Find the probability mass function of W. Let X and Y be independent Geom(P) random variables.
Let
\begin{equation}
   W=
      \begin{cases}
         0, & \text{if X<Y}\\
         2, & \text{if X=Y}\\
         3, & \text{if X>Y}
      \end{cases}
\end{equation}
Find the probability mass function of W.
My attempt:
$W\quad\;\;$ | $\quad\quad\quad\quad\quad\quad0\quad\quad\quad\quad\quad\quad\;\;\;\quad\quad1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad2$

$\mathbb{P}(W)\;\;$| $(1-p)^{k-1}p\left(1-\sum^{k-1}_{i=0}(1-p)^ip\right)\quad\left((1-p)^{k-1}p\right)^2\quad(1-p)^{k-1}p\left(1-\sum^{k-1}_{i=0}(1-p)^ip\right)$
(Sorry, I don't know how to draw tables on stack exchange.)
I've been looking at this problem and it seems really similar to this problem: Find the probability mass function of $V=\min(X,Y)$
I think the difference is that we do not sum up the values. My question then, is when do we know when to sum up the probabilities and when to leave them when calculating a pmf. I know this is a stupid question, but I've seen pmfs expressed both ways in my probability class.
 A: You need to actually perform the summation.  In particular, $$\begin{align}
\Pr[X = Y] &= \sum_{k=0}^\infty \Pr[X = k]\Pr[Y = k] \\
&= \sum_{k=0}^\infty \left( (1-p)^k p \right)^2 \\
&= p^2 \sum_{k=0}^\infty ((1-p)^2)^k \\
&= \frac{p^2}{1-(1-p)^2} \\
&= \frac{p}{2-p}. \end{align}$$
By symmetry, $\Pr[X < Y] = \Pr[X > Y]$ because $X$ and $Y$ are independent and identically distributed.  Therefore, $$2 \Pr[X < Y] + \Pr[X = Y] = 1,$$ or $$\Pr[X < Y] = \frac{1}{2}\left(1 - \frac{p}{2-p}\right).$$
A: Your answers are right except you should take the sum from k=1 to infinity for all three cases. It's possible to do the summation to find $P(X<Y)$:
$$P(X=1)P(Y>1)+P(X=2)P(Y>2)+...=$$
$$\sum_{i=0}^\infty P(X=i+1)\underbrace{\sum_{j=i+1}^\infty P(Y=j+1)}$$
The underlined portion is
$$1-\sum_{j=0}^i P(Y=j+1)\\
=1-p\sum_{j=0}^i (1-p)^j\\
=1-p\frac{1-(1-p)^{i+1}}{p}\\
=(1-p)^{i+1}$$
Now the whole thing is
$$p\sum_{i=0}^\infty (1-p)^{2i+1}\\
=p(1-p)\sum_{i=0}^\infty \left[(1-p)^2\right]^i\\
=p(1-p)\frac 1 {1-(1-p)^2}\\
=\frac{1-p}{2-p}$$
Now of course $P(Y<X)=\frac{1-p}{2-p}$ and $P(X=Y)=1-2\frac{1-p}{2-p}=\frac{p}{2-p}$
