How to define probability of a random variable while calculating expectation? 
I have a well shuffled deck of $52$ cards, and I can not see their faces.
  I am going to take one card at a time and check its face.
  If it is not the Ace of Spades, then keep it aside and pick another card.
  I am going to repeat this process until I get the Ace of Spades.
  What will be the expected number of cards required to be turned over?

I solved this question in two different approaches, but I got two different answers.    

Let $X$ be a random variable that denotes the number of cards we have to turn over.     
So $X$ can take any integral value from $1$ to $52$.
We have to calculate $E[X]=\sum\limits_{x=1}^{52}\left(x\cdot P[x]\right)$.    
Here are the two ways I tried to find the Probability function:


*

*$P[X=i]=\left(\dfrac{51}{52}\right)^{i-1}\cdot\dfrac{1}{52}$

*$P[X=i]=\dfrac{\binom{51}{i-1}}{\binom{52}{i-1}}\cdot\dfrac{1}{52}$    


Which one is the right method and why?
 A: Let $X$ be the number of cards you have to reveal in total, including the ace of spades. It can also be interpreted as the position of the ace of spades in the deck. 
As we all know, when the deck is well shuffled, the ace of spades is equally likely to be in any position, so we have $$P(X=i)=\frac{1}{52}$$ for any natural number $i\leq 52$.
Now we can calculate the expectation:
$$
E(X)=\frac1{52}+\frac2{52}+\cdots+\frac{52}{52}\\=\frac{52\cdot53}{2\cdot52}=\frac{53}2=26.5
$$
A: The probability that the first card you pick is the ace of spades is 1/52.  Since, if the card is not an ace of spades, you "put it aside", there are then 51 cards so the probability the second card you choose is the ace of spades is (51/52)(1/51)= 1/52 again.  (The "51/52" is the probability the first card chosen is NOT the ace of spades.)  Similarly, in order that the third card you choose be the ace of spades, you must NOT choose the ace of spades the first two times so the probability of that is (51/52)(50/51)(1/50)= 1/52 again.  It follows easily that the probability that you pick the ace of spades on the "nth" draw, for any n from 1 to 52, is 1/52.  The expected value of that number, x, on which draw we get the ace of spades, is 1/52+ 2/52+ 3/52+ ...+ 52/52= (1/52)(1+ 2+ 3+ ...+ 52)= (1/52)(52)(53)/2= 53/2, the "middle" value which might have guessed from the start.
A: To answer your specific question, neither of your formulas is correct. The correct formula is
$P[X=i]=\dfrac{\binom{51}{i-1}}{\binom{52}{i-1}}\cdot\dfrac{1}{53-i}$
since the particular card is among the $\;\;52-(i-1) = 53-i$ cards remaining
However, since cards don't have preference for positions, we can conclude directly that the probability $= \frac1{52}$ for each position. 
A: Consider symmetry. For any order of the cards in the deck, there also exists an order where they are the other way around.
For example, if in the first case the Ace is found after 11 cards (including the Ace), in the other order you would find the other 42 cards (again including the Ace) first. On average, that is 26.5 cards.
As this holds for every pair of an order and its reverse, it is also the expected amount for all possible orders.
A: $$
\Pr(X=i \mid X\ge i) = \frac{1}{52-(i-1)}, \text{ but } \Pr(X=i) = \frac 1 {52}.
$$
Thus
$$
\operatorname{E}(X) = \sum_{i=1}^{52} i\cdot\frac 1 {52} = \frac{1+ 2 +3+\cdots+52}{52} = \frac{53} 2 = 26+ \frac 1 2.
$$
