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Assume a coin has a probability p to get a head H. Suppose a coin is tossed until the partern T,T appear in the last 2 tosses. Once he got T,T then the game is finished. What is the expected number of flips is expected for the game?

I tried to find out the distribution of $N$ where $N$ stands for the expected value of the no. of tosses before getting T,T, but i don't know what kind of distribution is that, seems need another apporaches

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The distribution is not needed. Call $n$ the expected number of tosses needed to reach TT and $m$ the expected number of tosses needed to reach TT starting from T. Conditioning on the value of the first toss, one gets $n=1+(1-p)m+pn$ and $m=1+pn+(1-p)0$. Which yields $$ n=\frac{2-p}{(1-p)^2}=\frac1{1-p}+\frac1{(1-p)^2}. $$ Sanity checks: $n=2$ for $p=0$ (why?), $n\to\infty$ when $p\to1$ (why?) and the function $p\mapsto n$ is nondecreasing (why?).

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Hmm, wouldn't this be the same problem? But this is yielding a different result. –  Stefan Hansen Nov 3 '12 at 14:13
@StefanHansen Here the probability of Heads is $p$ and he looks for TT, in the other example, he looks for $HH$ for which heads has prob $p$. Substitute $q=1-p$ in the above to get what you have on the other page. –  Jean-Sébastien Nov 3 '12 at 14:26
@Jean-Sébastien: Thank you for the clarification. –  Stefan Hansen Nov 4 '12 at 11:07
i have a question, why $m=(1-p)+pn$ why it isn't correct? As i have p probability has to restart and 1-p probability to have an expectation of 1. Thx –  Mathematics Nov 7 '12 at 23:36
If one needs to restart, the duration is 1+n, not n. This happens with probability p hence m = (1-p)1 + p(1+n) = 1 + pn. –  Did Nov 7 '12 at 23:43

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