hoshi911:
By the way, in case black is the underdog, then of course there are three inside liberties counting for black (solution is correct here). However, as said in my last message, when white is the underdog, then there is only one inside liberties counting for white because of black's forcing move.

hoshi911:
The solution counts 3 inside liberties for the underdog. This is wrong. If at all, as it stands, I can count only two inside liberties. However, if for instance black as the favorite plays J10, this is a forcing move and white has to answer with K9 (otherwise black plays K9 next and lives with two eyes). So as a matter of fact, the three points J10, K9 and K10 count only as one inside liberty for white, the underdog. The results are also wrong according to the wrong counting.