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To build on my understanding from this question, I was wondering what is the expected number of tosses required to obtain the sequence H-T-H-T-T-H, given that the coin is fair and independent?

Edit: The way I was thinking about it, which I am not sure makes sense. Let $e$ be the expected number of coin tosses. Then, if we get a Tail immediately, expected number is $e+1$. If we get a H and another H, then the expected number is $e+2$. For the third toss, if we have a H,T,T then it is $e+3$ and for the fourth, fifth and sixth it is $e+4$, $e+5$ and $e+6$ respectively.

So, we have the equation

$$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{64}(e+6)+\frac{1}{64}(6).$$

I get that $e = 126$ which appears to be wrong compared to @Did's Answer.

Community
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Jojo
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2 Answers2

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If you start with $H-H$, you don't go all the way back to the start - you have already begun your second try with the second $H$.
I think it is a little messy: $$e=\frac12(e+1)+\frac12(f+1)\\ f=\frac12(g+1)+\frac12(f+1)\\ g=\frac12(h+1)+\frac12(e+1)\\ ... $$ then solve the collection of equations.

Empy2
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  • So, does the equation need to be $$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+1)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{64}(e+6)+\frac{1}{64}(6).$$? – Jojo Sep 11 '15 at 13:11
  • Thank you very much. I apologize, but could you please tell me what $f$,$g$ and $h$ are? – Jojo Sep 13 '15 at 02:28
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    The average after you have H; the average after you have H-T; the average after you have H-T-H – Empy2 Sep 13 '15 at 09:10
  • Just one last question. Could you explain what the equation $$e=\frac12(e+1)+\frac12(f+1)\$$ means in words possibly and I can attempt to deduce the other equations correspondingly? In the first equation I don't understand why we add 1 in $\frac12(f+1)$? The way I think about it is $\frac12(e+1)$ is the expected number of throws to get the sequence after getting a $T$ and hence, we've wasted a throw and add a 1. However, if we get a $H$ I don't get why we add a 1? – Jojo Sep 13 '15 at 18:17
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Probability is (1/2)^6. Approximately 64 attempts (of a sequence of 6 tosses in a row) would be required to gain the desired result.

user264985
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