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
1 Answer
$\begingroup$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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