Explanation of "Neutrality" of Non-Ranked Games on a Ballot (sticky)[edit source]

According to the rules, you can't rank your game at all. The modified Condorcet Ranked Pairs method that we use doesn't penalize an un-ranked game. Think of it like pluses and minuses: not ranking is like getting a zero, not a minus (nor a plus of course). In the grand scheme of things, one vote won't matter at all (hopefully! Otherwise, we have too few judges!). --David Artman 11:54, 7 September 2007 (EDT)

The main thing that matters is the margin, i.e. how many people preferred A to B or vice versa. If you assume that the designer of A prefers it to B and vice versa, then they cancel each other out and the margin stays the same. This assumption is the same as discarding all self-votes. I'm pretty sure this also does not affect the tiebreaker, but I'll leave that as an exercise for the reader. --Dougo 12:44, 7 September 2007 (EDT)

From Condorcet Ranked Pairs method : "unstated candidates are assumed to be equally worse than the stated candidates". Doesn't this mean leaving off your own game counts it as "worst" on that ballot, "penalizing that un-ranked game"? It's not a big deal, and "winning" is not the issue, but if "it is assumed that you'd give [your own game] #1", why not just allow such a vote in future? Such an assumption may even turn out to be incorrect! The hard work in setting all this up has been yours, David, and I am happy to accept any decision. Just wanted to point out something that struck me as "off". --Nycavri 14:12, 7 September 2007 (EDT)

I think it's the "equally worst" part that's sticking you, but that's not how it was explained to me. Consider the explanation above, by Dougo: We assume Your Game > My Game for you but My Game > Your Game for me; thus, each of the M>Y and Y>M get 1 point in the tally: a push. Repeat this across all other games (i.e. My Game > Every Game) and you get 1 point in every tally box (M>1, M>2, M>3... M>7) and, likewise, every other game gets 1 point in the tally boxes of It > Any (I>1, I>2, I>3... I>7). SO the difference, there, is again moot: all those tally boxes offset each other. I think. --David Artman 14:59, 7 September 2007 (EDT)

I just talked to Zarf (who ran the first four contests) and he says he used a modified Ranked Pairs algorithm that doesn't penalize missing entries. So that sentence on the Wikipedia page doesn't apply, because it's a slightly different algorithm. --Dougo 16:24, 7 September 2007 (EDT)

This discussion now mooted by a change in scoring method. Stored here because the new method was adopted for this IGDC.