joint probability of two hypergeometric trial sets Let's say I have one experiment where I'm drawing 50 balls from a bag of 1000 without replacement. The bag contains 50 white balls - the rest are black.  It's trivial to calculate the cumulative probability of drawing at least x white balls. 
Now, I repeat the experiment again and draw another 50 balls from a new bag of 1000 (again, 50 white, 950 black). 
I'm interested in the probability of drawing at least 2x balls in both of these experiments.  I thought that should be roughly equivalent to one pooled experiment, drawing 100 balls out of 2000 (basically combining the 2 bags together), but it doesn't appear to be the case from some simulations I ran.  This seems like a pretty simple question but I haven't seen any relevant discussions online.
 A: Since you are not replacing the balls, you start the second bag with a full set of 1000 balls whereas continuing to draw from the first you would start the second set of draws with only 950 with the distribution of black and white determined by previous draws as compared to 1000 distributed 950 black and 50 white independent of the first 50 draws.  See the difference? It has an effect on the probability of the various possible outcomes.
A: Consider the following scenarios to see why the probabilities you describe need not agree:
The Two Bag Scenario: We have two bags. Let's call them bag 1 and bag 2 for convenience. Each bag contains $N$ balls of which $K$ are white and $N-K$ are black. From each bag we draw $n$ balls without replacement. Let $X_1$ be the number of white balls drawn from the bag 1, and $X_2$ the number of white balls drawn from bag 2. The probability of drawing at least $2x$ white balls in total is given by
\begin{align*}
  P(X_1+X_2 \geq 2x) &= \sum_{k=2x}^{2n} P(X_1+X_2=k),\\
      &= \sum_{k=2x}^{2n} \sum_{i=0}^{k} P(X_1=i, X_2=k-i),\\
      &= \sum_{k=2x}^{2n} \sum_{i=0}^{k} P(X_1=i) P(X_2=k-i), &  \text{by assuming independence}\\
      &= \sum_{k=2x}^{2n} \sum_{i=0}^{k} \frac{\binom{K}{i} \binom{N-K}{n-i}}{\binom{N}{n}} \times \frac{\binom{K}{k-i} \binom{N-K}{n-k+i}}{\binom{N}{n}},
\end{align*}
where $\binom{K}{i}\binom{N-K}{n-i} / \binom{N}{n}$ is the probability of drawing $i$ balls in $n$ draws without replacement from bag 1, and $\binom{K}{k-i}\binom{N-K}{n-k+i} / \binom{N}{n}$ is the probability of drawing $k-i$ balls in $n$ without replacement from bag 2.
The One Bag Scenario: Now consider a third bag, call it bag 3, containing $2N$ balls of which $2K$ are white and $2N-2K$ are black. Let $X_3$ be the number of white balls drawn from bag 3 in $2n$ draws without replacement. The probability of drawing at least $2x$ white balls is given by
\begin{align*}
  P(X_3 \geq 2x) &= \sum_{k=2x}^{2n} P(X_3=k),\\
      &= \sum_{k=2x}^{2n} \frac{\binom{2K}{k} \binom{2N-2K}{2n-k}}{\binom{2N}{2n}},
\end{align*}
where the summand is the probability of drawing $k$ white balls in $2n$ draws without replacement from bag 3.
A Simple Example: To get an intuition for why these two formulas need not generally agree, consider the simpler setting in which we are only interested in comparing $P(X_1 + X_2 = 2x)$ with $P(X_3=2x)$ when $n=1$ (i.e. this is like your $x=1$ case). In other words: Is the probability of drawing a white ball from bag 1 and drawing a white ball from bag 2 the same as drawing two white balls from bag 3?
The probability of drawing a white ball from bag 1 and drawing a white ball from bag 2 is given by
$$P(X_1+X_2 = 2) = \frac{K}{N} \times \frac{K}{N} = \frac{K^2}{N^2} $$
since, assuming independence, the probability of drawing a white ball from either bag is just $K/N$. On the other hand, the probability of drawing two white balls without replacement from bag 3 is given by
\begin{align*}
P(X_3 = 2) &= \frac{2K}{2N} \times \frac{2K-1}{2N-1},\\
&= \frac{K(2K-1)}{N(2N-1)}.
\end{align*}
So $P(X_1+X_2 = 2)$ and $P(X_3 = 2)$ will differ in general. The intuitive reason being that it makes a difference when one neglects to put the first drawn ball back into bag 3 after having drawn it. This exemplifies why one cannot combine hypergeometric trials in the manner described in the question.
