By definition, there are seven rounds in an 8-player round robin and 14 rounds in an 8-player double round robin. If that's not how many rounds there are with eight players, then you don't have a round robin.
How many rounds are in an 8 player round robbin?
It probably does not matter how the competition is done unless you are in open section. The fact is there are artificial rating brackets. You are paired with people a bit stronger than you, and if you are the strongest, hopefully you are not too strong you are put in the next bracket. People go to tournaments for the environment, as well as the opportunity to really challenge their brain in OTB deep chess.
If someone is going for the purpose of winning, their strategy is simple: study secretly until you are 400 points stronger than your current rating, and hope no one else there did the same thing. People who follow that stategy will either soon find themselves in the open sections, or discover that independent study does not work as well as OTB exercise, and that lower rated sections are more about exercise than an actual goal.
I agree with the second poster - a round robin means you play everyone, by definition.
If you're thinking of a swiss, those are paired by ratings. 1v5, 2v6, 3v7, and 4v8 is the first round. Let's say the player 5 guy is underrated and is actually the second strongest player, and draws #1. After 2 rounds, player 2 will have a score of 2, and players 1 and 5 will both have a score of 1.5. But if there is a third round, either player 1 or player 5 will then play player 2. So if they beat him, player 2 will have a score of 2, and players 1 and 5 will have a score of 2.5. So it all works out assuming you have at least 3 rounds.
If you don't have at least 3 rounds for your 8 players, that was kinda silly of you. Otherwise, any 2 players could have just both won their first 2 games and never played each other, and you'd end up with a tie. For something like a candidate's match, they'd almost certainly at LEAST have enough rounds for this to not be a problem. Lots of times they do a true round robin - even if you lost the first 6, you gotta play that 7th game.
It would be nice if everyone played everyone. In fact, it would be nice if everyone played everyone as white and black. But since that would take too many rounds, often tournaments just have to assume that if Player A knows how to exploid Player B's weaknesses while playing as Black, and Player B was able to exploit Player C's weaknesses while playing as Black, then Player A probably is better than Player C, and we can save time by not matching them.
What if the first and second best players play each other in the first round and draw? There are two 1/2's, three 0's, and three 1's. Most likely next round they will match the 1's, match the zeros, and have a 0 and 1 play each of the 1/2's. The strongest two would then have 1.5's, but it is still possible for two 2's to emerge from that round as well.
Would they have 4 rounds? Or would they just have a second round and pair the two 2's and give the winner first place, and pair the orginal strongest players who have 1.5's, and let them fight for 2nd place?
I was reading about the world championship. Kasparov was very picky about who could challenge him. Most matches had several draws, and were won by a few games lead. Who won against whom seemed to go around in a circle, making the current champion not want to play everyone, but only play the winner of a pre-run off.