Borda Count vs Average Ranking

Solution 1:

They are equivalent.

As pointed out in the comments, the lowest average rank always has the highest Borda score. If there are $N$ voters and $M$ candidates then a candidate with average rank $r$ will have a Borda score of exactly $N(M-r)$.

So why use the count over the average?

First off, note that neither is much easier to compute, as we can quickly compute either of these measures if we have the other, because they are in a direct relation.

Computationally the count is perhaps a little simpler. It allows you to work only with natural numbers, which is nice. Also, ranked preference ballots tend to allow a voter to leave alternatives unranked. If we want to compute an average we have to tally each unranked alternative as having rank $M$, whereas the count can simply ignore them (since they receive 0 points).

Personally I also find the count simpler to understand and explain than the average, but of course you may very well disagree.