You are playing with a short pack of just aces, kings, queens and jacks.
The Rule is make 21 and you win.

So there are 48 ways to win (4*12)
There are 6 ways to lose with 2 aces (4*3)/2

You have your back to the dealer and he informs you your first card is an ace!
Great your chance of a win is 48/(48+6) = 8/9 = 89%

He then tells you that your card is the Ace of Hearts.
Alas.
There are 12 winning cards left for you and 3 losing cards (the other aces).
Your chances of winning are now only 12/15 = 4/5 = 80%

Winning combination is A+K or A+Q or A+J or K+A or Q+A or J+ A. All other combinations lose, right? or is K+K+A (QQA, JJA) also winning?

The actual Point of your question is that you compare apples to oranges. Before you know that the Ace is the ace of hearts you have 15 Cards left and 12 of those win. If you know the Suite nothing changed.

The probability to win if the first Card is an ace is always 12/15

The probability to win if the first Card is NOT and ace and you Need an ace is 4/15.

If however the ace counts one in case of a three card combination, then the probability to win is 4/15 for the second and 4/14 in the third draw.

The number of winning combinations given by you with 48 from 54 outcomes is false to begin with.

Originally posted by Ponderable So I Need clarification:

Winning combination is A+K or A+Q or A+J or K+A or Q+A or J+ A. All other combinations lose, right? or is K+K+A (QQA, JJA) also winning?

The actual Point of your question is that you compare apples to oranges. Before you know that the Ace is the ace of hearts you have 15 Cards left and 12 of those win. If you know the Suite ...[text shortened]... The number of winning combinations given by you with 48 from 54 outcomes is false to begin with.

Just two cards (keep it simple)

Hands containing at least one ace are:
A & picture card (4*12) = 48
2 aces (4*3)/2 = 6
You must have one of these two types if you know you have one ace.
No other possibilities.