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A man decides to play the lottery every week for 30 years. That would be a total of $52\times30$ weeks. The lottery is selecting 6 digits out of 48. The order of the numbers are not important. I'm looking for the probability of winning at least once during those 30 years. First I calculated the probability for winning one week.

$\frac{6!(48-6)!}{48!} = \frac{1}{12271512}$

From there I just assumed that one could multiply by the number of weeks. But that doesn't give the correct answer. The complement would be the probability of not winning a single time during those 30 years, in case it makes sense to go at it that way.

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    sorry! still don't understand. ^ is power.2014-03-11

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Yes, reasoning by complements makes your life much easier.

Probability of not winning this week as you claimed is: $1-\frac{1}{12271512}$.

Assuming independence, probability of not winning for 1560 weeks is then just:

$\left(1-\frac{1}{12271512}\right)^{1560}$

Hence, the probability of winning at least once is:

$1 - \left(1-\frac{1}{12271512}\right)^{1560} \approx 0.00013$

The chances are pretty slim...

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    You're right, there's a 0 zero missing. Duly noted.2011-07-14