Probability of winning the lottery twice I'm struggling to get the answer to this question. I believe I have almost solved it but am having trouble getting the final answer.
What is the probability of someone winning the lottery twice?
It is broken down as follows:
a) Given a lottery consisting of 49 numbers where 6 are drawn without replacement what is the probability of any specific ticket (or person) winning the lottery?
b) What is the probability of that same person winning a second time on the same lottery assuming they take part in 1000 lottery draws?
c) Assuming a population of 60 million people, who all play the lottery, what is the probability of any one person winning the same lottery twice?
Part a):
I used the combination formula to calculate the possible combinations of 6 numbers from 49:
$$ \frac{49!}{6!(49-6)!} = 13, 983,816 $$
Giving a probability of a specific ticket winning as:
$$ \frac{1}{13,983,816} $$
Part b):
I use the binomial equation to calculate the probability of two tickets in 1000 winning:
$$ \binom{1000}{2}\left(\frac{1}{13,983,816}\right)^2\left(1-\frac{1}{13,983,816}\right)^{998} $$
Which gives me a value of:
$$2.554\times10^{-9}$$
Part c):
This is the part I am struggling with. I want to multiply the probability calculated in part b by 60,000,000 which gives me ~0.153 but this doesn't feel right.  
Please could someone help point me in the right direction for solving this?
 A: As saulspatz says, your approximation is already good, but if you want an exact answer:
For (b), the probability that a single person wins at least twice (because if they win 3 or more times, they have still won twice) can be calculated as the total probability minus the probability of losing every time or winning only once:
$$1-\dbinom{1000}{0}\left(\dfrac{1}{\dbinom{49}{6}}\right)^{0}\left(1-\dfrac{1}{\dbinom{49}{6}}\right)^{1000} - \dbinom{1000}{1}\left(\dfrac{1}{\dbinom{49}{6}}\right)^{1}\left(1-\dfrac{1}{\dbinom{49}{6}}\right)^{999} \approx 2.55\times 10^{-9}$$
which is basically what you found, as well.
For (c), if we let $p$ be the probability of winning at least twice found above, we can do the same thing. The probability of at least one person winning at least twice is one minus the probability of no one winning at least twice gives the probability of at least one person winning twice.
$$1-(1-p)^{60000000} \approx 0.142089$$
So, your estimation was very close.
A: It depends on how accurate you need to be.  In the case where you calculated the probability that one person wins the lottery twice, what you calculated was the the probability that he wins exactly twice.  This ignores the possibility that he wins $3$ or more times.  To write an exact formula, you need to use exclusion and inclusion.  On the other hand, the probability that he wins $3$ or more times is so small that if what you want is an approximation to a reasonable number of decimal places, you can ignore this possibility.
Similarly, in calculating the probability that someone among $60$ million players wins twice, you have ignored the possibility that two or more people win twice.  You would have to use inclusion and exclusion to account for this.  I would guess that this possibility is so small it can be ignored.
If you want more information about how good these approximations are, look up the Bonferroni inequalities. 
