Let $X$ be the number of draws needed to draw the ace of spades. Find the pmf of $X$ if we draw with replacement I'm learning about PMF at the moment and it's a bit hard to grasp. Watched a bunch of videos but I want to clarify using a problem example assigned from my homework.


  
*Cards are drawn one by one at random from a deck of cards. Let $X$ be the number of draws needed to draw the ace of spades. Find the pmf of $X$ if we draw
  (a) with replacement and
  (b) without replacement.
  

So I tried working this out by hand.
A) Drawing for the ace of spades with replacement means the card is put back each time it's drawn. So the chance for drawing ace of spades is consistently $1/52$. I doubt, however, the PMF is $1/52$ then, since a property of PMF is that all the probabilities must total to $1$. 
I saw an example in my textbook using dice, which is arguably a replacement problem as well and the equation they gave me was this: 

Example 2.7. If we roll a die and let $X$ denote the number of rolls until the first $6$ appears, then Example 1.41 reveals that the pmf of $X$ is 
  $$p(k) = P(X = k) = \frac{1}{6}\left(\frac{5}{6}\right)^{k-1}, k = 1,2,\dotsc$$

I understand here that there is a $1/6$ chance of getting a $6$ on a roll. But why does the second part of the equation exist: $\left(\frac 56\right)^{k-1}$? Is this supposed to show that as the number of rolls goes higher and higher, the chance of getting a 6 at that specific roll grows smaller and smaller? 
And if that's the case, then the graph itself as we go to infinity gets closer and closer to $0$. However, with the probability I calculated for part a I'll just end up getting a straight line.
Does that mean my equation then should be: $\frac{1}{52} \left(\frac{51}{52}\right)^{k-1}$?
B) I'm actually pretty lost here. I understand that the denominator shrinks ... so
$P(1) = 1/52 $
$P(50) = 1/2$
How do I show this in PMF?

Regarding the dice problem, notice that it is talking about the number of trials until you get a six. Another way of saying this is the number of trials until the first success. Then ask yourself, what is the probability that I roll a six on the first try? Clearly, $1/6$. We look to the forumla
$$P(X = 1) = \frac{1}{6}\left(\frac56\right)^{1-1} = \frac{1}{6}$$
and it matches. What about the probability that it happens on the second trial? It must fail the first trial and succeed on the  second, which is $(5/6)(1/6) = 5/36$ using independence. Again, the formula gives
$$P(X = 2) = \frac{1}{6}\left(\frac56\right)^{2-1} = \frac{5}{36}$$
which matches.
So the formula in the example is saying that you fail the first $k-1$ times and the succeed on the $k$th trial. 
Using similar logic, part (a) should be 
$$P(X = k) = \frac{1}{52}\left(\frac{51}{52}\right)^{k-1}\quad k\geq 1.$$
Now, the same logic applies for part (b) except, we no longer have independence since we do not put the card back. Having that in mind, what should the probabilities be? Consider $X= 1$. This means get the ace of spades on the first try. Then 
$$P(X= 1) = \frac{1}{52}.$$
If we consider $X = 2$, then this means fail on the first trial and then succeed on the second, which gives
$$P(X = 2) = \frac{51}{52}\cdot\frac{1}{51}.$$
Can you finish? What are the possible values of $k$?
