Minimalist example of voter profiles yielding different outcomes Let the set of alternatives be $\left\{a,b,c,d\right\}$.

Let there be $n$ types of voter profiles, each type with a different strict ordering over the set of alternatives.

The number of voters belonging to the profile types need not be the same, let us denote these by $v_1,v_2,...,v_n$.

What is the minimal number of $n$, for which you can construct the voter profiles in such a way that the plurality rule (a.k.a. first-past-the-post), simple run-off (election with two rounds) rule, elimination rule and the Borda count all yield different outcomes? (Assume non-strategic voters. Outcomes should not be ties.)



EDIT: 
Plurality rule has one round of voting and whoever has the most votes wins.  Example: United Kingdom parliamentary elections 

The simple run-off rule has two rounds of voting. The two candidates with the most votes in round 1 advance to round 2. In round 2 voters vote for their favourite remaining candidate, and whoever has the most votes wins. Example: French presidential election 

The elimination rule is similar to the run-off rule but has three rounds (number of alternatives minus one). In each round the candidate with the least votes is eliminated (figuratively). Others proceed to the next round.  Example: Academy award (Oscar) vote



Minimalist guesses ("here are some profiles with $n = 4$") are fine, ideal solution would provide a proof that $n$ is minimal.
First, let me state that this is a beautiful problem!

Here is a proof that $n=4$. To prove that $n \leq 4$, consider the following example:


$m+4$ voters have preference $a \succ d \succ b \succ c$
$m+3$ voters have preference $b \succ d \succ c \succ a$
$m+2$ voters have preference $c \succ d \succ b \succ a$
$m$ voters have preference $d \succ c \succ a \succ b$


where $m \geq 2$. I leave it to you to check that $a$ is selected by the plurality rule, $b$ by the simple run-off, $c$ by the elimination rule and $d$ by the Borda procedure.

Now, let us prove that $n&gt;3$. The cases $n=1$ and $n=2$ are trivial, so let us assume $n=3$ and write $v_1,v_2,v_3$ for the number of voters in  the three voter profiles. Without loss of generality, let us assume that $v_1 \geq v_2 \geq v_3$, that $a$ is selected by the plurality rule, that $b$ is selected by the run-off election, that $c$ is selected by the elimination rule and that $d$ is selected by the Borda count.

The outcome of the plurality rule shows that $a$ is most preferred by voters of group 1 (since it wins with no ties). If $a$ were also preferred by voters of group 2, $a$ would win the run-off election, which is a contradiction. Thus $a$ is most preferred by voters of group 1, $b$ is most preferred by voters of group 2, and $v_1&gt;v_2 \geq v_3$, the latter inequality being strict if $b$ is not most preferred by voters of group 3. But then it is clear that both $a$ and $b$ survive the first two rounds of the elimination rule, i.e. that $c$ and $d$ are eliminated, which is a contradiction.