There are two methods of election that I often hear about: first past the post (whichever candidate gets the most votes wins) and the two-round system. In the two-round system, the top two candidates advance to the next round, where a simple majority then decides the winner. I am looking for the probability that these two systems will elect different people.
I am making several assumptions:
$1$. Voters choose who to cast their vote for randomly.
$2$. Voters will vote for the same candidate in both systems.
$3$. If the candidate a voter votes for advances into the second round of the two-round system, then they will cast their vote for the same person. If their candidate does not advance, then they will randomly choose one of the two remaining candidates.
Let's say there are $n$ candidates and $m$ voters, $m > n > 2$, although I am most interested in the limit of the probability as $m$ approaches $\infty$ for a fixed $n$. The first round for both systems will necessarily have to be the same. Let $P_i$ be the candidate with the $i$th most votes in the first round and $v_i$ be the associated number of votes. The elected person will be different if and only if $(1)$ $v_1 < \frac{m}{2}$ and $(2)$ $P_2$ wins overall (i.e. beats $P_1$) in the two-round system.
Each of the rest of the votes ($N = m - v_1 - v_2$) will be randomly parceled out to $P_1$ and $P_2$. $P_2$ will win if $v_2 + a > \frac{m}{2}$, where $a$ is the number of votes that $P_2$ gets from the "kicked out" candidates from the first round. Rearranging the inequality yields that $$a > \frac{m}{2} - v_2$$
Since $a$ follows a binomial distribution, the probability for this happening is $$\sum_{a = \frac{m}{2} - v_2}^{N} \binom{N}{a} \left(\frac{1}{2}\right)^{N} \tag 1$$
This means that $P_{n, m}(\mathbb{different} | v_1, v_2) = \sum_{a = \frac{m}{2} - v_2}^{N} \binom{N}{a} \left(\frac{1}{2}\right)^{N}$. This can be approximated by the normal distribution.
Now I need to find the joint distribution of $v_1, v_2$, which is the part where I am getting stuck. Once I have that joint distribution $f_{n, m}(v_1, v_2)$, I think I can do $$\sum_{v_1=m/n}^{m} \sum_{v_2 = \frac{m-v_1}{n-1}}^{\min(m-v_1, v_1)} \left( f_{n, m}(v_1, v_2)\sum_{a = \frac{m}{2} - v_2}^{m - v_1 - v_2} \binom{m - v_1 - v_2}{a} \left(\frac{1}{2}\right)^{m - v_1 - v_2} \right)$$ in order to get the required probability, but I am not completely sure. Any guidance on solving this problem?
Edit: As $m \to \infty$, I can use the normal distribution to get $$\frac{1}{2}\left(1+\operatorname{erf}\left(\frac{v_2-v_1+1}{\sqrt{2(m-v_1-v_2)}}\right)\right)$$ for the inner sum.
Let $V_i = v_i/m$ be the portion of voters $P_i$ gets. I can rewrite the double sum using integrals to get $$\int_{\frac{1}{n}}^{1}\int_{\frac{1-V_{1}}{n-1}}^{\min\left(1-V_{1},\ V_{1}\right)}f_{n}\left(V_{1},\ V_{2}\right)\ \cdot\frac{1}{2}\left(1+\operatorname{erf}\left(\frac{mV_{2}-mV_{1}+1}{\sqrt{2\left(m-mV_{1}-mV_{2}\right)}}\right)\right)dV_{2}dV_{1}$$
As $m \to \infty$, the argument of the $\mathbb{erf}$ function will approach $-\infty$, so $\mathbb{erf}$ will approach $-1$. This seems to suggest that the whole integral will be $0$, which doesn't make sense to me. I was instead expecting a positive probability. Can anyone explain whether I messed up somewhere or if $0$ is the actual probability for all $n$ as $m \to \infty$?