# Given two iid geometric random variables with p=1/5, what is the probability that variates from them equate?

Given two iid geometric random variables with $p=\frac{1}{5}$, what is the probability that variates from them equate?
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Plutbantonavv
Step 1
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 ${\left(\frac{1}{5}\right)}^{2}$
- The probability that both variates are 1 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2}$
- The probability that both variates are 2 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{4}$, etc.
Step 2
Thus the probability that both variates are equal is
$\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2n}$
$=\frac{1}{25}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{16}{25}\right)}^{n}$
$=\frac{1}{25}\cdot \frac{1}{1-\frac{16}{25}}=\frac{1}{9}$
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Vincent Norman
Step 1
Think of it this way: You want to find

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 $P\left(X=Y\right)=P\left(X=1,Y=1\right)+P\left(X=2,Y=2\right)+P\left(X=3,Y=3\right)+...$
Step 2
Now, we use the fact that the two variates were considered IID. As such, each term $P\left(X=n,Y=n\right)={P}^{2}\left(X=n\right)$ and we ultimately end up with the infinite sum $P\left(X=Y\right)={P}^{2}\left(X=1\right)+{P}^{2}\left(X=2\right)+{P}^{2}\left(X=3\right)+...$