Intuitive reason why the variance of a hypergeometric variable is smaller than the variance of the corresponding Binomial variable I am looking for the most cleanest and most intuitive possible description of why the variance of the number of successes when sampling without replacement is smaller than the variance of the number of successes when sampling with replacement.  I desire a verbal explanation, perhaps combined with an illuminating example.
Notes:  I can derive the formulas for both variances and see that one is obtained from the other by a correction factor.  I only want a clear description of the "intuition" behind the fact that this happens.   
 A: Consider the sequence of the $k$ trials for a hypergeometric r.v. $X$. As each success occurs, the probability of a success in the next trial is reduced. Similarly, each failure that occurs reduces the probability of subsequent failures.
Thus, outcomes that deviate by large amounts from the mean are made less probable compared to the corresponding experiment using "sampling with replacement", where $X$ has a binomial distribution.
Since variance is a measure of the expected deviation from the mean,  this means the hypergeometric distribution has a smaller variance than the corresponding binomial distribution.
Example:
An urn contains $7$ red balls and $3$ blue balls and we draw $2$ balls from it.
Hypergeometric (sampling without replacement):
$$
\begin{array}{ll}
P(R_1) = 7/10 \qquad\text{ and } & P(R_2\mid R_1) = 6/9 \lt 7/9=P(R_2\mid B_1) \\
P(B_1) = 3/10 \qquad\text{ and } & P(B_2\mid B_1) = 2/9 \lt 3/9=P(B_2\mid R_1) \\
\therefore\quad P(R_1,R_2) = \dfrac{7}{10}\dfrac{6}{9} = \dfrac{42}{90} \\
\quad\quad P(B_1,B_2) = \dfrac{3}{10}\dfrac{2}{9} = \dfrac{6}{90}. \\
\end{array}
$$
Binomial (sampling with replacement):
$$
\begin{array}{ll}
P(R_1) = P(R_2) = 7/10 \qquad\text{ and } & P(R_1, R_2) = \left(\dfrac{7}{10}\right)^2 \gt \dfrac{42}{90} \\
P(B_1) = P(B_2) = 3/10 \qquad\text{ and } & P(B_1, B_2) = \left(\dfrac{3}{10}\right)^2 \gt \dfrac{6}{90}. \\
\end{array}
$$
