Q 7.33 - A First Course in Probability

Question

A coin that lands on heads with probability p is continually flipped. Compute the expected number of flips that are made until a string of r heads in a row is obtained. Hint: Condition on the time of the first occurrence of tails to obtain the equation E[X] = (1 − p) r i=1 pi−1(i + E[X]) +(1 − p) q i=r+1 pi−1r Simplify and solve for E[X]. 

Step-by-Step Solution

Answer

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

Step 1 Given Information

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.  

Step 2 Explanation

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,

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

$=\sum _{i=1}^{\infty }E[X\mid Y=i]P(Y=i)$

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

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

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

$E[X\mid Y=i]=r$

 

Step 3 Final Answer

${E\left[X\right]=(1-p)\left(\sum _{i=1}^{r}i{p}^{i-1}\right)+E\left[X\right]\left(1-p^r\right)+r{p}^2}^{}$

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

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

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