Basic ? from Sklansky's Tournament Poker book

[VeMan]

On page 124 of Tournament Poker for Advanced Players Sklansky says that the number of hands to move in with, given that no-one has raised in front, using his "System" equals 13% of all two card combinations. That would be any pair, any other Ace-suited, ace-king (suited or off), or two suited connected card, except 43s, or 32s.
He then writes that if I can't figure out how he got that number I should close the book, that I don't know squat about poker, and deserve to lose.
I have gone through many of Sklansky's other books, especially about hold-em, and I have yet to find any reference to the number of two card combinations in Hold-Em, including in his beginning book- Hold-Em Poker. So if this is such an important number as to determine winners from losers, HE'S certainly not telling.

I'm not a math genius, but I guessed that the number of possible two card combination is 52 x 51= 2652- is that wrong??

The number of hands I calculated from his System is 174 which is 13% of something other than 2652. But I can't tell from his text if he means that I should count "Aces" as 6 possible pairs, or just 1 possible hand with 6 permutations. Likewise with the other hands.

Obviously I've been reading the wrong books or don't know enough basic math, but if Sklansky says I don't deserve to win unless I can figure this out- and apparently I can't, then I'd like some help.

Can someone help me with this math, and/or point me to some online or print reference to basic poker numbers that I'm supposed to know before I even look at a deck of cards?

Thanks

VeMan

[jb9]

When counting starting hand combinations, you don't count every combination of suits, so there are only 169 starting hands.

AKs (suited) counts as 1 hand not 4 hands (A[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/diamond.gif[/img]K[img]/images/graemlins/diamond.gif[/img] A[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/heart.gif[/img] A[img]/images/graemlins/club.gif[/img]K[img]/images/graemlins/club.gif[/img]).

[VeMan]

[ QUOTE ]
When counting starting hand combinations, you don't count every combination of suits, so there are only 169 starting hands.

AKs (suited) counts as 1 hand not 4 hands (A[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/diamond.gif[/img]K[img]/images/graemlins/diamond.gif[/img] A[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/heart.gif[/img] A[img]/images/graemlins/club.gif[/img]K[img]/images/graemlins/club.gif[/img]).

[/ QUOTE ]

OK, so we have:

13 different pairs
1 AKs
1 AKo
11 Axs
9 suited connectors (KQs - 54s)

Not counting suit variations that adds up to 35 playable hands when no one has raised (again according to Sklansky's "System" in Tournament Poker for Advanced Players.

35 hands is nearly 21% of possible two card combinations. But Sklansky said that it should be 13% which should be around 22 hands.

I don't know how this adds up. David Sklansky would be very disappointed with me. Please correct me where I'm wrong.

V

[jb9]

Hmmm, it's early in the morning, but I can't see a problem with your math, which must mean I deserve to lose too (which certainly fits well with the evidence I have from last night [img]/images/graemlins/frown.gif[/img]).

I re-read Tournament Poker recently, but, to be honest, I skipped the section about The System. Sorry I can't help...

[tek]

[ QUOTE ]
He then writes that if I can't figure out how he got that number I should close the book, that I don't know squat about poker because I'm Christian and deserve to lose.

[/ QUOTE ]

FYP

[AKQJ10]

[ QUOTE ]
When counting starting hand combinations, you don't count every combination of suits, so there are only 169 starting hands.

AKs (suited) counts as 1 hand not 4 hands (A[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/diamond.gif[/img]K[img]/images/graemlins/diamond.gif[/img] A[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/heart.gif[/img] A[img]/images/graemlins/club.gif[/img]K[img]/images/graemlins/club.gif[/img]).

[/ QUOTE ]

They're not all equally likely, though, so calculating using 169 as your denominator will lead you astray.

For your denominator, the OP was in fact right if we use permutation (however, that complicates things): There are 52 possible first cards and 51 possible second cards (given the first card). However, A [img]/images/graemlins/diamond.gif[/img] 3 [img]/images/graemlins/club.gif[/img] and 3 [img]/images/graemlins/club.gif[/img] A [img]/images/graemlins/diamond.gif[/img] are functionally equivalent and equally likely, so we'll simplify things by using combinations instead of permutations. The number of combinations, C(52, 2), is 52 * 51 / 2. That's your denominator.

As for the numerator: For each hand that Sklansky names, you need to count the number of combinations. Pocket aces, for example, are six different combinations, C(4, 2) = 4 * 3 / 2 because we already said order does not matter. AQs is four different combinations (the four suits). So you'd have to sum this across each of the hands Sklansky names. I'm too lazy to do the math, but that's the simplest way to do it.

Note that AA is 6 combinations but AQs is only 4, so AA is more likely than AQs. This is why you can't just count the number of conventional hold 'em hands and divide by 169; you'd be assuming they're all equally likely, but they're not.

[jb9]

OK, math isn't my strong suit, but according to this web page there are 1326 starting hand combinations, and then the hands we are looking at are (I hope I got this right...):

78 different pair combinations
4 AKs combinations
12 AKo combinations
44 Axs combinations
36 suited connector (KQs - 54s) combinations

For a total of 174 combinations.

(174/1326)*100 = 13.1%

Right?

Does this mean I'll deserve to win tonight [img]/images/graemlins/cool.gif[/img]?

[VeMan]

[ QUOTE ]
OK, math isn't my strong suit, but according to this web page there are 1326 starting hand combinations, and then the hands we are looking at are (I hope I got this right...):

78 different pair combinations
4 AKs combinations
12 AKo combinations
44 Axs combinations
36 suited connector (KQs - 54s) combinations

For a total of 174 combinations.

(174/1326)*100 = 13.1%

Right?

Does this mean I'll deserve to win tonight [img]/images/graemlins/cool.gif[/img]?

[/ QUOTE ]

It means we ALL deserve to win!!!
So my 174 number was right, but my total number of hands was twice too much- so I'm only half an idiot. That gives me an even chance to beat total idiots. Wait...

Is my math right on that??

V

[AKQJ10]

[ QUOTE ]

So my 174 number was right, but my total number of hands was twice too much-

[/ QUOTE ]

Yes -- this was because you were calculating the number of permutations (distinct hands where 9 [img]/images/graemlins/club.gif[/img] 3 [img]/images/graemlins/heart.gif[/img] is a different hand than 3 [img]/images/graemlins/heart.gif[/img] 9 [img]/images/graemlins/club.gif[/img]) but calling it combinations, and using combinations in your numerator. Note that with two cards, the ratio of permutations to combinations is exactly 2, meaning there are only two different "orders" in which to deal a specific hold 'em hand: XY or YX.

If you were doing the same for three-card starting 7-stud hands (and leaving aside the fact that order does in fact matter, because opponents see your third card),

P(52, 3) = 52*51*50
C(52, 3) = 52*51*50 / (3 * 2 * 1)

The ratio of the two is six because there are six different orders to deal a three-card hand: XYZ, XZY, YXZ, YZX, ZXY, ZYX.

You can do Omaha and draw as an exercise. [img]/images/graemlins/grin.gif[/img]