# Borda Count vs Average Ranking

Recently, I was explaining to my high school class what the Borda Count was. We had taken a class survey on something and everyone ranked their choices in order of preference. I calculated the Borda Count scores and showed the class the official class ranking of their choices. One student asked "That seemed a bit complicated...couldn't you have just taken the average ranking of each option?"

I went back and did her suggestion, and just looked at each choice's average ranking, and it turned out to lead to the same outcome as just doing the Borda Count. So now I don't know if this is equivalent to the Borda Count, or is there some obscure situation where these different methods lead to different societal ranking. Are they equivalent or is there some counter example? I feel like if they were just equivalent, then this would be the standard definition of the Borda Count process, right?

• Yes, they are equivalent. Let $B$ be the Borda count of any one among $n$ candidates and let $r$ be that candidate's average ranking, where $m$ votes have been cast. Then $rm = nm - B$. So the lowest value of $r$ (best average rank) corresponds to the highest $B$. Dec 21, 2021 at 3:13
• Thanks for the short and easy reply. It does seem obvious, but my only hesitation was my last sentence. If they are equivalent, then why not use this as the definition instead of having to tally how many 1st place votes, 2nd place votes, 3rd place votes, etc. a candidate got? Dec 21, 2021 at 3:33
• You would need those numbers to compute the Borda count, wouldn't you? Dec 21, 2021 at 14:08
• @HansEngler I'm saying: if these methods are equivalent, then why not say the Borda Count is the average ranking instead of saying the Borda Count is a longer calculation involving counting individual rankings each candidate got? Actually, it makes me wonder why Borda Count even exists now if we could just report the average rank instead... Dec 21, 2021 at 20:19
• To compute the average rank of each candidate, you need to look at each ballot. It's the same computational effort as computing the Borda count. No difference whatsoever. Try it out with a simple case, say 3 candidates and 5 voters. Dec 21, 2021 at 20:40

## 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.

• I see now, thank you! To me, either calculation seemed the same, but you bring up a good point about 1) staying with real numbers and 2) handling unranked candidates. Jan 24 at 20:30