Flip a coin three times. What is the probability of getting a sequence of HTT as compared to a sequence of HTH? The answer is obvious. Both probabilities are the same.

Let us now tweak the question and the situation.

Suppose you would flip a coin until you hit the sequence HTT. You manage to do this with n flips. Suppose also that you would flip a coin until you hit the sequence HTH, and you do this with m flips.

On average then, how would n compare to m? Would n be larger, equal or smaller than m?

The answer surprisingly is on average, n would be smaller than m. Ie, it would take less flips of the coin to achieve the sequence HTT as compared to the sequence HTH. The explanation is no rocket science and in fact is very logical and simple.

Suppose you are aiming to obtain the sequence HTH. You flip the coin and it gets into the sequence HT. If your third flip is H, you would have gotten the required sequence HTH and the game would be over. If your third flip is T, you would have to continue flipping and start the sequence of HTH only the next time you obtain a H.

Suppose now you are aiming to obtain the sequence HTT. You flip the coin and it gets into the sequence HT. If your third flip is T, you would have gotten the required sequence HTT. If your third flip is H, you would have to continue flipping but the key difference is this time, you are already one third into the sequence HTT.

I ran a basic Matlab script (see below) and obtained the value of n to be 7.00 and the value of m to be roughly 8.97. I wonder how this can be solved using probabilistic theory.
Somebody extremely smart out there reading this.. please provide me with the proof!! =)