# Sucker Bet - Coin Flipping Stochastic Process

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.

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 $$\frac{2}{3}$$?
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 $$\frac{1}{3}$$.