Why can't I use the solution for birthday paradox here? I was given the task:
A ticket manager from a cinema is asking each visitor for their birthday,
the first person whose birthday is already told to the manager, is given a free ticket. The birthdays of the people in the queue is evenly distributed and independent. 
Question: 
What is the probability to get a free ticket, if there is $k$ amount of people in the queue in front of you? 
And on which position of the queue should you be in for the free ticket? 
My solution for the probability would be : 
$$ \Bbb P(A)=1-\frac{365!}{(365-k)!*365^k}$$ 
And I approximated with a calculator that on position 68 I would have around 99,999% probability to get a free ticket. 
Thanks for helping! 
 A: Without the 1- part, that formula is the chance of the first k people not sharing a birthday among them ... Which will be close to 0 as you realized .... But notice that this event is what does need to happen if you want a chance at the free ticket! So, don't subtract that from 1! In fact, you in position k+1 need to have this happen and need to share a birthday with one of them to win the ticket. So, multiply that almost 0 chance with 68/365 and you have your chance of winning ... Which is even closer to 0.
As far as what the best position is ... If you look at the graph of the birthday problem that shows the chance of having a shared birthday among the k people, it seems steepest around 23 just around where that chance becomes about 50%. So, I would calculate your chances of winning (using the same formula as just discussed (if you are in position k+1, multiple the chance of k people not sharing a birthday with k/365) for values of k around 23.  ... You should find a number around there where if you go either lower or higher, the chance of winning will decrease. Maybe it is 23 itself, but maybe it is 22 or 24, I don't really know ... Just try!
EDIT 
@JyrkiLahtonen in the comments says the best chance of winning is to be in position is 19 and that it is a little over 3% for all positions 15 to 24 ... Yeah, that is certainly compatible with the graph
A: For you to win a free ticket you need:


*

*The $k$ people in front of you to have distinct birthdays (otherwise
one of them will get the free ticket instead of you). This happens with probability $P(k)$ that you should be able to reproduce from the birthday paradox calculation.

*You need to share a birthday with one of those $k$ people. Given that they were all distinct, this happens with probability $k/365$ (ignoring leap days here).


So your probability of winning is
$$
Q(k)=P(k)\frac k{365}.
$$
As Bram28 correctly explained, $P(k)$ is that fraction in OP's formula (sans that complementing $1-$). We see that
$$
Q(k+1)>Q(k)\Longleftrightarrow (1+k)(365-k)>365k.
$$
I leave it to you to solve this quadratic inequality and conclude that
$Q(k+1)>Q(k)$ if and only if $k<(-1+\sqrt{1461})/2\approx 18.61$. This means that the winning chances max out at $k=19$. Do you see why?
Here's a Mathematica plot of $Q(k)$ vs. $k$. The latter is on the horizontal axis.


We further see that:


*

*there isn't much change in your winning probability when $k$ is between 15 and 25, so you probably shouldn't bother wrestling with the next guy, if you are approximately there.

*my formula passes the lithmus test that $\sum_{k=0}^{365}Q(k)$ seems to come to exactly $1$ (checked it with Mathematica using 12 digit accuracy, but didn't bother with an exact verification).

