I'm not sure the conditional probability argument explains it. Specifically, given A=H, the probability of (H,H) is the same as the probability of (H,T) so that

I'm not sure the conditional probability argument explains it. Specifically, given A=H, the probability of (H,H) is the same as the probability of (H,T) so that alone shouldn't inform your guess. I believe the core mechanism at play here is that coordination eliminates the possibility of one player being right and the other being wrong, because one can partition the coin outcomes into "We have the same result" and "We have different results"; by agreeing in advance to only guess "we have the same result" every time you eliminate the "I'm right, you're wrong" and "You're right I'm wrong" failure modes (50% of the probability space under random guessing). One could accomplish the same by always guessing the opposite of what you obtain (i.e. always bet on "we got different results" which again has a 50% probability). This is related to the "guess your own hat's color" riddle: "You and a friend are on a game show. The host sets you facing each other at a table, blindfolded. A hat, known to be either white or black, is placed on each of your heads. The host removes the blindfolds, and asks each of you to write down the color of your own hat without communicating with each other. If either of you guesses correctly, you both win a prize. How do you guarantee success by pre-communicating a strategy?" The answer here uses the same partitioning into "either we're the same, or we're different". One friend agrees to always guess his own hat to be the same as the friend's, while the other always guesses that his hat is different.