Coin tosses, A wins if HTT, B wins if TTH, is this fair? I found this riddle on the web so I am not sure of what it is actually asking. I think it is saying:
A and B toss the coin and the game ends whenever the following two exact combinations appear: HTT (A wins) and TTH (B wins).
My approach:
I am assuming that it depends on the fact that the expected number of flips $E(n)$ of having a T followed by a H is not equal to the $E(n)$ of having a T followed by a T.
So no, the game is not fair.
$E(TT) = 1 + P(T)E(TT|T) + P(H)E(TT|H) $
$E(HT) = 1 + P(H)E(HT|H) + P(T)E(HT|T) $
Following this reasoning and solving the equations:
$E(TT) = 6$ and $E(HT) = 4$
From here I don't know how to calculate the expected value though. Suggestions?

If the first toss comes up H, then HTT is going to win; eventually T will come up twice in a row, and the first time that happens, the game is over and HTT wins.
If the first two tosses come up TH, then HTT wins for exactly the same reason as before; the initial T is irrelevant.
If the first two tosses come up TT then TTH wins as soon as the first H comes up.
The point is that the game is completely decided by the first two tosses, and HTT wins in 3 cases out of 4.
That's all there is to it; no expectations, no infinite series.
