Estimate probability of selecting the same card more than once Suppose I have a deck of N=1,000 cards where each card is a unique number from 1 to 1,000. 



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*Draw 1: draw n=10 cards at random.  Put them back

*Draw 2: draw n=10 cards at random.  Put them back

*...

*Draw i: draw n=10 cards at random.  Put them back

On the ith draw, what is the probability that any of the 10 cards was selected at least once in previous draws?
I've solved this problem through simulation. But I can't figure out the probabilistic approach.

Update
Thank you for all your contributions. My original question was not properly articulated, resulting on two different interpretations.
What I'm seeking is 1-P[drawing distinct cards on each draw]
 A: Let's say we are in the $i$-th draw, and chose 10 cards. In the previous $i-1$ draws, we want any of them to be selected at least once. We'll find the probability of those cards not being selected in any of the previous $i-1$ draws, and subtract from $1$. The probability of choosing $10$ cards that doesn't contain any of the $10$ cards we specified is $p={990 \choose 10}/{1000 \choose 10}$.
If we assume independent draws as usual, not choosing any of the cards we specified in total $i-1$ draws become $p^{i-1}$. The probability we ask for is $1-p^{i-1}$.
Numerically, we have $p\approx 0.904$, and if we are talking about e.g. $5$ draws, the probability we want is $\approx 0.332$.
A: The answer provided by "gunes" is a good start. If  the question is to determine the probability that only the ones at draw $i$ to be repeated from the previous ones the answer would be correct. However, if we want to see the probability that draw $i$ and any draws before have any repeated cards, the solution will be slightly different. At $i = 2$ the solution will be the same as above:
$$ 1-\left [ \frac{\binom{N-n}{n}}{\binom{N}{n}} \right ]$$
However, for $i=3$ it will be 
$$ 1-\left [ \frac{\binom{N-n}{n}}{\binom{N}{n}} \right ]. \left [ \frac{\binom{N-2\times n}{n}}{\binom{N}{n}} \right ]$$ 
because during the third draw, we should exclude the ones that have been selected in first two draws and not repeated. Thus, the choices we will have are among the ones that have not been selected before ($N-2\times n$). With these, the general formula after pulling $i$ samples will be 
$$ 1- \Pi_{j=1}^{j=i-1} \frac{\binom{N-n\times j}{n}}{\binom{N}{n}}  $$
A: Below is my code that regenerates the results of my analytical solution:
$$ 1- \Pi_{j=1}^{j=i-1} \frac{\binom{N-n\times j}{n}}{\binom{N}{n}}  $$. The results of simulation and the analytical formula are close enough. For example at i= 6, the simulation gives 0.789 and the analytical formula gives 0.785.
n_runs <- 5000
N <- 1000
d <- data.frame(id = 1:N)
n_draws <- 6
n_samples <- 10

draw_sim <- c()

for(j in 1:n_runs){
  draw_all <- c()

  for (i in 1:n_draws){
    temp_draw <- sample_n(d, n_samples, replace = FALSE)
    draw_all <- bind_rows(draw_all, temp_draw)

  }

  all_repeats <- draw_all %>% 
    group_by(id) %>% 
    mutate(freq = n()) 

  repeats <- all_repeats %>% 
    filter(freq>1)   


  repeats <- length(unique(repeats$id))  %>% 
    as.data.frame() %>%
    rename( Freq = '.')

  draw_sim <- bind_rows(draw_sim,repeats)

}

# Probability: 
prob_draw <-  nrow(filter(draw_sim,Freq>0)) / nrow(draw_sim)

