<s>[INST] 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?
 [/INST] 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) + ....
$S = \frac{2}{3}(1) + \frac{4}{7}\frac{2}{3}(2) + (\frac{4}{7})^2\frac{2}{3}(3) +\cdots$
$S = \frac{2}{3}\dfrac{1}{(1-\frac{4}{7})^2} = \frac{98}{27}$
</s>