Please explain Probability Puzzle 3: Flipping Coins

If you flip a coin 3 times, the probability of getting any sequence is identical (1/8).

There are 8 possible sequences: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Let's make this situation a little more interesting. Suppose two players are playing each other. Each player choses a sequence, and then they start flipping a coin until they get one of the two sequences.

We have a long sequence that looks something like this: HHTTHTTHTHTTHHTHT.... We continue until one of the two wins.

Do you think this is a fair game, and that under these rules each sequence has an equal chance to appear first?

Think again! If you chose HHH and I chose THH, I have a much higher chance that you do!

The only way that you win is if the first three tosses are HHH. In any other event, I win.

Agree? Do you see why?

For the sequence HHH to appear anywhere except the first three flips, it must come after a T, right? So, the actual sequence for you to win is THHH.

But if there is a sequence of THHH then I already won before that sequence is over (because my sequence was THH).

So, THH will win 7 times out of 8. HHH will only win if the first three are HHH (a one in eight chance).

Suppose you are going to flip a coin until you get the sequence HTH. Say this takes you x flips. Then, suppose you are going to flip the coin until you get the sequence HTT. Say this takes you z flips. On average, how will x compare to z? Will it be bigger, smaller, or equal?