Working out expected steps of absorbent Markov Chain with more than one sink

So I’ve seen answers regarding the relative probabilities of winning given that two people are tossing coins until one wins, with one person winning with HHT and the other with HTT. (2/3 HHT wins, 1/3 HTT wins) and I understand that you can use the transition matrix to work out the fundamental matrix to work this out.

Similarly, to work out the expected length of this game before someone wins, you can sum the top row of the fundamental matrix (5.33).

However, a related question – if HHT (or HTT) won, what is the expected number of tosses? Can this be worked out from the matrix? The solutions are given here (4.67 for HTT and 5.66 for HHT). Note that these are not the same as working out the expected number of tosses needed, in general, to get HTT or HHT (which are 10 and 8) because this is conditioned on HHT/HTT being the winning sequence.

I notice that the expected duration of the game is just the weighted sum of these two expected durations weighted by the probability of winning (4.67 * 0.33 + 5.67 * 0.67), but I don’t know how to work out the two durations of 4.67 and 5.67 (other than numerically iterating the probability that the win occurs after 1, 2, 3, 4, 5 … N tosses).