Tournment poker

[TxSmoozer]

I just completed Mr. Sklansky's book Tournment Poker. For those that make have the book it (page 123) he refers to a system he developed (for a young lady new to poker)in that "if no one else has raised in front of your, she was to move all her chips in with any pair, any ace-other suited, ace king (suited or off suit, or two suited connected cards, except for four -trey or trey duce. He further states that this comprises 13 per cent of all the two card combinations. I am fairly new and would like to know how this was caculated and how many total combinations are there in hold em. Any assistance appreciated

[TrainHardDieHard]

This is wrong forum section for your post. But I read the book a long time ago and I'm pretty sure he said if you don't know how to get that number then you shouldn't be reading the book. [img]/images/graemlins/smile.gif[/img]

[Rob Usto]

[ QUOTE ]
This is wrong forum section for your post. But I read the book a long time ago and I'm pretty sure he said if you don't know how to get that number then you shouldn't be reading the book. [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]

I remember that exact passage, and I remember thinking it was far too strong a statement.

[slik]

52 * 51 combinations = 2652 (since the first card you're dealth is out of a full deck. the second card is out of a deck-1). divide by 2 for symmetry (AsAd = AdAs) = 1326.

Next there are 12 suited aces (A2s-AKs) * 4 because can be any suit = 48 combos
1 AKo (AdKs, AdKc, AdKh) * 4 for any suit = 12 combos
13 pairs * 6 (each pair has 6 combos .. 2s2c 2s2d 2s2h 2c2d 2c2h 2d2h) = 78,
9 suited connectors 56s+ other than AKs * 4 =36.

So that comes to 174 combinations.
174/1326 = .131 = 13.1%

PS, these calculations are annoying. Google for "poker calculator" or something of the sort, and most will have this sort of function in them for you.

[slik]

also, these type of posts should go in the probability forum