Does partitioning a sample space change probability of events? Suppose I have 10 white balls and 1000 black balls.
Considering that picking each ball is equally likely, the probability of picking a white ball comes out to be $\frac{10}{1010}$, now if I keep all the white balls in box-1 and all the black balls in box-2, will this constitute as valid partitioning, as now the probability of selecting a white ball becomes $\frac{1}{2}$ i.e. equal to the probability of selecting box-1?
Although sample space and measure of that space (probability) are different concepts, I feel that this view of partitioning is wrong, as it violates the following equation*:

If U=[A1,...,An] is a partition of sample space,S and B is an arbitrary event then, 
  P(B)=P(B|A1)P(A1)+...+P(B|An)P(An)

So according to this the probability of withdrawing a white ball should have remained the same and partitioning would have given us just another way to calculate it.
$*$ from page number 32, of Probability, Random Variables and Stochastic Processes ($4^{th} edition$) Book by Athanasios Papoulis and S. Unnikrishna Pillai.
 A: The probability of selecting box-1 is not 1/2 but 10/1010. This corresponds to the factors $P(A_1),\ldots,P(A_n)$ in the quoted expression.
Using the correct probabilities for selecting box-1 or box-2, we get "probability of selecting a white ball" = 1 * 10/1010 + 0 * 1000/1010 = 10/1010, as expected.

The important point is that, when partitioning a sample space, the parts of the partition are weighted by their probabilities.
A: @jflipp gave the correct answer, but I'll offer another perspective since this has confused me in the past as well.  When you physically divide the white and black balls into two separate boxes, you have gone way beyond partitioning the sample space.  You have actually created new sample space where the elements are Box 1 and Box 2.  Now the probability of selecting Box 1 = probability of selecting Box 2.  Since your sample space is now different, the outcome of the experiment is different.  Since you've essentially created a new "experiment", the outcomes will be as you stated (50% chance of picking white or black).
