Creating a PDF for a discrete random variable with a countably infinite set of values? I am unsure how to transition from discrete random variables with a finite set of values to ones with a countably infinite set of values. 
The question that spawned this problem:
A bucket has two white and one black marble. You will continuously draw marbles from the bucket until you get a white, but if you draw a black you put it and an extra black back into the bucket. If we let X = # of draws, find P(X=k) and show the function is a PDF.
X seems to me to be a discrete random variable as the set of values it can take on is {1, 2, 3 ... }, the set of positive integers. However, the set of values is countably infinite, which is preventing me from using the simple steps to show the probability distribution when the set finite.
 A: The simple thing to do is to calculate small cases, detect a pattern, and generalize.
For instance, the probability that you would only need to draw once is $$\Pr[X = 1] = \frac{2}{3},$$ because the initial state of the bucket is $2$ white and $1$ black. Thus the chance you get a white on the first draw is $2/3$, and the drawing stops upon the first draw of a white marble.
Next, the probability that you would need to draw exactly twice is $$\Pr[X = 2] = \frac{1}{3} \cdot \frac{2}{4} = \frac{1}{6},$$ because the only way to stop at 2 draws is to first fail to draw a white with probability $1/3$, and then to draw a white on the second try, which occurs with probability $2/4$ (as there are now two black marbles as a result of the first failed draw).
And now we can also easily calculate $$\Pr[X = 3] = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{2}{5} = \frac{1}{15},$$ and $$\Pr[X = 4] = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{2}{6}.$$  And now the pattern is evident:
$$\Pr[X = n] = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \ldots \cdot \frac{n-1}{n+1} \cdot \frac{2}{n+2}.$$  Can we simplify this product?  Sure:  $$\Pr[X = n] = \frac{4(n-1)!}{(n+2)!} = \frac{4}{n(n+1)(n+2)}.$$
And this makes sense, because the probability that exactly $n$ draws are required to obtain the first white marble is to note that we must have $n-1$ successive draws of a black marble, followed by drawing a white.  The $k^{\rm th}$ draw of a black marble, conditioned on the previous $k-1$ failures, occurs with probability $\frac{1}{3} \frac{2}{4} \frac{3}{6} \cdots \frac{k}{k+2}$.
The only thing remaining to check is that $$\sum_{n=1}^\infty \Pr[X = n] = 1;$$ in other words, that this probability mass function is in fact a valid one.  This is not terribly difficult to do--here is a hint:  $$\frac{4}{n(n+1)(n+2)} = 2 \left( \frac{1}{n} - \frac{1}{n+1} - \frac{1}{n+1} + \frac{1}{n+2} \right).$$
A: \begin{align}
\Pr(\text{W on 1st trial}) & = \frac 2 3 \\[10pt]
\Pr(\text{W on 2nd trial}\mid \text{B on 1st}) & = \frac 2 4 \\[10pt]
\Pr(\text{W on 3rd} \mid \text{B on 1st & 2nd}) & = \frac 2 5 \\[10pt]
\Pr(\text{W on 4th} \mid \text{B on 1st 3}) & = \frac 2 6 \\[10pt]
\Pr(\text{W on 5th} \mid \text{B on 1st 4}) & = \frac 2 7 \\
& \,\,\,\,\vdots 
\end{align}
So for example,
$$
\Pr(X=6) = \left(1 - \frac 2 3\right)\left(1 - \frac 2 4 \right)\left(1 - \frac 2 5 \right) \left(1 - \frac 2 6\right)\left( 1- \frac 2 7 \right)\cdot\frac 2 8.
$$
Next you might wonder if there's a closed form.
A: Just to get you on track:


*

*$P(X=1)=\frac23$

*$P(X=2)=\frac13\frac24$

*$P(X=3)=\frac13\frac24\frac25$

*$P(X=4)=\frac13\frac24\frac35\frac26$


Do you see why?
Do you recognize a pattern that enables you to find a closed form for $P(X=k)$?
You are also asked to show that $\sum_{k=1}^{\infty}P(X=k)=1$ wich shows that you are dealing with a probability mass function (a PMF), and not a probability density function (a PDF).
That means that almost surely a white marble will be drawn once.
