Combinatorics: how many hands have at least one suited ace in them?
Posted by OMGredline
Posted by
OMGredline
posted in
Low Stakes
Combinatorics: how many hands have at least one suited ace in them?
I had a quick look on PPT and can't figure out how to input this.
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hmm I had a go and got 16.95%, is this right?
I got 25.24%. I inputted Ax:xx and used the count function, think this should be correct.
So that means 'ace + random card: both same suit'?
Axx is fine and 16.95%. This includes all AA and AAA combo's. If you want all unpaired Axx combo's add $np so:
Axx$np 12.03% of all hands
OK, I understand now. I'm asking because I play all suited aces without trips on the button, and wanted to know how much of my range on the button would be made up of these hands.
Just to be clear
Axx - 16.95% hands includes suited AA and AAA combo's
Axx$np - 12.03% no AA or AAA combo's
Axx$nt - 16.84% no AAA combo's
That is interesting, just started to use PPT again but I would have thought that the Ax would restrict my results to a suited ace not an Ace and any flush draw.
Correct which is why I thought that Ax:xx would translate to Ax(a suit) :(and) xx(two of a suit). But I guess in PQL it assumes Ax(any suit) and xx(two of any suit). I know that this is not a bug but I think that once we use x, it should be a suit, not any suit. In math when we use variables you should be able to solve for them. If you asked me to list all Ax:xx hands and write out all hands where x=spades would expect me to write Ad2s5s8c?
I agree this is confusing. Intuitively, one would think that Ax:xx would mean the same thing as Axx, but it doesn't. It means the same thing as A:xx though. Apparently suit variables (wxyz) are not remembered when combining two hand ranges (using ':' for example).
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