Probability of equality of variates from two iid geometric random variables 
Given two iid geometric random variables with $p=\frac15$, what is the probability that variates from them equate?

The joint mass function is just the product of the mass functions I know that, but I suck at the rest! Can someone give me a hint?
 A: How can it happen that $X=Y$?
It could happen if $X=Y=0$ (provided you have the kind of geometric distribution in which $X=0$ is a possible outcome).
It could happen if $X=Y=1.$
Or if $X=Y=2.$
Or if $X=Y=3,$ and so forth.
Moreover each of these events is disjoint from all the others.
You can express the event $X=Y$ as the union of a collection of
disjoint events of this type. How?
What is $P(X=Y=0)$? What is $P(X=Y=1)$? What is $P(X=Y=2)$?
What is $P(X=Y=3)$?
What about all the other events of this kind?
A: No need for the joint mass function. If the variables count the number of failures before the first success:


*

*The probability that both variates are 0 is $\left(\frac15\right)^2$

*The probability that both variates are 1 is $\left(\frac15\right)^2\left(\frac45\right)^2$

*The probability that both variates are 2 is $\left(\frac15\right)^2\left(\frac45\right)^4$, etc.


Thus the probability that both variates are equal is
$$\sum_{n=0}^\infty\left(\frac15\right)^2\left(\frac45\right)^{2n}$$
$$=\frac1{25}\sum_{n=0}^\infty\left(\frac{16}{25}\right)^n$$
$$=\frac1{25}\cdot\frac1{1-\frac{16}{25}}=\frac19$$
A: Think of it this way: You want to find
$P(X=Y) = P(X=Y=1 \text{ or } X=Y=2 \text{ or } X=Y=3 \text{ or } ...)$.
These are clearly disjoint cases, so we break apart into a sum of probabilities while also realizing that each such statement $X=Y=n$ is equivalent to saying $X=n$ AND $Y=n$. Thus we have 
$P(X=Y) = P(X=1,Y=1) + P(X=2,Y=2) + P(X=3,Y=3) + ...$ 
Now, we use the fact that the two variates were considered IID. As such, each term $P(X=n,Y=n) = P^2(X=n)$ and we ultimately end up with the infinite sum 
$P(X=Y) = P^2(X=1) + P^2(X=2) + P^2(X=3) + ...$
Should be pretty straight-forward from there.
