What will be the mean time of an active customer based on the probability of a 2% monthly drop off? We run a subscription service. During any given month, about 2% of our customers quit the subscription.
My intuition tells me that $1/0.02$ = $50$ months will be the average time a customer stays on our site, however, if I compare this to a deck of cards that I pull cards out of, the probability that I will draw for example queen of hearts, is also ~$2$% (actually about $1.92$%). The average number of draws, I guess, is $26$ which is about $0.5/0.02$. Which one is it?
 A: If $N_0$ is the initial number of customers, the number of remaining customers after $t$ months is $N=N_0 (1-0.02)^t$. The average time is obtained by integrating this function between zero and infinity to get the total number of customer-months, and then dividing it by $N_0$. This leads to
$$ \int_0^{\infty} N_0 \, 0.98^t dt=N_0 [-\frac{1}{\log(0.98)}]$$
which divided to $N_0$ gives an average time (in months) of
$$-\frac{1}{\log(0.98)}\approx 49.5 $$ 
Note that this value only refers to the average customer time, a concept that is different from the expected value of draws to get the queen of hearts in your example. Keeping the similarity with the above integral, this last probability is given by the value of $N$ that identifies half of the area of the integral. Taking a deck of $52$ cards, and considering $51/52 \approx 0.9807$, this can be calculated by solving 
$$ \int_0^{N} N_0 \, 0.9807^t dt=N_0 [-\frac{1}{2\log(0.9807)}]$$
which leads to
$$\frac{0.9807^N}{\log(0.9807)}- \frac{1}{\log(0.9807)}
= -\frac{1}{2\log(0.9807)}$$
$${0.9807^N}-1
= -\frac{1}{2}$$
$$N=\log_{0.9807}( \frac{1}{2})=\frac{\log(\frac{1}{2})}{\log(0.9807)} \approx 36$$
in accordance with the expected number reported in the OP.
