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What's the probability of $HTHT$ occuring before $HHTT$ in a stream of $H$'s and $T$'s (both equally likely) that will stop if either of those occur? What's the mean number of throws such that $HHTT$ occurs?

Hello, I figured out that this is an instance of Penney's game.

This question is also related. What makes this question hard for me is the fact that I have to somehow account for the fact that $HTHT$ occures before $HHTT$. I have to solve this using Markov chains, does someone have an idea?

Analysis
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    I suggest you make a state transition diagram for a Markox chain with states $$\big{\emptyset,H,HT,HH,HTH,HHT,HTHT,HHTT\big}$$ with $HTHT,HHTT$ absorbing. You're trying to find the probability of being absorbed into the state $HTHT.$ –  May 29 '23 at 15:45
  • Thanks, I followed your advice. However, I fail to compute the probability of being absorbed into $HHTT$. I guess I could follow every path and use Chapman-Kolmogorov ... Is there a trick for computing the absorption probability? – Analysis May 29 '23 at 16:43
  • I've seen several questions that compute the expectation by "divide et impera" using conditional expectation. I'll try approaching this too. – Analysis May 29 '23 at 18:58

1 Answers1

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This solution is an elaboration on the comment by @MatthewH.

Let $P_i$ denote the probability of hitting $HTHT$ first when in state $i$. The equations relating the probabilities are

\begin{align*} P_{\emptyset} &= \frac 12(P_H + P_{\emptyset}) \\ P_H &= \frac 12 (P_{HH}+P_{HT}) \\ P_{HT} &= \frac 12 (P_{HTH}+P_{\emptyset}) \\ P_{HH} &= \frac 12 (P_{HH} + P_{HHT}) \\ P_{HTH} &= \frac 12 (P_{HH} + P_{HTHT}) \\ P_{HHT} &= \frac 12 (P_{HTH} + P_{HHTT}) \\ P_{HTHT} &= 1 \\ P_{HHTT} &= 0. \end{align*} This is a large linear system, but it can be solved fairly easily by substitution. The solution I found is \begin{align*} P_{\emptyset} = P_H &= \frac 49 \\ P_{HT} &= \frac 59 \\ P_{HH} &= \frac 13 \\ P_{HTH} &= \frac 23 \\ P_{HHT} &= \frac 13, \end{align*} but please double check to make sure I didn't make a mistake.

user6247850
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  • Using $P_{\emptyset}=a$, $P_H=b$, $P_{HT}=c$, $P_{HH}=d$, $P_{HTH}=w$, $P_{HHT}=x$, $P_{HTHT}=y$ and $P_{HHTT}=z, your result is verified by: https://www.wolframalpha.com/input?i=solve+a%3D1%2F2%28b%2Ba%29%2C+b%3D1%2F2%28d%2Bc%29%2C+c%3D1%2F2%28w%2Ba%29%2C+d%3D1%2F2%28d%2Bx%29%2C+w%3D1%2F2%28d%2By%29%2C+x%3D1%2F2%28w%2Bz%29%2C+y%3D1%2C+z%3D0 – Analysis May 29 '23 at 18:56
  • How can I combine these probabilities to obtain the probability in question? @user6247850 – Analysis May 29 '23 at 18:57
  • @Analysis The probability in question is $P_{\emptyset}$: The probability of reaching $HTHT$ when not conditioning on being in any state. – user6247850 May 29 '23 at 19:25