It depends on the mechanism that reveals you only have one heads coin.
Situation A:
You flip two coins and look at the first one, keeping the second hidden. You learn that one is heads because you see the first one is head. This observation happens with probability 0% in the TT world, probability 50% in the HT and TH worlds, and probability 100% in the HH world. Prior to the observation, you would have equal reason to believe you lived in either of the four worlds. So your prior odds are 1:1:1:1 for TT:TH:HT:HH, and the odds ratio of your observation is 0:50:50:100 or just 0:1:1:2. Multiply these together, and you have your posterior odds, namely 0:1:1:2. That means your likelihood of having only one heads (TH+HT) is the same as your likelihood of having two (HH). And hence, you should expect with 50% probability that, once you lift up your other hand, the second coin will also be a heads. This is expected, as the first bit in a random sequence should say nothing about the second.
Situation B:
You flip two coins, but you can't see the outcome whatsoever. You have a trusted friend look at the two coins, and ask him if the outcome contains at least one heads. He says "Yes." This may feel like the same information you got earlier, but it actually tells you something different. Let's consider the four worlds again. Your prior odds are still 1:1:1:1 for TT:TH:HT:HH. But what's the likelihood that your friend says "Yes." in the different worlds? Obviously, TT predicts this outcome with 0% chance. But in the other three worlds, TH, HT, and HH, since your friend has seen both coins, he will correctly tell you that there's at least one heads 100% of the time. (So there's the difference. In situation A, you were using a noisy observation method (looking at only one coin), which doesn't always tell you about "at least one head" cases.) We can now compute the odds 0:1:1:1 and conclude a probability of 1/3 that you're in a HH world, and 2/3 that you're in TH or HT.
