If you take a peek at the results here, ( http://warlight.net/MultiPlayer.aspx?TournamentID=1315 )it seems like in the 'North vs South' distribution, South tends to win more often.

I'm curious if there are any ways (for fizzer, perhaps) to pull statistics for distribution modes in general, or for this one, to see if my hunch is accurate.

you are right a bit the south has more territories to choose from and higher total power.South definately wins in full distribution or unlimited distributions

If I counted well:
North: 21 wins
South: 41 wins

H0: pN = pS = 0.5, H1: pN =\= pS =\= 0.5

Under H0, the number of wins follows a binomial distribution: ~B(62,0.5)
The p-value (probability that a situation like this would appear when H0 is true)
= 0.01513408 (1.5%)

Conclusion: we don't really have "proof" that the distribution is skewed (the p-value is not unreasonably small), it is however unlikely that it isn't skewed (i.e. we expect this situation would only appear in 1.5% of the tournaments with this distribution if there really was no skewed distribution, which is relatively small).

Remarks: theoretically, this would only be true if in every game both teams were equally strong, which is not the case. As a result, it is statistically more difficult to actually disprove H0 (the distributions aren't skewed).

Final interpretation: Most likely the distributions are actually skewed, it would be better though if there are more tournaments with this distribution (I know at least 1) so there is more data.

The fact that some teams are stronger doesn't change anything.
If north/south were balanced, you would expect each to win half the time (if only because the stronger team is assigned to north half the time).

FD, it does change a lot, since the distribution under H0 is not a perfect binomial.

(We do not have 62 independent bernouilli experiments each with p=0.5, but rather 62 experiments, each bernouilli with a different p (with average of all p's expected to be 0.5) and which are also slightly dependent (same teams))
Therefore the p-value (calculated to 1.5%) is not exact, but assumingly an upper bound.

(statistician talking here, you may believe me)