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You have a unfair coin with $70\%$ chance of getting a head($H$). You toss it and it returns a tail($T$). You will then toss until there is equal number of heads and tails. What is the expected number of tosses before you stop? Do not count the first toss where the result is tail.

I ran a simulation and it requires 2.5 additional toss on average. Yet I do not have an approach to arrive at the answer.

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  • $\begingroup$ What are your thoughts on this problem? $\endgroup$ – saulspatz Aug 8 '18 at 13:47
  • $\begingroup$ I ran a simulation and it requires 2.5 additional toss on average. Yet I do not have an approach to arrive at the answer. $\endgroup$ – user122049 Aug 8 '18 at 13:49
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    $\begingroup$ You should put all the context in your question, so that people browsing will see it. Often, people will just vote to close a question with no context, without reading the comments. $\endgroup$ – saulspatz Aug 8 '18 at 13:51
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Let $E$ be the expected number of tosses it will take to get you back to even (so $E$ is the desired result).

Consider the first toss (not counting the initial $T$). If it is an $H$ you are done (in one toss).

Suppose your first throw (not counting the initial $T$) is again a $T$. Then you expect it will take you $E$ tosses to get back to where the game started, and another $E$ tosses to get to even. Thus your expected number from that state is $2E$.

It follows that $$E=.7\times 1+.3\times (2E+1)\implies E=\frac 52$$

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