Number of days until day I am looking for a formula that describes the following probability:
There are $x$ important days in one year. What is the probability that the next day will be $t$ days from now? (If we are at the end of the year, then the number of days until the next event in the next year should be tested. Therefore the probability should be the same of every day in the year)
It is important that the event needs to be exactly $t$ days from now, not more not less (therefore no summed probability)
Since $x$ is going to be at least $10$, the number of possible chosen days is very large and therefore just computing all possibilities is not an option here. ( the number of possibilities of chosing $10$ out of $365$ is $\frac{(365+10-1)!}{364!10!}$, which is very big ;) )
Any idea how to solve this?
EDIT:
The $x$ days are chosen at random and can overlap. You can think of $x$ as describing the number of friends whose birthdays we are waiting for. Two (or more) of your friends can have the same day as birthday.
$t$ is the number of days until the next event, therefore $p(t)=0$ for $t>365$
You can assume that the number of days in a year is 365.
 A: If I understand the question correctly, then the answer is
$$\sum_{k=1}^x{x\choose k}\left({1\over 365}\right)^k\left({364-t\over365}\right)^{x-k}$$
That is, some positive number of friends $k$ must have birthdays coming up in exactly $t$ days, and none of the others can have a birthday today, tomorrow, the day after, etc., through day $t$ from now, which means their birthdays are restricted to a range of $364-t$ days.
Added later:  A much nicer, much simpler, answer was given by drhab.  If neither today nor the next $t-1$ days is anyone's birthdays, then the $x$ birthdays are all in a range of $365-t$ days; if the $t$-th day is also not a birthday, then they're all in a range of $364-t$ days.  So the probability of the next birthday being in exactly $t$ days is the difference
$$\left({365-t\over365}\right)^x-\left({364-t\over365}\right)^x$$
This is clearly much easier to both explain and compute than the formula I gave.
A: Set yourself on day $0$. 
To be calculated is 
$P\left[\min\left(T_{1},\ldots,T_{x}\right)=t\right]$ where the $T_{i}$
are iid rv variables taking values in $\left\{ 0,1,\ldots,364\right\} $
uniformly.
Here: $$P\left[\min\left(T_{1},\ldots,T_{x}\right)=t\right]=P\left[\min\left(T_{1},\ldots,T_{x}\right)\geq t\right]-P\left[\min\left(T_{1},\ldots,T_{x}\right)\geq t+1\right]=\left(\dfrac{365-t}{365}\right)^{x}-\left(\dfrac{364-t}{365}\right)^{x}$$
Note that overlap is possible. You can have $T_i=T_j$ while $i\neq j$.
Here it is possible that $t=0$ (your are on a 'birthday'). If you don't want that then let the $T_i$ take values in $\left\{ 1,\ldots,364\right\} $
A: Taking the birthdays of $10$ people (assuming uniformity, independence, etc.) and that the time to the next important day is always positive (never $0$)   then 


*

*the probability that a particular friend has a birthday in the next $t$ days is $\frac{t}{365}$

*the probability that  that friend does not is $\frac{365-t}{365}$

*the probability that  none of the $10$ friends has a birthday in the next $t$ days is $\left(\frac{365-t}{365}\right)^{10}$

*the probability that  at least one of the $10$ friends has a birthday in the next $t$ days is $1-\left(\frac{365-t}{365}\right)^{10}$

*the probability that first of the friends' birthdays will be $t$  days from now is  $\left(\frac{366-t}{365}\right)^{10} - \left(\frac{365-t}{365}\right)^{10} $
This final expression is a decreasing function of $t$, so the probability for $t=1$ is about $0.027$ (slightly less than $\frac{10}{365}$), while the probability for $t=365$ is about $2.4 \times 10^{-26}$ (i.e. $\frac{1}{365^{10}}$)
A: Take a day $0 \leq d<365$ chosen at random, and denote $X=\{x_n:0 \leq x_n<365\}$ the set distinct important days ($|X|=x$).
The set of least waiting times is then $\{(x_n-d)\mod{365}\}=\{t_n:0 \leq t_n<365\}$ and has the same number of elements as the set of important days. Hence the probability of least waiting a correct amount of time is equal to the probability of choosing an important day.
We also want to exclude waiting times such as $t \geq 365$ because they obviously are not least with regards to a given day. So:
$$P(t)=\frac{x}{365} H[365-t]$$
Where $H$ is the unit step function.
