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An unbiased coin is tossed repeatedly and outcomes are recorded. What is the expected no of toss to get HT ( one head and one tail consecutively)?

My approach-- e=expected number of tosses e=[if first tail comes] + [if first is head and second is head] + [required probability of first head and second tail] e= 1/2(e+1) + 1/4(e+2) + 1/4

is it correct?

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If the first comes down heads, the expected value is $3$ since you get $HT$ the next time you flip tails, which will take an average of two more flips. If it comes up tails, then you start over. So the recursion can be written $$T = 3/2 + (T+1)/2 $$ which has the solution $T=4.$

The problem with yours is the middle term $(T+2)/4.$ You don't have to start all the way over in the case of $HH,$ you could get $HHT.$

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Let $x$ be the expected number of tosses to get $HT$ if the last flip was $H$ and $y$ be the expected number if the last flip was $T$ or nothing. If the last flip was $H$ you either flip $T$ and are done or flip $H$ and are in the same situation, so $$x=\frac 12\cdot 1+\frac 12 \cdot (1+x)\\x=2$$If you haven't flipped heads, you have $\frac 12$ chance to flip heads and then expect $x$ and $\frac 12$ to flip tails and be in the same state, so $$y=\frac 12(1+x)+\frac 12(1+y)\\y=2+\frac y2\\y=4$$

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Just to add another method:

There is a $(1/2)^n$ chance of getting any arbitrary configuration of $n$ coins. For these $n$ coins, there are $n-1$ ways for it to end in $HT$ without having $HT$ appear earlier. For example, at $n=5$, we could have

$HHHHT \\ THHHT \\ TTHHT \\ TTTHT$

This implies the expected values will be equal to: $$\sum_{n=2}^\infty n(n-1) (1/2)^n = 4$$ Where $n$ is the number of flips to get a consecutive $HT$ (it cannot be below $2$ and has no upper bound) and $(n-1)(1/2)^n$ is the probability that a chain of $n$ coins ends in $HT$ but does contain $HT$ anywhere else.

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