Is this reasoning correct?

1. For the TH case, there is some number greater than or equal to zero of heads until the first tails, then some number greater than or equal to zero tails, until the next heads. That will be the first case of TH. The first number has a geometric distribution with parameter p=1/2, as does the second. So it's the sum of 2 r.v.s, each with mean 1/p=2 for a total of 4.

2. For the HH case, say there are some number of tails until the first heads. That would also be geometric. If the next toss is heads, it's done. That would be 1 + X where X has geometric distribution as above. But if the toss after heads turns up tails, then this needs to be repeated, until the toss after the heads is another heads. So it's like a random (geometric) number of these sequences, each of which is like 1 + X with X also geometric. So the expected number would be 2 * (2+1) = 2 * 3 = 6.