Probability for lottery games related to the number of players. Suppose we have a lottery game with e.g the player having to choose 6 different numbers between 1 and 49. And the draw has 6 different numbers from 1 to 49.
The number of combinations of different 6-numbers is of course: N =
$\left( {\begin{array}{*{20}{c}}
   {49}  \\
   6  \\
\end{array}} \right)$
So to make it simpler of what i'm trying to say, let's have the completely equivalent game of the player having to choose 1 number from 1 to N. And the draw to have just one number drawn from 1 to N also.
Now let's have K players, each of them having selected 1 number(not necessarily different between them, we just don't know if they are different or not and how many are different etc) from 1 to N and to participate in the game.

So they have selected K numbers(from 1 to N).
And the lot has drawn 1 number(from 1 to N).

My question is what is the probability at least one player to win? Expressed differently:
What is the probability at least one of the K numbers played, to match the drawn numbers?

Let's have an example:
N = 3 numbers(1,2,3)
K = 2 players(numbers)
Without loss of generality let's suppose the number 2 is being drawn. And we have the following situations about the numbers the players had chosen:
•1)All numbers are different. 
Possible combinations({player-1 , player-2}):
{1,2} , {2,1} , {1,3} , {3,1} , {2,3} , {3,2}
•2) 2 similar numbers. 
Possible combinations({player-1 , player-2}):
{1,1} , {2,2} , {3,3} 
It's obvious that in 5 cases there is at least one win(the drawn number 2 is observed in player's choices) so the asked probability is 5/9.
I just want the general case with N and K. The probability has always the form X/(N^K). I just want to know X. It's not that easy.....
 A: If all the numbers are different, $K$ out of $N$ have been selected, so the chance that somebody wins is $\frac KN$.  If you know there is exactly one matched pair, then $K-1$ numbers have been selected and the chance that (at least) person wins is $\frac {K-1}N$  A similar argument applies to any known distribution of numbers-just count how many of the $N$ have been chosen.  
Added after comment:  If you are assuming the $K$ numbers are chosen independently, the chance that a given player loses is $1-\frac 1N$.  The chance that all the players lose is then $(1-\frac 1N)^K$, so the chance that at least one wins is $1-(1-\frac 1N)^K$
A: 
My question is what is the probability at least one player to win? Expressed differently:
What is the probability at least one of the K numbers played, to match the drawn numbers?

Assume the players choice of numbers are independent and identically uniformly distributed. (Each player is equally likely to pick any number and they don't influence each other's choice.)
The probability that any one player does not select the winning number is: $(N-1)/N$.
The probability that all $K$ players do not select the winning number is: $(N-1)^K/N^K$
Therefore probability that at least one player does select a winning number is: $1-(N-1)^K/N^K$
$$\mathsf P(X\geq 1) = \frac{N^K-(N-1)^K}{N^K}$$
EG: $K=2,N=3 \implies \mathsf P(X\geq 1) = \dfrac{3^2-(3-1)^2}{3^2} = \dfrac{5}{9}$

Remark Sometimes it is easier to work out the probability of the complement of the event you want.
