the sweepstakes and the total number of owners is 222,600

-each owner can enter the sweepstakes up to 6 times

-there is one grand prize and three runner-up prizes

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What does "estimated odds of winning are 1 in 13,000" mean? Does it mean 1 in 13,000 per entrance? And does it mean the odds of winning at all, or the odds of winning the grand prize? –  Justin L. Oct 23 '10 at 5:30

1 Answer 1

If the estimated odds of winning are $1$ in $13,000$ per entrance, then if you enter once, your odds of winning are $1:13000$, of course.

However, entering more than once, it's a bit more complicated. It's easier to calculate the odds of you not winning, and subtract it from $1$.

The odds of you not winning once is $12999/13000$. The odds of you not winning twice is $(12999/13000)^2$, and three times is $(12999/13000)^3$, etc.

We could then say we can calculate your odds of not winning $n(x)$, where $x$ is the number of entries, as $n(x) = (12999/13000)^x$.

However, you want to calculate the odds of winning $p(x)$

We know that $n(x) + p(x) = 1$ (Think about it. The sum of the odds of not winning and the odds of winning equal up to 1, because there is a 100% chance that one or the other will happen). So, $p(x) = 1 - n(x)$

So to calculate your odds of winning, you would use

$p(x) = 1 - (12999/13000)^x$

where $x$ is the amount of times you've entered.

If you enter 6 times, then your total odds are $\frac{2227329628937465077999}{4826809000000000000000000}$, which is about $0.0004614497$.

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Thank you for your answer; it makes perfect sense. This question is based on real contest, and they literally state that the odds of winning are 1:13,000 without indicating whether that applies per entrance or to which prize(s). They probably should have been more specific, but again, thanks! –  Anonymous Oct 24 '10 at 17:30

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