TPFAP Question

[Godfather80]

Okay, so I have been reading Tournament Poker for Advanced Players and have become very confused regarding a statement David makes on page 124. He writes the following regarding his No Limit Tournament "System":

"Notice that the hands that she was to move in with (again, when no one has raised in front of her) comprised about 13% of all the two card combinations. (If you don't know how I got that, stop reading this book right now. You are not ready for it. You don't know enough about poker. And, you deserve to lose."

Yipes. The problem I have with this statement is that on page 123 David writes:

"If no one else has raised in front of you, move in all your chips with any pair, any ace-other suited, ace-king (suited or offsuit), or two suited connected cards, except for four-trey or trey deuce."

So let me spell it out. David would have his "System" player go all in if no one raised in front with any pair (13 hands), any ace-other suited (12 hands), ace-king suited or off-suit (1 hand, no double counts allowed), or two suited connected cards, except for 43s and 32s (9 hands).
13 + 12 + 1 + 9 = 35 hands.
The problem is that 35 hands out of a possible 169 two card combinations actually comes out to over 20%.

So according to the book, because I cannot see how 13% = 35/169, I deserve to lose. Maybe I already have lost by spending $29.95 on a writer to arrogant to check his math. Or maybe I'm an idiot. Thanks in advance for the help.

Godfather

[Sheriff Fatman]

The 169 hand combinations are a summary of each hand type but they don't represent the total number of starting hands you can be dealt, as there is more than 1 way of being dealt each of these combinations. To do the maths you need to use these individual 2 card starting hands, which requires a basic understanding of permutations and combinations. Taking suits into account there are actually 1326 starting hands you can be dealt.

For example, there are 6 ways of being dealt AA due to the combinations of the suits. This is the same for any pair. Unsuited cards have even more possibilities available for each of the AKo type descriptions.

Therefore the whole foundation of your calculation is flawed as you cannot use the 169 combinations as the basis for it.

If you want help in actually doing the calculations I'd suggest posting a query on the probability forum where the maths gurus lurk! They will probably be able to explain it further better than I can.

A good book for this type of calculation is Holdem Odds by Mike Petriv which starts from a very basic level and takes you through this types of calculations. It may help your understanding if you're still struggling.

Sheriff

[William]

Maybe I already have lost by spending $29.95 on a writer to arrogant to check his math. Or maybe I'm an idiot. Thanks in advance for the help.

I guess you kind of answered your own question.
You have spent 29.95 in a book you apparantely don't understand (don't blame David, the book IS called FOR ADVANCED PLAYERS )and it seems that you're right as well about the second half of your statement.

Life's brutal, I know [img]/images/graemlins/frown.gif[/img]

William

[Paul2432]

I'll give a helpful response.

Each of the 52 cards in the deck can combine with any of the other 51 cards to form a starting hand. Hence there are 52*51=2652 starting hands. However, the order of the cards does not matter (7c6h is the same hand as 6h7c) so this number should be divided by two giving us 1326 starting hands.

Each pair can be made in six different ways. For example a pair of aces can be made by AsAh, AsAd, AsAc, AhAd, AhAc, AdAc.

Each suited connector or suited ace can be made in four different ways. One for each suit.

Finally each off suited hand can be made in twelve ways. (each of the four suits can combine with three other suits).

So going back to the list of hands to play.

Any pair: 13 pairs x 6 each = 78 hands
Any ace-suited: 12 hands x 4 each = 48 hands
Any AKo: 12 hands
Suited Connectors 54s to KQs: 9 hands x 4 each = 36 hands

Total hands = 78 + 48 + 12 + 36 = 174 hands

174/1326 = 13 27/221% = ~13.122%

Paul

[DrPhysic]

As much as i love william, the sheriff, and paul, they will never make a living teaching math. let me try it simple:

Yes there are 169 possible hands.
But that counts AA as one.

There are 6 ways to deal any pair. Lay out 4 aces on the table and count up the ways to make any 2 of them. it's 6.
There are 4 ways to deal any suited connector (AKs)
There are 12 ways to deal any non suited connector (AKo)

Figured that way there are 1320 possible starting hands.

And that is the way you have to figure possible starting hands, because there are 16 ways to get AK and only 6 to get AA.

Now go back and recount. The total number of hands he has her use is about 13% of 1320. NOT 13% of 169.

Does that make it simpler?

Doc

(edit: note Anytime you find something that Sklansky is doing wrong with the math, go back and find out what you didn't understand. His nickname isn't Einstein by accident)

[M.B.E.]

Sklansky is arrogant (not many would disagree with that), but he does usually check his math. He has admitted that his statement "you deserve to lose" etc. was over the top.

[Al_Capone_Junior]

hoss, you have to do the weighted average of how often you get each of those hands. there's more ways to get AK than AA.

al

[William]

And where, may I ask, have I made a math mistake?
I didn't even write a number, I just implied that he couldn't do the math himself.

Doc, you need a new pair of glasses [img]/images/graemlins/grin.gif[/img]

William

[M.B.E.]

[ QUOTE ]

Figured that way there are 1320 possible starting hands.

[/ QUOTE ]
DrPhysic -- it's 1326, not 1320. Paul was correct.

[DrPhysic]

you're right, so is paul. I shouldn't write things from memory late at night.

D