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A coin that lands on heads with probability p is continually flipped.

We have to find the expected number of flips that are made until a string of$r$ heads in a row is obtained.

A coin that lands on heads with probability is continually flipped. Compute the expected number of flips that are made until a string of $r$heads in a row is obtained.

Let $X$denote the number of flips are made until a string of r heads in a row is obtained.

Let $Y$denote the number of flips until the first occurrence of tails. Then,

$E\left[X\right]=E\left\{E\right[X鈭�Y\left]\right\}$

$=\underset{i=1}{\overset{鈭�}{鈭�}}E[X鈭�Y=i]P(Y=i)$

$\text{Note}P(Y=i)=p^{i-1}(1-p)$

For$i鈮�r$, then on the $i+1$flip, the$r$consecutive heads is,
$E[X鈭�Y=i]=i+E\left[X\right]$

For $i>r$, then the first $r$ flips are heads is

${E\left[X\right]=(1-p)\left(\underset{i=1}{\overset{r}{鈭�}}i{p}^{i-1}\right)+E\left[X\right]\left(1-p^r\right)+r{p}^2}^{}$

$=(1-p)\left(\frac{1-(1-p)r{P}^{\text{'}}-p^{\text{'}}}{(1-p{)}^2}\right)+E\left[X\right]\left(1-p^{\text{'}}\right)+r{p}^{\text{'}}$

$=\frac{1-p^{\text{'}}}{1-p}+E\left[X\right]\left(1-p^2\right)$

$=\frac{1-p^{\text{'}}}{p^{\text{'}}(1-p{)}^2}$