Statistical samples of different sizes I have 2 samples of different sizes, and want to compare them. I want to know if it's a problem that one of the samples gets too big.
Suppose initially both of them have a sample size of 500.
If i collect more data and one of them becomes 1000, and the other 10000.
I was told that it would be better to take 1000 from the larger one and compare them. 
Can anyone provide me clarification on why I can't compare them?
Obs: initially I was thinking in terms of measure. If ($A \pm \Delta$A) - ($B \pm\Delta$B) = $C \pm \Delta$C
, then no matter how small $\Delta$B gets, it shouldn't be a problem, even if we "waste" it's precision.
 A: 
I was told that it would be better to take 1000 from the larger one and compare them. 

That is bad advice.  So long as the data were collected properly, you should not discard perfectly good data.  Statistical models and tests can accommodate cases where the sample sizes in two groups are different, so there is no reason to discard perfectly good data just to reduce the sample size of the larger group to the same size as the smaller group.  If you discard data you will simply make a less accurate inference in your problem.
If you have not yet done your data collection then it is worthwhile learning a bit about optimal sample size allocation in statistical problems.  For statistical analysis where you are trying to estimate the difference in means between two groups, it is optimal to collect data in such a way that the sample sizes for the two groups are proportional to the standard deviations of values in the two groups (see below).  This means that you should generally collect more data from the group that you think will be more variable.  (Of course, you usually don't know the standard deviations until you collect data, but you sometimes have an a priori idea of which will be larger.)

Optimisation of sample size: If you are forming a standard confidence interval (at level $1-\alpha$) for a difference in population means $\mu_X - \mu_Y$ then this will have length:
$$L(n_X, n_Y) = 2 \cdot t_{\alpha/2, DF} \cdot \sqrt{\Big( \frac{1}{n_X} - \frac{1}{N_X} \Big) s_X^2 + \Big( \frac{1}{n_Y} - \frac{1}{N_Y} \Big) s_Y^2},$$
where $n_X$ and $n_Y$ are the sample sizes, and $N_X$ and $N_Y$ are the corresponding population sizes.  If we want to minimise the length of this confidence interval (i.e., maximise its accuracy) subject to the overall sample size constraint $n_X + n_Y = n$, then it can be shown that we get optimising values:
$$\hat{n}_X = \frac{s_X}{s_X + s_Y} \cdot n \quad \quad \quad \quad \quad \hat{n}_X = \frac{s_Y}{s_X + s_Y} \cdot n.$$
(These are not generally integers, so we would check the integer values around them for the optimising values over the integers.)  This tells us that in cases where we have a fixed overall sample size to allocate to two populations (at the same cost) we should set the sample sizes proportional to the expected sample standard deviations from those populations.
