When playing the lottery you have to pick 6 numbers out of 45 possibilities. Since the order of the numbers don't matter, the number of possible combinations for the jackpot (and hence the 6 correct numbers) is given by:

$
\dbinom{45}{6} = C^6\_{45} = \frac{45!}{(45-6)!6!} = 8145060
$

Assuming the numbers are picked at random out of a uniform distribution the chance of winning the jackpot is given by:

$P(win) = \frac{1}{8145060}\approx 0.0000123 \%$

Now I can figure two possible scenarios:

1. You play every week with the same numbers hoping that one day you'll get the jackpot.
2. You play every week with a different set of numbers hoping that one day you'll get the jackpot.

Now I was wondering if there is a difference in the chance of winning between the two methods. I was thinking that in the first case the chances of winning are larger because if you stick to your number the alternatives might run out. While if you switch you don't have the other options that become eliminated. But then I started wondering, because in the [Monty Hall paradox](http://en.wikipedia.org/wiki/Monty_Hall_problem) the chances become larger if you switch.

For the first case I believe the chance of winning in the $n^{th}$ drawing is simply calculated as:

$1-(1-P(win))^n$

But in the second case I wouldn't know how to succeed.