14th card is ace clarificarion. There are 2 questions. 

I need to find probablity that 14th card is ace and that first ace occurs on 14th card.

For the first part, I though it like this.

Choose any of 1 ace in ways = ${4 \choose 1}$

And choose 13 cards from remaining 51 cards= ${51 \choose 13}$

Choose 14 cards from 52 = ${52 \choose 14}$

Probability = $\frac{{4 \choose 1}{51 \choose 13}} {{52 \choose 14}}$

For Second Part

I removed all four aces, cards remaining  = 48.
Choose 13 from these = ${48 \choose 13}$
And any one ace = ${4 \choose 1}$
Probability = $\frac{{4 \choose 1} {48 \choose 13}} {{52 \choose 14}} $

Both answers do not match.
First part:
$$\frac{\binom 41 \binom {51}{13} 13!}{\binom {52}{14}14!}$$

Second part:
$$\frac{\binom 41 \binom {48}{13} 13!}{\binom {52}{14}14!} $$

NOTE:


Order matters.($13!$ and $14!$)


Your answer consider these two(and many more) cases the same:
$$[2,3,\color{green}{4,5},6,7,8,9,10,J,Q,K,2,A]\leftrightarrow[2,3,\color{red}{5,4},6,7,8,9,10,J,Q,K,2,A]$$
because you only select those $13$ cards, not arrange them before the $14$th ace card
For (1), your denominator is wrong, because of the $14$ cards chosen order is important (for at least one of the cards).  (1) would be easier if you imagined picking the $14$th card first (or if you prefer, laying out the first $13$ cards face down, and then turning up the $14$th card.

For (2) your denominator is again wrong.  You might try doing this one with order being important throughout the problem.  (I.e., choose a non-ace for slot 1, choose a non-ace from the remaining cards for slot 2, etc.)