What is the probability of marking the same object twice or more while keeping the same ratio of objects and marks? I have a simple but a bit more complex probability question.
If i have 8 different kinds of hats and want to mark half of them (4), what is the probability ill mark ANY of them twice (or more)? What if there are 1000 hats and 500 picks, and what is the relationship between increasing the number of hats and picks and the probability of marking any of them twice, WHILE keeping the ration the same, half. Is there any formula? Would infinite number of hats be aproaching 100 percent probability of any of them repeating? What if i changed the ration to 1/3, would infinite number of kinds of hats still aproach 100 percent probatility?
 A: This is actually the generalized birthday problem.  $n$ is the number of days.  The number of items you have to mark to get a $50\%$ chance of a match is about $\sqrt {2 \ln 2} \sqrt n$.  For other probabilities the constant $\sqrt {2 \ln 2}$ changes, but the $\sqrt n$ behavior does not.  Roughly speaking, the chance of any pair of marks going on the same item is about $\frac 1n$ so you need the number of pairs of marks to be about $n$.  The number of pairs is about half the square of the number of marks.  
This means that if you choose a fixed fraction of $n$ to be the number of marks, eventually the probability of a remark will approach $1$ as $n$ gets large.
A: If you choose $n$ of $2n$ hats without replacement then there are $2n$ ways to choose the first hat, $2n-1$ ways to choose a second hat, $2n-2$ ways to choose a third hat different from the first two, and so on.  There are $${(2n)!\over n!}$$ ways to choose $n$ distinct hats.
There are $(2n)^n$ way to choose $n$ hats without replacement.  So the probably that no hat is chosen twice is $${(2n)!\over n!(2n)^n}$$  This goes to $0$ as $n$ goes to infinity, as you guessed.
If you choose $n$ of $3n$ hats, then by similar reasoning, the probability that none is chosen twice is $$(3n)!\over (2n)!(3n)^n$$ and again, this goes to $0$.
