Probability of 'same lottery numbers drawn twice'? In some lottery one can buy a ticket by choosing seven distinct numbers each of them from numbers ${\{1, 2, \dots, 45}\}$ (so $1/(45379620)$ is the probability to win the first prize). 
Let this week the numbers draws be $a_1,a_2,\dots, a_7$. What is the probability that the next draw would be the same numbers, regardless of their order? 
Thank you. 
EDIT - Balls of lottery do not have memory of course, but why appearance of different 7-number-s happen rather than same 7-number? Suppose that we throw a coin 1000 times; is the probability of 1000 times only tail same as probability of any other mode?
 A: Given that this week's draw has already happened and that it doesn't influence the next one, the probability of the numbers being the same will be (again) 
$$\frac{1}{45\choose 7}=\frac{1}{45379620}$$.
A: As man_in_green_shirt has explained,
probability remains $\dfrac{1}{45\choose 7}=\dfrac{1}{45379620}$
You have remarked "..intuitively it seems to be much more difficult appearance of same numbers."
Please understand that what the result means is:
the same set of #s is expected to come up only once in 45379620 further lotteries !
A: As man_in_green_shirt notes, the probability remain the same. But the probability of picking two sets $a_1, a_2, \ldots, a_7$ and $a_1', a_2', \ldots, a_7'$ that are identical over two draws is equal 
\begin{equation}
\frac{1}{45379620}\frac{1}{45379620} = \frac{1}{2059309911344400},
\end{equation}
a much smaller probability. The first situation concerns picking one unique set of integers, the probability above is repeating them twice, given that they are independent. 
