how to apply 'margin of error' in opinion poll to small answer buckets? Australia has preferential, aka alternate voting, at Federal elections. Opinion polls about voting intention ask two questions: which party do you intend to vote for, and if that is a 'minor' party, which of the two major parties do you intend to preference? Therefore, results of these polls indicate a "two-party preferred vote", and also results for the individual parties.
If a properly-conducted, random-sample opinion poll of a population of 20m has a quoted 'margin of error' of 3%, and if respondents are asked to choose from one of ten political parties but also their preference for one of two major parties, the margin of error is usually applied to the two party vote.
How do we apply the margin of error to a result that minor party A will attract 10% of the vote? Or 1% of the vote?
 A: I don't know about Australian pollsters, so I'll discuss common practices in the US. For polls that sample a thousand prospective voters or more the 95% 
confidence interval
$$\hat p \pm 1.96\sqrt{\frac{\hat p(1 - \hat p)}{n}},$$
is good enough. Here $\hat p = X/n,$ where $X$ counts favorable responses for
a particular party and $n$ is the sample size.
If $p$ is in in the neighborhood of  $1/2,$ (say between .3 and .7) some
pollsters just use $\hat p \pm 1/\sqrt{n}.$ If $p = .5$ then the margin of
error $M$ simplifies by conflating $\sqrt{4}$ with $1.96:$
$$M =1.96\sqrt{\frac{\hat p(1 - \hat p)}{n}} = 1.96\sqrt{\frac{1}{4n}} \approx \frac{1}{\sqrt{n}}.$$
This is not far off if $p = .3$, because then $p(1-p) = .21$ (instead of $.25$ for $p = .5$).
Because most US elections have $.3 < p < .7,$ this gives rise to the estimating
the required sample size as $n = 1/M^2.$ For example if you want $M = 3\% = .03$
you have $n = 1/(.03)^2 = 1111.$ 
The largest $M$ for given $n$ results from $p = .5.$
So, for values of $p$ farther from 1/2, $M$ is smaller. US pollsters usually
use the $M$ for values of $p \approx 1/2$ even for very small (or large)
values of $p,$ probably because they understand nonsampling error may be relatively more serious for extreme $p$.

Notes: (a) In the above I have not always distinguished between $p$ and $\hat p$ because
under the square root sign one really would like to be able to use $p$, not $\hat p$. And for large $n$,
there is not much difference.
(b) All of this applies to sampling error only. It does not take into account
nonsampling error such as selection bias in sampling, nonresponses, etc.
(c) As long as the sample size $n$ is less than about 10% of the population
size, no adjustment for population size is necessary. This is almost always
the case in nationwide polling.
(d) If you are looking at the difference between two candidates or parties,
then you should double the margin of error. If $M = 3\%$ then poll
percentages 48% vs. 52% are a 'statistical dead heat' because they
are within 6% of each other. (Especially, if there are only two meaningful parties, every person for A
is also one person against B. So the percentages 48% and 52% are not
independent values.)
