Question: Given a fair coin, there are $4$ possible outcomes if we toss the coins twice, they are $HH, HT, TH, TT.$ If we obtained one of $HH,HT,TH,$ then we stop tossing. If we obtained $TT,$ we continue tossing. Find the expected number of tosses to stop tossing the fair coin.
Answer: Let $N$ be the number of tosses required to stop tossing. Then we have $$E[N] = \frac{3}{4}\times 2 + \frac{1}{4}(2 + E[N]).$$ Solving the equation above leads to $$E[N] = \frac{8}{3}.$$
I understand that $$E[N] = \frac{3}{4}\times 2 + \frac{1}{4}(2 + E[N])$$ is due to the law of total expectation. But I would like to prove the equation above using the law. But to no avail.
Any help is appreciated.
$p(2m+2) = \left( \frac{1}{4} \right)^m \frac{3}{4}$, for $m \geq 0$.
So $\operatorname{E}(N) = \sum_{i=0}^{\infty} (2i+2) \left( \frac{1}{4} \right)^m \frac{3}{4}$
Now, see that $\left( \operatorname{E}(N) - 2 \left( \frac{3}{4} \right) \right) \times 4 = \operatorname{E}(N) + \operatorname{E}(2)$.
Your distribution is geometric, so you are just solving for the expected value of a geometric random variable.