A fair die is tossed until the sequence “44” is seen. Let $N$ be the number of tosses this requires. Find $E[N]$.
I have my own solution which I need someone to verify. And this problem has to be solved with the property $E[N]$ = $E \left[E[N|X]\right]$. Below is my solution:
Let $X$ be a random variable such that $$ X = \left\{ \begin{array}{lr} 1 &, \text{getting a "4" after a "4" }\\ 0 &, \text{not getting a "4" after a "4"} \end{array} \right. $$ And \begin{align} E[N] &= E \left[E[N|X]\right]\\ &= E[N|X=1]P[X=1]+E[N|X=0]P[X=0]\\ \end{align} For $E[N|X=1]$, since we condition on getting the first "4", we know that once we get the first "4", the next roll will be a "4" as well. Therefore, we only need to care how many rolls we need to get the first "4" in this situation. And this becomes a "geometric distribution". A geometric random variable has a expectation of $\frac{1}{p}$, in this case, $\frac{1}{p}=\frac{1}{\frac{1}{6}}=6.$ And adding the next roll on which we will get the next "4", $E[N|X=1]=6+1=7.$
For $E[N|X=0],$ let random variable $\tilde{N}$ denote the number of tosses required to get sequence "44" after getting a first "4", given that the next roll is not a "4". Since the expected value of getting the first "4" is 6, $E[N|X=0]=E[6+\tilde{N}]=6+E[\tilde{N}].$ And it is not hard to see that $E[N]=E[\tilde{N}]$ since both scenarios need to get the "44" sequence and each roll is independent.
To sum up: \begin{align} E[N] &= E \left[E[N|X]\right]\\ &= E[N|X=1]P[X=1]+E[N|X=0]P[X=0]\\ &= 7P[X=1]+\left[6+E[\tilde{N}]\right]P[X=0]\\ &= 7\cdot\frac{1}{6}+\left[6+E[\tilde{N}]\right]\cdot\frac{5}{6}\\ &= \frac{7}{6}+5+\frac{5}{6}E[\tilde{N}]\\ E[N] &= \frac{7}{6}+5+\frac{5}{6}E[N]\\ \frac{1}{6}E[N] &=\frac{37}{6}\\ E[N] &=37 \end{align}
I have typed this for so long and any input would be greatly appreciated!
Actually i is not hard to see that $E[\tilde{N}]=E[1+N]$ since, when evaluating $\tilde N$, one assumes that the first toss after getting a first "4" is not a "4" hence one needs a supplementary toss when compared to $N$.
After that, everything flows smoothly and one gets $E[N]=42$.