EV of probability given outcome Michael has a crush on a girl. Every night, he texts her and asks to go on a date. There is a 1/7
chance that the girl
says yes, a 2/7
chance that the girl says no, and a 4/7 chance that the girl asks Michael to text her again tomorrow, which
Michael does. Given that she said no, what is the expected number of days it took her to decide?
 A: First, notice the following equality
\begin{equation}
\sum_{n=0}^\infty\left(\frac{4}{7}\right)^n\left(\frac{1}{7}+\frac{2}{7}\right)=1,
\end{equation}
where you can use the geometric series to prove that.  Clearly the first element of the sum is the probability of yes, and the second is the probability of no.  We can also see that the probability of yes is $\frac{1}{3}$, and no is $\frac{2}{3}$.  We know she said no, so we normalize the probability by multiplying by $3/2$, and the expected number of days is 
\begin{equation}
\frac{3}{2}\frac{2}{7}\sum_{n=0}^\infty (n+1)\left(\frac{4}{7}\right)^n=\frac{3}{2}\frac{2}{7}\left(\frac{7}{3}\right)^2=\frac{7}{3},
\end{equation}
where we have used the sum $\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$, found in wikipedia on geometric series.  We have assumed that the first day would mean one day.  If you assumed that a no on the first day is $0$, then you would subtract another geometric series, giving $\frac{4}{3}$.
A: I have assumed that the No the first day is assumed to be 1.  If that were to be the case, then the expected no of days on we which she decided would be the following:
Probability of a decision made = Probability of a No + Probability of a Yes
$P(Decide) = \frac{2}{7}+\frac{1}{7} = \frac{3}{7}$
USing Bayes' theorem:
Probability of $P(No/Decided) = \dfrac{\frac{2}{7}}{\frac{3}{7}} = \frac{2}{3}$.
$P(Yes/Decided) = \dfrac{\frac{1}{7}}{\frac{3}{7}} = \frac{1}{3}$.
$P(Decided/No) = \dfrac{\frac{2}{3}}{\frac{2}{3}+\frac{1}{3}} = \frac{2}{3}$
On each day it is this probability of decision that she has to make.
Thus S = Porbability of a Decision/No + Probability of not making decsion the previous day * Porbability of a Decision/No (2) + (Probability of not making decision the previous day)^2 *Porbability of a Decision/No (3) + ....
$S = \frac{2}{3}(1) + \frac{4}{7}\frac{2}{3}(2) + (\frac{4}{7})^2\frac{2}{3}(3) +\cdots$
$S = \frac{2}{3}\dfrac{1}{(1-\frac{4}{7})^2} = \frac{98}{27}$
