Help me out a bit, dont know how to

[stokken]

Say I am in the SB and hold A9os.

There are then 3 A`s left and 4 ranks higher then my 9

There are 3 ways to be delt AA, and twelwe ways of Ax x= T-K

So so 51 combinations of A and the higer ranked cards?

There are 52x51/2= combinations of any two cards, but I hold 2 so 50x49/2= is the number that I will compare my feared combinations towards?

What then if I am on co adding 3 players to the situation.
Likelyhood of any of the combinations I fear then? Simply add? What about if there are more then one Ax hand there?

Please help, new to this

[donkeyradish]

Thats not the way to look at it

If there is a raise before you then think carefully if that player likely has an A with a better kicker. Some players raise with only that or big pairs. Consider folding.

But if there isn't then you are fairly likely to have the best ace.

[Galois]

There is a small error in your computation as far as the number of combinations of cards is concerned.
There are 3 A's left and 16 cards from T-K, so the number of ways to choose the first A is 3 times the remaining 18 cards, the 16 cards from T-K plus the two other aces. So the number of combinations is 54.

[AaronBrown]

[ QUOTE ]
There is a small error in your computation as far as the number of combinations of cards is concerned.
There are 3 A's left and 16 cards from T-K, so the number of ways to choose the first A is 3 times the remaining 18 cards, the 16 cards from T-K plus the two other aces. So the number of combinations is 54.

[/ QUOTE ]

Almost. 3x16 = 48 ways to get Ax, x>9, is correct. But there are only 3 ways to get AA (any one of the three remaining Aces can be left out). The formula for that is 3*2/2 because the Aces are indistinguishable, rather than 3*2 as you effectively did. So 51 is the correct number.