Do multiple polls with identical margins of error decrease the total margin of error? Is it possible to take multiple polls and combine their margin of error to produce a single, overall margin of error?
For an example, let's say we have two populations with the following characteristics, obtained from similar polls:


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*Population A prefers X over Y by 2% with a margin of error of 2%. The population size is 1000 people.

*Population B also prefers X over Y by 2% with a margin of error of 2%. The population size is also 1000 people.


Can we conclude that the 2000 people surveyed prefer X over Y by a margin of error less than 2%, or is the margin of error still just 2%? If possible, provide a mathematical proof of either conclusion.
My intuition collides with my friend's intuition, so I'm looking for an objective, conclusive answer. I'm searching through some of my old statistics material and general information on margins of error, but I'm not finding an answer to this specific question. (If it's not obvious, I'm interested in political polls, but I'm trying to keep the question objective.)

Depends on what you mean by margin of error. Usually this is some multiple $m$ of the standard error to achieve a pre-determined level of confidence (95%, 99%, whatever). In this case, for two outcomes, margin of error is a function of the number of votes $n$ and proportion of votes for one outcome which you can simply call $p$ (because $p_x=1-p_y$):
$$margin=m\sqrt{\frac{p(1-p)}{n}}$$
If you combine two polls with the same $p$ and $n$, your new margin will be:
$$m\sqrt{\frac{p(1-p)}{n+n}}=m\frac{1}{\sqrt{2}}\sqrt{\frac{p(1-p)}{n}}=\frac{1}{\sqrt{2}}margin$$
Which makes the combined margin of error 1.4%.
