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Having a lot of trouble working out this exercise. I have tried constructing the 8x8 matrix with all possible combinations of three flips of the coin {HHH, HHT, HTH, ... , TTT} and then calculating an exit distribution and trying to find the P(going to player 2's strategy < going to player 1's strategy) but I keep getting the 1 vector when solving. (Using the method out lined in Durrett of (I-r)^-1 * v = h). Any advice would be greatly appreciated.

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For the first strategy: if player $1$ picks $HHH$ and player $2$ picks $THH$, then the only way player $1$ can win is if $HHH$ comes up immediately: as soon as a $T$ appears, player $1$ is doomed, because in order for $HHH$ to appear after $1$ or more $T$'s, you have to first get to $THH$, and thus player $2$ is bound to win. So, player $1$ can pnly win with a probability of $\frac{1}{8}$, meaning that player $2$ wins with a probability of $\frac{7}{8}$.

Likewise, for the second strategy: if the coin flips start with $HH$ (which occurs with probability $\frac{1}{4}$) then player $1$ is a guaranteed winner ... but as soon as a $T$ appears in the first two flips, player $2$ is bound to win, since in order to get 'back on track' for $HHT$, the initial $HH$ will need to occur after one more $T$'s have been thrown, and hence player $2$'s sequence must occur before player $1$'s sequence can occur. So, player $2$ has a $\frac{3}{4}$ chance of winning.

The key to player $2$'s advantage in all the other cases is to likewise take the first two entries of player $1$'s chosen sequence, and then strategically put a $H$ or $T$ in front of that, so that if player $1$ sequence gets 'broken', player $2$ ends up with an advantage.

So, from that perspective, can you now analyze why the probability for player $2$ winning for the last two strategies is $\frac{2}{3}$?

Answer:

Note that as you start flipping the coin, you can ignore any $T$'s that the resulting sequence begins with. Things start to get interesting when you get the first $H$. At that point, if another $H$ occurs (probability $\frac{1}{2}$), player $2$ is guaranteed to win the game. If a $T$ occurs, then player $1$ will immediately win if the coin flip after that has the right outcome (so player $1$ will win if the right two coin flips happen after the first $H$, i.e. with a probability of $\frac{1}{4}$. But if after the $T$ player $1$ does not get the right outcome, it's back to square one and we'll have another go-around. But since player $2$'s $\frac{1}{2}$ of the first go-around is twice that of player $1$'s $\frac{1}{4}$, that means that player $2$ has twice the chance of eventually winning as the chance of player $1$ winning, meaning that player $2$ wins with probability $\frac{2}{3}$, and player $1$ with probability $\frac{1}{3}$.

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