Help understanding pot odds calc from Sklansky's book

[jsthomas64]

I'm re-reading Sklansky's Theory of Poker and for some reason can't re-calc one of his examples - someone PLEASE help me or I will not fall asleep tonight.

In Chapter 5 relating to pot odds Sklansky writes the following relating to a seven card stud game:

You start with (5 [img]/images/graemlins/spade.gif[/img] 5 [img]/images/graemlins/club.gif[/img]) A [img]/images/graemlins/diamond.gif[/img]
on your first three cards in seven card stud. You have seen seven other cards. Chances for aces up or three of a kind are as follows, based on total number of 5's and A's seen on other players boards.

0 - 41.0%
1 - 34.1%
2 - 26.5%
3 - 18.3%
4 - 10.5%

Can someone please explain the math because I'm not getting it correct.

[jay_shark]

Since you've seen 7 of your opponents exposed cards and you have 3 yourself , there must be 42 unknown cards remaining .

What you really want to know is what's the probability your hand improves with 4 more cards to come .

Lets assume you get to see the next 4 cards ; and that no aces or 5's have been exposed .
The probability you hit trips with 5's is 2*40c3/42C4= .176
The probability you hit quads with 5's is 40c2/42c4.
The probability you hit exactly two pair is 3*(42-3-2)C3/42c4=0.208
The probability you hit 3 aces with 2 fives is 3*(42-2-3)C2/42C4=0.0178
The probability you hit quads with aces is 39/42C4

Add up all these events and you should get 41 % .

You have to work these things into cases and make sure you don't over count your probabilities .

[jsthomas64]

[ QUOTE ]
Since you've seen 7 of your opponents exposed cards and you have 3 yourself , there must be 42 unknown cards remaining .

What you really want to know is what's the probability your hand improves with 4 more cards to come .

Lets assume you get to see the next 4 cards ; and that no aces or 5's have been exposed .
The probability you hit trips with 5's is 2*40c3/42C4= .176
The probability you hit quads with 5's is 40c2/42c4.
The probability you hit exactly two pair is 3*(42-3-2)C3/42c4=0.208
The probability you hit 3 aces with 2 fives is 3*(42-2-3)C2/42C4=0.0178
The probability you hit quads with aces is 39/42C4

Add up all these events and you should get 41 % .

You have to work these things into cases and make sure you don't over count your probabilities .

[/ QUOTE ]

Thanks for the response, but could you do me one more huge favor and remind me what mathematical function 'C' indicates? I can't remember and still can't re-calc. I searched for a glossary or common terms for this forum but couldn't find it.

[jay_shark]

Most calculators have this function and it's pretty easy to do . You'll probably see a nCr function either on the button itself or above it . If it's located above , then you use the second function after you hit n . So to figure out 40c2 , you hit the number 40 , then hit the second function button on your calculator and press the button that's labeled nCr followed by a 2 .Your calculator may also use different notation to describe this such as C(n,r).

In general C(n,r) = n!/((n-r)!r!)
so 40C2= 40!/(2!38!)= 40*39*38!/(2!38!), the 38!'s cancel out and you're left with 40c2=40*39/2

Likewise 40C3= 40*39*38/(3*2*1)

[BruceZ]

[ QUOTE ]
Thanks for the response, but could you do me one more huge favor and remind me what mathematical function 'C' indicates?

[/ QUOTE ]

C(x,y)

[jsthomas64]

Fantastic - thanks!