If it's known that the 4 cards have different ranks, what is the probability that the first card is a spade? First, I choose the spade; 13 ways. Then throw away the other 3 cards of that rank. From the remaining 48 cards I choose 1, and throw away the other 3 cards of that rank. From the remaining 44 cards I choose 1, and throw away the other 3 cards of that rank. From the remaining 40 cards I choose 1. The size of the sample space is C(52,4). Therefore,
$P(E) = (13)(48)(44)(40)/C(52,4)$
but this is larger than 1.
Second, I tried a conditional approach. P(1st card is a spade|the 4 cards are of different ranks). So the probability that the four cards are of different ranks, is P = C(13,4)/C(54,4). Then the probability that the first card is a spade is P = C(13,1)C(12,3)/C(52,4). But again the conditional probability is larger than 1.
What am I doing wrong?
 A: Your numerator counts ordered draws of four cards, while the denominator counts the unordered draws so there is roughly a factor $24$ in excess.  The approach is wrong because you are given that the ranks are different, not calculating the chance that from a random draw you get a spade first and three other ranks.  
The simple answer is to argue that the three other cards do not change the chance that the first card is a spade at $\frac 14$ or to argue symmetry between the suits for the same result.
A: The answer is "obviously" $\frac14$ but if you want to do the detailed calculation then:


*

*The number of ways of choosing $4$ cards in order is $52 \times 51 \times 50 \times 49$

*The number of ways of choosing $4$ cards of distinct ranks in order is $52 \times 48 \times 44 \times 40$

*The number of ways of choosing $4$ cards of distinct ranks in order with the first being a spade is $13 \times 48 \times 44 \times 40$, as you have already said


So if it is known that the $4$ cards have different ranks, the probability that the first card is a spade is $$\dfrac{ 13 \times 48 \times 44 \times 40}{52 \times 48 \times 44 \times 40}= \dfrac{13}{52} = \dfrac14$$
One of your errors is to have the numerator as a count of four cards in order and the denominator as a count of four unordered cards
A: You're double-counting.  Picking a club, then a diamond, then a heart, say, gives the same hand as picking a heart, then a diamond then a club.  But all this is unnecessary.  The first card is equally likely to be a spade as it is to be a club, or a heart of a diamond, so the probability is $\frac14.$
In case I didn't make my meaning plain, if the question were "What is the probability that the first card is a diamond?" what would you do differently?
A: 
The size of the sample space is C(52,4). 

No, it's not. The sample space is the all sets of 4 cards with different ranks. There are P(13,4) different ways to choose the ranks (since you are distinguishing between the cards in the numerator, you should do so in the denominator, so it's permutations, not combinations) and once you choose the ranks, you have to choose among four suits four times, so that's 4^4. This gives 1/4 as the final answer.

So the probability that the four cards are of different ranks, is P = C(13,4)/C(54,4). 

Even assuming the 54 is a typo and taking it to be 52, this is not correct. C(13,4) gives you the number of different ways to choose different ranks. But choosing ranks is not the sames as choosing cards. Given a rank, there are four different cards corresponding to that rank, so you need to multiply it by 4^4. 

Then the probability that the first card is a spade is P = C(13,1)C(12,3)/C(52,4). 

You don't present what logic you're using, so we can't say exactly what went wrong, but one issue here is that you seem to be distinguishing between the "first" card in the numerator, but in the denominator you are using combinations, which means that you aren't distinguishing between cards. So, for instance, (2S, 3D, 4H, 5S) and (5S, 3D, 4H, 2S) would be counted as separate hands in the numerator, but the same hand for the denominator.
