Compare two indepedent random variables? Suppose X and Y are two independent variables with $$P(X=k) = 2^{-k} $$ and $$P(Y= k) = 2\cdot 3^{-k}.$$ Find $$P(X<Y).$$
My approach is to break the probability into the sum of disjoint probability, but I failed to work this out. Thanks
 A: We can break it up into cases. 
1) If $X=1$, then we will be OK if $Y=2,3,4,\dots$. The probability that $Y\ge 2$ is the sum of an infinite geometric series, first term $\frac{2}{9}$, common ratio $\frac{1}{3}$, so sum $\frac{1}{3}$. Thus the probability that $X=1$ and $Y\gt 1$ is $\frac{1}{2}\cdot \frac{1}{3}$.
2) If $X=2$, then we will be OK if $Y=3,4,5,\dots$. The probability that $Y\ge 3$ is the sum of an infinite geometric series, first term $\frac{2}{27}$, common ratio $\frac{1}{3}$, so sum $\frac{1}{9}$. Thus the probability that $X=2$ and $Y\gt 2$ is $\frac{1}{4}\cdot \frac{1}{9}$.
3) Similarly, the probability that $X=3$ and $Y\gt 3$ is $\frac{1}{8}\cdot \frac{1}{27}$.
4) And so on.
Finally, add up. We need to find the sum of a geometric series.
A: We have
$$
\begin{align*}
\mathbb P(X<Y) &= \sum_{k=1}^\infty \mathbb P(X<Y, Y=k)\\
&= \sum_{k=1}^\infty \mathbb P(X<Y|Y=k)\mathbb P(Y=k)\\
&= \sum_{k=1}^\infty \mathbb P(X<k)\mathbb P(Y=k)\\
&= \sum_{k=1}^\infty \sum_{j=1}^{k-1}\mathbb P(X=j)\mathbb P(Y=k)\\
&= \sum_{k=1}^\infty 2\cdot 3^{-k}\sum_{j=1}^{k-1}2^{-j}\\
&= \sum_{k=1}^\infty 3^{-k}\sum_{j=0}^{k-1}2^{-j}\\
&= \sum_{k=1}^\infty 3^{-k}\frac{1-2^{-k}}{1-2^{-1}}\\
&= 2\left(\sum_{k=1}^\infty 3^{-k} - \sum_{k=1}^\infty 6^{-k}\right)\\
&= 2\left(\frac1{1-3^{-1}} - \frac1{1-6^{-1}}\right)\\
&= 2\cdot \frac3{10}\\
&= \frac35.
\end{align*}
$$
