Imagine that we line up the students at random, and assign the first $4$ to one group, the next $4$ to another, and the last $4$ to another.
We can analyze as follows. We have $3$ copies of the letter $H$, and $9$ copies of the letter $M$. We make a $12$-letter word. What is the probability there will be an $H$ among the first $4$ letters, and an $H$ among the next $4$, and an $H$ among the last $4$?
There are $\binom{12}{3}$ equally likely ways to place the $3$ $H$'s.
The number of choices with one $H$ in the first $4$, another $H$ in the next $4$, another in the next $4$ is $4^3$.
Our required probability is therefore $\dfrac{4^3}{\binom{12}{3}}$.
Remark: Your number $\binom{12}{8}\binom{8}{4}\binom{4}{4}$ counts the number of ways to divide our $12$ people into three uniformed teams, the Blues, the Whites, and the Reds. That's fine.
Now we count the number of ways to divide into uniformed teams with an Honours student on each team. The Honours student of the Blues can be chosen in $\binom{3}{1}$ ways. The rest of the team can be chosen in $\binom{9}{3}$ ways. For every such choice, the Honours student in the Whites can be chosen in $\binom{2}{1}$ ways, and the rest of her team in $\binom{6}{3}$ ways. Now we are finished, although we can tack on a decorative $\binom{1}{1}\binom{3}{3}$ for a total of $\binom{3}{1}\binom{9}{3}\binom{2}{1}\binom{6}{3}\binom{1}{1}\binom{3}{3}$.
Now for the probability, divide.
After briefly reviewing sampling distribution, as K.defaoite suggested, I dismissed it because I got the feeling that it either was not going the right way to answer my question, or the road to get there would be too long. But it pointed me to binomial distribution, what pointed to hypergeometric distribution, and after some thinking, I believe I got the right answer. I would very much appreciate if you could verify it.
I know I framed the question in $n$ and victory margin $m$, but it is easier to answer considering the number of votes in favor (call it $x$) and the number of votes against (call it $y$), so:
$$
x = n * m \\
y = n - x \\
x \ge y
$$
A voter drawn at random from $P$ has a probability $p$ of supporting the winner. This means that the result is correct only if $p > 0.5$, otherwise the other candidate would have won if everybody was allowed to vote.
We don't know the probability $p$, but we know, from the election result, that out of $n$ voters drawn at random, $x$ voters supports the winner and $y$ voters do not.
Let $f_{x,y}(p)$ be the probability of a given $p$ yielding the known $x$ and $y$ election result. For every possible $p$, the ones with greatest $f_{x,y}(p)$ are the most likely to be the real proportion of voters in $P$ that supports the winner.
If we assume $f_{x,y}(p)$ is proportional to the likelihood of $p$ being the real support for the winner (and I think it is, I am just not sure how to prove it), then the probability of the election being wrong is given by:
$$
l(x,y) = \frac{\int_{0}^{0.5}f_{x,y}(p)\, dp}{\int_{0}^{1}f_{x,y}(p)\, dp}
$$
Which is to say the more cases of $p < 0.5$ are capable of producing the know result of $x$ and $y$, bigger the chance of the election results being wrong, because it is only correct if $p > 0.5$.
Now we only need to know $f_{x,y}(p)$ to be able to calculate $l(x,y)$.
The case for $|P| \gg n $:
It wont make much of a difference for the proportion of supporters if 10 or 10,000 voters are removed out of 300,000,000, thus, for these cases, we can approximate the chance of each one of the $n$ voters to be a supporter of the correct winner as independent of each other, and use the binomial distribution, in which case $f_{x,y}(p)$ is given by:
$$
f_{x,y}(p) = \frac{(x+y)!}{x!y!} p^x (1-p)^y
$$
By plugging into the definition of $l(x,y)$, we get:
$$
l(x,y) = \frac{\int_{0}^{0.5} p^x (1-p)^y\, dp}{\int_{0}^{1} p^x (1-p)^y \, dp}
$$
The numerator bears striking resemblance for the incomplete beta function, as the denominator to the full beta function, so much that it can be written as:
$$
l(x,y) = \frac{B(0.5; x+1,y+1)}{B(x+1,y+1)}
$$
Now we can see a striking resemblance to the regularized incomplete beta function, so much that it can be written as:
$$
l(x,y) = I_{0.5}(x+1, y+1)
$$
Turns out that the regularized incomplete beta function is the CDF for the beta distribution, thus I think we can say the probability of a given result in a sampled election have its voters chosen from a population with some support level to the winner is given by the beta distribution.
In practice, $l(x,y) = 0.5$ for $x = y$, and quickly drops to 0 as the difference between $x$ and $y$ and their magnitude increases.
The case for small $|P|$
If you want to consider the statistical dependency between draws from $P$, you can instead use the PMF of the hypergeometric distribution as $f_{x,y}(p)$:
$$
f_{x,y}(p) = \frac{\binom{p|P|}{x} \binom{(1-p)|P|}{y}}{\binom{|P|}{n}}
$$
Notice that this function is now discrete, and is only defined for where values of $p$ where $p|P| \in \mathbb{N}$. This means that you'll need to replace the integrals for summation over every valid discrete value of $p$ within the integration interval.
Best Answer
Yes, your explanation is absolutely correct. However, I would like to present even a simpler explanation for this problem.
Imagine having a dinner table with 60 seats. The table is divided into 2 sections of 20 seats and 40 seats. See the diagram below
Your problem essentially is to place a particular candidate to 20-seat-section. And you don't care how the rest of other candidates are seated.
Probability of the person getting seated to 20-seat-section is simply $\frac{20}{60}$ = $\frac{1}{3}$