Suppose I flip a fair coin four times, and the result is HTHT. What is the probability that the following four flips will also result in HTHT? Would it be 1/16,
(1/16)^2, or 1/(2^8)

The answer is none of the above. Since you don't really care what the first four flips are, and just need to match the last four flips in precise sequence to the first four, here's how to think it through:

For the first of the "matching flips" (flip 5 in your example), you win if it's a head - the probability of that happening is 1/2. For the next flip you win if it's tails, which again has a probability of 1/2. So the probability of getting both a heads on the first flip and a tails on the second is (1/2) x (1/2) = (1/2)^2. Now consider the third flip - you must get a heads to win, which has probability of 1/2. So the odds of matching the first three is (1/2)^3. I/m sure by now it's pretty clear what the odds are of matching all four.

Another way to do this is to write out all possible combinations of four flips in a row - there are (1/2)^4 = 16 possible combinations. And since only one precisely matches HTHT the odds of getting HTHT is 1/16. Note that if the first four flips had been something else - say HHHT - there is again only one sequence of four flips that matches that as well. So you see it doesn't matter what the first four flips are - all that matters is that there is one out sixteen ways to match whatever they were.