RobertJasiek 2005-01-21: I have offered a prize (a free copy of all volumes of my yet to be written Go Rules Encyclopedia) for the previous question if answered by a mathematical proof. It has been unclaimed for ca. 8 years now.
Background: Experience seems to suggest that under area scoring games with an odd number of not scored intersections (in sekis) in the game end position are scarce. Is this also so in theory? Due to this experience, komi for area scoring are 1.5, 3.5, 5.5, 7.5, 9.5, etc. but not 0.5, 2.5, 4.5, 6.5, 8.5, etc. (in even games on odd boards without or with an odd handicap) and usually in practice closest scores differ by 2, while under territory scoring they differ by 1. Can this experience be justified by theory or is it just a sign of how weak human play is?
Herman Hiddema: I would conjecture that the number of legal games with an odd number of non-scoring points is a lot higher than what we see in practice (and that result is pretty much meaningless with regard to weak human play). This is because legal games are not necessarily finished games (quite the contrary, actually). The vast majority of legal games will have many neutral points. Not because they are points in seki, but because both players have passed in an unfinished position. It would seem logical that the number of unfinished legal games is vastly larger than the number of finished legal games, and that the number of neutral points in an unfinished game has roughly a 50% probability of being odd. Perhaps these questions would be more meaningful if "finished" is added?
RobertJasiek: Maybe. One would like to see studies for either. (With "finished" you mean "both players want to pass in perfect play"?)
Herman: Yes, although "want to" is a little vague and might be better as "should" or "would".
Slarty: The question as posed is too horrific. It seems to me this is the same as asking about the percentage of "final positions" in well over 99.9% of cases.
RobertJasiek: It is not quite as tough because one can try to treat together whole subtrees or even classes of subtrees with the same winner. The question is very difficult for sure though. If it were less difficult, I would try to solve it by myself.