Is there a way to find the odds of an A high flush getting beat by a straight flush.

the str8 flush doesn't occur enough for you to worry laying down your Ahigh flush

LOL.
No, of course not. Who does that? That wasn't even the question. But thanks for the advice.

Maybe I'm in the wrong forum.

Any math guys here know how to calculate this?

yea wait for some Carpals and veterans, they'll help you if they won't find your question not useless. GL

100%. An Ace high flush is never not beat by a straight flush.

Sigh.

It really highly depends on the whole situation.
Is it on the river? or on the flop? How does the board look like?
Can you make any assumptions about villain's range?

Here is one easy example:

It is on the river. Board looks like: 3h4h5h6hTd
You have Ah5d
Now there are 52-5-2= 45 unknown cards.
Your opponent holds two. You can't make any assumption about his range, so he can have any two (This won't be true at this point but it makes it easier to calculate)

Now what is the probability that he is either holding a 2h or a 7h.
The possibility that he doesn't hold at least one of them is: 43/45 * 42/44
So 1 - (43/45 * 42/44) is the possibility that he holds at least one of these cards
=> 8.787878 %

But practically the probability will be different. Sometimes higher, sometimes lower.
Lets say you vbet the river with your nutflush and you get raisedshove all in for 3times the pot. No he is super likely to have a flush. Otherwise he wouldn't do this. So he could have the Khigh flush, sometimes the Q high flush or the straightflush.
Depending on his preflop handselection it is much more likely for him to have highcards but on the other hand we can't be sure that he valueshoves always the Qhigh flush.

All you doing now is like flipping if you are beat or not. (Sometimes people have a bluffingrange here. Then it is less likely they have you beat, obv.)

You should only worry about a straight-flush when there are 4 of the same suit on board. Let's say you hold the A and a non-diamond card and the board contains 5 6 7 8 . If your opponent has one diamond only he would have a straight-flush 2 out of 8 times, that is 25%. If he holds 2 diamonds and it is equally likely he holds any two 13 out of the 28 combinations would make him a straight-flush so it is nearly 50%. But in practice he wouldn't play any 2 suited cards and he will have straight-flush less often. It is likely though and since you will only get action by someone holding a diamond you need to be more careful.

If there are 3 suited cards on board, lets say 5 6 7
and you hold A 2 and you believe that your opponent holds 2 diamonds there are 3 out of the 21 2-diamond combinations that makes him a straight flush. If he only plays suited connectors, 34 and 89 will make him a straight flush and 9T,TJ,JQ,QK won't. He could also play more combinations of suited high cards that would make the straight - flush less likely. Also on a board such as 5 6 8 you need to have the 7 so it even less likely. In most boards it is not even possible.