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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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Yes it does (make sense to go at it that way). Call $w$ the probability to win one given week, then the probability not to win one given week is $1-w$ and the probability not to win $n$ given weeks is $(1-w)^n$ hence the probability to win is $1-(1-w)^n$ (note that this is always strictly less than $nw$). You already computed $w$ and your $n$ is something like $365.25\cdot 30/7\approx1565$ hence you have everything you need. – Did Jul 14 '11 at 11:28
Note that multiplying by the number of weeks cannot be the right approach for the following simple reasoning: if he would play for $12271513$ weeks, the probability would be greater than 1... – N. S. Jul 14 '11 at 15:33
@N.S. Please explain why p = (1 - 1/n)^n is greater than 1. – Mick Mar 10 '14 at 18:34
Edit: p = (1 - 1/n)^(n + 1) – Mick Mar 10 '14 at 18:42
@Mick That is power, not multiplication ;) – N. S. Mar 11 '14 at 3:36
up vote 2 down vote accepted

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:


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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Could the problem be asked like this: what is the probability that at least one of 1560 people playing one week will win? – Algific Jul 14 '11 at 12:03
Careful: not if these 1560 people do not bet independently. For example, if they choose to all bet the same, the result is not the one you ask for. Likewise if they all bet differently. – Did Jul 14 '11 at 13:15
You're right, there's a 0 zero missing. Duly noted. – johnny Jul 14 '11 at 22:57

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