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?

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)
