A lottery will be held. From 1000 numbers, one will be randomly chosen as the winner. A lottery will be held. From 1000 numbers, one will be randomly chosen as the winner. A lottery ticket is a random number between 1 and 1000 with replacement. 

How many tickets do you need to buy for the probability of winning to be at least 50%?

I am having trouble starting this problem and was told to find the probability of no winning tickets out of n tickets.
If there wasn't replacement then the probability would just increase by a thousandth with every new ticket, but I am unsure of how the possibility of buying two tickets that are the same affects the increase in probability from having multiple tickets
 A: You need $500$ distinct numbers to have a $50\%$ chance of winning.  This is the coupon collector's problem except that you do not need to collect all the coupons.  We can compute the expected number to get $500$ distinct numbers.  The first coupon is guaranteed to get a new number.  To get a second new number takes on average $\frac {1000}{999}$.  Once you have two, the third takes $\frac {1000}{998}$ and so on.  We can compute the expected number to get $500$ distinct numbers as  as $$\begin {align} 1+\frac {1000}{999}+\frac {1000}{998}+\ldots \frac {1000}{501}&=1000(H_{1000}-H_{500})\\ &\approx 1000(\log(1000)-\log (500))\\&=1000\log (2)\\ &\approx 693 \end {align}$$
This gives the expected number of days to get $500$ different tickets.  It is close to, but not guaranteed to be the same as, the number of days to get your chance of $500$ distinct tickets over $50\%$.  I don't have a good way to calculate the second of these.  I suspect the $693$ is an overestimate of the number of days to have a $50\%$ chance to have $500$ different because there will be a long tail.
