The first coin toss is the "$2/3$-$1/3$" load coin. The probability the he sees two heads in a row before he sees two tails in a row is?
let H signify the occasion of landing a head and T because that a tail. So there room some possibilities: HH...,THH..TT...,HTHH...TT yet how to execute further?
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edited Jul 2 "20 in ~ 11:42
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$\begingroup$ perform you average exactly 2 heads in a row? $\endgroup$
Jul 2 "20 at 12:04
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Denote the two coins by $A,B$. After ~ the an initial toss, every energetic state the the game can be defined by two parameters: what worth did you simply toss and also which coin space you about to toss? We brand the four energetic states together $(H,A),(H,B), (T,A), (T,B)$. We represent by, say, $p_H,A$ the probability that you"ll check out $HH$ before you check out $TT$ assuming girlfriend are at this time in state $(H,A)$. The starting state can never be got to again, we"ll signify it by $\emptyset$. Of course, $p_\emptyset$ is the answer us seek.
Now, consider the feasible results that the an initial toss. V probability $\frac 23$ you acquire an $H$ and move come state $(H,B)$. V probability $\frac 13$ you acquire a $T$ and also move to state $(T,B)$. Thus $$p_\emptyset=\frac 23\times p_H,B+\frac 13\times p_T,B$$
Similarly we obtain $$p_H,B=\frac 14\times 1 +\frac 34\times p_T,A\quad \quad \quad p_T,B=\frac 34\times 0+\frac 14\times p_H,A$$ $$p_H,A=\frac 23 \times 1 +\frac 13\times p_T,B\quad \quad \quad p_T,A=\frac 13\times 0 + \frac 23\times p_H,B$$
Barring arithmetic error (always possible), this system means $$\boxed p_\emptyset=\frac 1333$$
Note: i am somewhat surprised the this is much less than $\frac 12$. ~ all, girlfriend are an ext likely to throw $H$ originally so I believed that would provide $H$ part advantage. I imply checking the arithmetic really carefully.