Probability of $2$ heads in a row before $2$ tails in a row? Frank alternates between flipping a weighted coin that has a $2/3$ chance of landing heads and a $1/3$ chance of landing tails and another weighted coin that has a $1/4$ chance of landing heads and a $3/4$ chance of landing tails.
The first coin toss is the "$2/3$-$1/3$" weighted coin. The probability that he sees two heads in a row before he sees two tails in a row is?

let H denote the event of landing a head and T for a tail. So there are some possibilities: HH...,THH..TT...,HTHH...TT  but how to do further?

 A: Denote the two coins by $A,B$.  After the first toss, every active state of the game can be described by two parameters:  what value did you just toss and which coin are you about to toss?  We label the four active states as $(H,A),(H,B), (T,A), (T,B)$.  We denote by, say, $p_{H,A}$ the probability that you'll see $HH$ before you see $TT$ assuming you are currently in state $(H,A)$.  The starting state can never be reached again, we'll denote it by $\emptyset$.  Of course, $p_{\emptyset}$ is the answer we seek.
Now, consider the possible results of the first toss.  With probability $\frac 23$ you get an $H$ and move to state $(H,B)$.  With probability $\frac 13$ you get a $T$ and move to state $(T,B)$.  Thus $$p_{\emptyset}=\frac 23\times p_{H,B}+\frac 13\times p_{T,B}$$
Similarly we get $$p_{H,B}=\frac 14\times 1 +\frac 34\times p_{T,A}\quad \quad \quad p_{T,B}=\frac 34\times 0+\frac 14\times p_{H,A}$$ $$p_{H,A}=\frac 23 \times 1 +\frac 13\times p_{T,B}\quad \quad \quad p_{T,A}=\frac 13\times 0 + \frac 23\times p_{H,B}$$
Barring arithmetic error (always possible), this system implies $$\boxed {p_{\emptyset}=\frac {13}{33}}$$
Note:  I am somewhat surprised that this is less than $\frac 12$. After all, you are more likely to throw $H$ initially so I thought that would give $H$ some advantage.  I suggest checking the arithmetic very carefully.
A: You can model this as a Markov process. With the following states:

*

*just flipped heads and is flipping the 1/4 coin

*just flipped tails and is flipping the 1/4 coin

*just flipped heads and is flipping the 2/3 coin

*just flipped tails and is flipping the 2/3 coin

*has already gotten two heads in a row

*has already gotten two tails in a row

I will call the first four states $x_{h1}, x_{t1}, x_{h2}, x_{t2}$. And denote by $p_{h1}, p_{t1}, p_{h2}, p_{t2}$ the probabilities that you will get two heads first when you are in the respective states.
If you are in state $x_{t1}$ you will either get two heads in a row with probability 0.25 or move to state $x_{t2}$. So you get the following equation
$$p_{h1} = \frac{1}{4} \cdot 1 + \frac{3}{4} \cdot p_{t2}$$
You get a similar equation for all the other states. Then you can set up the system of equations and solve it. Lastly you just need to apply this to the first coin toss. The probability of getting two heads in a row first is $\frac{2}{3}p_{h1} + \frac{1}{3}p_{t1}$.
A: I assume you mean the event is $HHTT$. Here's a stronger result using Borel-Cantelli lemma.
If you have an infinite coin toss experiment, i.e. you don't stop after getting $HHTT$, you can define an event:
$$
E_{n} = \{E_{4n+1}=H \cap E_{4n+2}=H \cap E_{4n+3}=T \cap E_{4n+4}=T \}, n \in \mathbb{Z}
$$
then all events $E_n$ are independent, and $P(E_n) =\frac{2}{3} \frac{1}{4} \frac{1}{3} \frac{3}{4} = \frac{1}{24}$. Note this is a subsequence, as it only counts the events $HHTT$ that start with the first coin. Accorinding to Borel-Cantelli Lemma (#2), the probability of event happening infintely often is $1$ if the corresponding sum of probabilities diverges. In your case,
$$
P(E_n \ i.o) = \sum_{k=0}^{\infty}P(E_k) = \sum_{k=0}^{\infty}\frac{1}{24} = \infty
$$
Therefore, the probability to observe event $HHTT$ i.o. is $1$
