multinomial distribution sampling I was reading the Wikipedia example from the Multinomial Distribution article:
In it they say:

"Note: Since we’re assuming that
  the voting population is large, it is reasonable and permissible to
  think of the probabilities as unchanging once a voter is selected for
  the sample. Technically speaking this is sampling without replacement,
  so the correct distribution is the multivariate hypergeometric
  distribution, but the distributions converge as the population grows
  large."

I am confused because doesn't sampling without replacement, indicate that the probabilities will change once a voter is selected from the sample? 
 A: As said in comments, this is sampling without replacement, which we can model with the (multivariate) hypergeometric distribution, see https://en.wikipedia.org/wiki/Hypergeometric_distribution#Multivariate_hypergeometric_distribution.  Then if the total population size (and each of the subpopulation sizes) becomes very large compared to the sample size, this can be effectively approximated by the multinomial distribution, which models sampling with replacement.
This can be shown formally using a limit argument, where we need Stirling's approximation for the factorial:
$$
  n! \sim \sqrt{2\pi n} \left( \frac{n}{e}  \right)^n
$$ where the tilde ($\sim$) means asymptotically equal, that is, the quotient of the two sides converge to 1 when $n\to \infty$.  Let the total population size be $N$ and the subpopulation sizes be $N_i$, we have $\sum_i N_i =N$. The sample size is $n$ and the sample numbers from the subpopulations is $n_i$, $\sum_i n_i=n$.
Now we assume that the population sizes goes to infinity $N \to \infty, N_i \to\infty$ in such a way that $\frac{N_i}{N} \to p_i$, $\sum_i p_i=1$. Taking limits, the sample size $n$ is a constant.
We can write the (multivariate) hypergeometric mass function as
$$
\frac{\prod_i \binom{N_i}{n_i}}{\binom{N}{n}}
$$
First, write out the binomial coefficients as factorials, rearranging and using Stirling's formula, noting that the $\sqrt{2\pi}$ and $(\frac1e)^\cdot$ things will cancel immediately, we get
$$
  \frac{\prod_i \frac{N_i!}{n_i! (N_i-n_i)!} }{\frac{N!}{n! (N-n)!} } = \binom{n}{n_1 n_2 \dotso}\times \prod_i \frac{N_i^{1/2}N_i^{N_i} }{N^{1/2}}N^N \times  \prod_i \frac{(N-n)^{1/2} (N-n)^{N-n}}{(N_i-n_i)^{1/2}(N_i-n_i)^{N_i-n_i}}
$$
Now,taking limits, $\frac{N_i}{N_i-n_i} \to 1$ and $\frac{N}{N-n} \to 1$. So the square root terms will disappear in the limit. What we have left then is
$$
   \binom{n}{n_1 n_2 \dotso} \times \prod_i\left( \frac{N_i}{N}  \right)^{N_i} \times \prod_i \left( \frac{N-n}{N_i-n_i}  \right)^{N_i-n_i}
$$
Taking limits, rearranging and cancelling we get the mass function for the multinomial distribution
$$
   \binom{n}{n_1 n_2 \dotso} \prod_i p_i^{n_i}
$$
