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?
Best Answer
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.