Your are correct of course, I should have been more careful. Thinking about the pairs of players who play together, some pairs will play together only m times (or not at all in the case of m=0), while other pairs will play together as many as n times, where m<=n. Ideally we would like m=n for a fair tournament, but this is not possible. My solution makes m and n as close together as possible, while your solution maximizes m. They are two different ways of thinking about making it as fair as possible. The cost of having having m=1, is that n must be 4 for one pair, while the cost of having m and n as close as possible is that m=0 for three pairs. I don't believe that there is any other useful schedule in between, so you should go for one or the other. Hope that helps.
