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?
 A: It is a riddle.
Hint: Consider that there must be a first time $TT$ is flipped. Either the first $TT$ is flipped in the first two flips.... or the first  $TT$ is flipped after the first two flips.
If a $TT$ is flipped in the first two moves then player $B$ will always win.  Why? Because as soon as the very first $H$ is flipped it will have followed two $T$s and we will have just flipped $TTH$ and player $B$ wins.
If a $TT$ is flipped sometime after the first two flips then player $A$ will always win.  Why?  Because the flip before the very first $TT$ has to be an $H$.  Thus if the first $TT$ doesn't occur at the very beginning when we roll the $TT$ we will have rolled $HTT$ and player $A$ would have won.
So B wins if and only if the first two rolls are $TT$.  A will win every other time.  So $B$ has a 1 in 4 chance of winning and $A$ has a 3 in 4.
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Okay.... you seem to think the above was confusing.
Consider that you flip ".......xTTy.............."  and that $TT$ is the very first time that two $T$s occured in a row.
What is $x$?  Well it can't be $T$ because the $TT$ after the $x$ was the first two $T$s in a row.  But it could be that $TT$ were the very first to flips and $x$ is a non-flip.
Claim:  If $x$ is a $H$ then player A has won.
Why? Before the TT the can't have been any earlier $TTH$s so player $B$ will not have won yet.  But if $x = H$ then we just flipped $HTT$ and player $A$ just won.
Claim:  If $x$ is a non-flip the player $B$ will win.
Why?  eventually we will have to flip the first $H$.  When we do because the first two flips were $TT$ we will have $......  TTH$ and as we just rolled $TTH$ player B will win.  (And as this will be our very first $H$ we can't have had A play HTT before.)
So the probability of $B$ winning is exactly the same as the probability of the first two flips being a $T$. i.e. 1 in 4.
A: The key idea here is to "think backwards", let's start with assuming $B$ wins in this game, which means at some point he throws a pattern of "$TTH$", now if we think backwards, what could last throw be before the tail in "$TTH$"? Clearly it can't be head since otherwise we will have a "$HTT$" which gives the victory to $A$. By now we have "$TTTH$", again, what could last throw be? Same reason, it can only be tail, then we keep repeating this induction, we get a sequence "$TTT\ldots TTH$", which means in the first two throws, $B$ must throw two consecutive tails, otherwise he has no chance of winning, and this gives the probability of $\frac14$. What if the first two throws are not $TT$? Then $A$ must win since $B$ is definitely gonna lose the game, so $A$ can relax a bit, drinking a cup of tea while throwing, just waiting for his "$HTT$" to happen(it's definitely gonna happen as the probability is positive). Now you can see, the chance of $A$ winning is $\frac34$ while $B$ is $\frac14$, so the game is unfair.
(To make it clear about why the the probability of B wins is $\frac14$, you can sum up a geometric series)
A: 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.
