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A coin is flipped infinitely until you or I win. If at any point, the last three tosses in the sequence are $HHT$, I win. If at any point, the last three tosses in the sequence are $HTH$, you win. Which sequence is more likely?

Unfortunately, this configuration does not seem like the ones as "$HHT$ versus $THH$" (where clearly only $HHT$ wins iff the first two occuring $H$ are consecutive). Of course, here we can still assume that $TT$ does not occur (as after such a thing the game restarts), but it does not seem to help me enough.

Any help appreciated!

DesmondMiles
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    it seems like you already broke it down. make a flow chart with probabilities – Saketh Malyala Oct 05 '19 at 11:24
  • There are only finitely many states of interest...all you care about is the last two tosses. Figure out the transitions between those states and solve the linear system. – lulu Oct 05 '19 at 11:26
  • @lulu can you show a bit more details? I guess you mean for some sort of Markov chain but I cannot think of a suitable one. – DesmondMiles Oct 05 '19 at 11:34
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    Actually, you only care about $3$ states, as $HH$ is a guaranteed win. Label the three active states as $TT, TH, HT$. For each state $s$ let $p_s$ be the probability that $HHT$ wins given that you are in state $s$. Then, for instance, $p_{TT}=\frac 12\times p_{TT}+\frac 12\times p_{TH}$. since you effectively start in state $TT$ you just need to solve for $p_{TT}$. – lulu Oct 05 '19 at 11:39
  • If $TT$ is allowed then there is a chance that noone wins if the followin sequence was formed $HTTHTTHTTHTTHTT..$ is that okay? – Fareed Abi Farraj Oct 05 '19 at 11:44
  • Of course $TT$ is allowed, and true, there are possible stalemates. but the probability of getting a stalemate is $0$. – lulu Oct 05 '19 at 11:46
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    See https://math.stackexchange.com/questions/66670/probability-about-a-coin-games – Gerry Myerson Oct 05 '19 at 11:58

3 Answers3

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An informal argument as to why HHT is more likely:

After starting the game, we keep tossing the coin until we see a head. The next toss puts one of the players closer to winning, H for you and T for me. Suppose it’s the latter. If the next toss is H, then I win, otherwise the game starts over and we’re both effectively two steps away from winning again. On the other hand, if we had HH, when the next toss is an H the game remains in the same state: you’re still one step closer to winning than I am. So, whenever a non-winning coin toss comes up that doesn’t favor you, you never lose any ground, whereas I get put back to square one whenever the coin toss goes against me. “On average,” I’m usually two steps from winning, but you’re only one step away. I would even hazard a guess that you’re twice as likely to win this game.

This is borne out by a calculation. The game can be modeled by an absorbing Markov chain with transition matrix $$P = \begin{bmatrix}\frac12&\frac12&0&0&0&0\\0&0&\frac12&\frac12&0&0\\0&0&\frac12&0&\frac12&0\\\frac12&0&0&0&0&\frac12\\0&0&0&0&1&0\\0&0&0&0&0&1\end{bmatrix}$$ with state 5 representing the HHT win and state 6 the HTH win. The absorption probabilities work out to be $2/3$ and $1/3$, respectively.

amd
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Hm, I think I actually have an answer. Assuming there is no TT, consider the first occurence of HTH and suppose it is winning. Then just before it we can't have HH (else HHHTH has HHT in the beginning), we can't have TT by above, we can't have HT (else we get TT) and if we have TH, then in THHTH we have HHT in the middle. So HTH to win it must occur among the first four tosses. The possible ones are HTH, THTH, with total probability $\frac{1}{8} + \frac{1}{16} = \frac{3}{16}$. So HHT to be winning has $\frac{13}{16}$ probability.

DesmondMiles
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    It's not clear "we can't have $TT$" means. Of course $TT$ can occur. If you are saying, as I think you are, that $TT$ just restarts the game, that is correct, but then of course it changes the other probabilities. – lulu Oct 05 '19 at 11:40
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(1) If you started by $HH$ then the 1st person wins.

(2) If you started by $HT$ you have 2 equal probabilties that the game restarts (or it is a tie) if you got a $T$, and the 2nd wins if you got an $H$

(3) If you started by $TH$ then you have equal probabilities that the 1st person wins if you got an $H$, and if you got a $T$ then part (2) happens.

(4) If you started by $TT$ then the game restarts (or it is a tie).

Now you may have completed your tree diagram

Fareed Abi Farraj
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