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Given a poll, where $N$ people were polled, and $n_i$ people voted for party $i$, so that: $\sum{n_i} = N$ If there are M parliament seats in total we can expect: $m_i = M\cdot\lim_{N\rightarrow\infty} (n_i/N)$ To be the number of parliament seats party $i$ will have.

My question regards the error involved in this prediction:

What is it's error distribution and what is it's variance?

If the number of seats were not finite, I'd say that the distribution would be Poisson and each poll value should be $n_i \pm \sqrt{n_i}$, but since the sum is given, It would seem the errors must be correlated in some way.

Any ideas?

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    @Henry - of course. The actual system used in my country is a variant of the [D'Hondt method](http://en.wikipedia.org/wiki/D%27Hondt_method), And I wouldn't even try writing the formula down. I just was trying to understand what the actual polling error is, as the papers don't seem to be very keen on explaining what "4.5% error" refers to.2012-12-29

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If the polling population is infinite, the $n_i$ follow a multinomial distribution. If the population is finite, the $n_i$ follow a multivariate hypergeometric distribution. The Wikipedia pages have the variances and covariances for both cases.

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    Now that I think of it, the multivariate hypergeometric distribution gives me the opposite of what I'm looking for - It gives the chances of getting $n_i$ voters in the poll who will vote $i$ given $N_i$ total $i$ voters. What I'm looking for is the opposite: The chances of there being $N_i$ total voters, given $n_i$ voters in the poll who will vote $i$. How can I "invert" this distribution?2012-12-29