Pot Odds

[oldvarmint]

In chapter 5 (Pot Odds)of Sklanskys "The Theory of Poker". He gives an example of counting all the "seen" cards. He starts out with a 7 card stud open ended straight draw (6,7,8,9)after 4 cards have been dealt. The number of "other" seen cards are 8. He then gives the odds if none of the seen cards are a 5 or a 10; to hit the straight as 49.8 percent. Could someone tell me where I'm missing this. I see a total of seen cards (counting my 4) as 12, total cards left unseen 40. So I see it 8 cards can help me hit the straight and 32 won't. Which means to me you have a 25% chance of hitting it (I know I'm probably doint this all wrong). How does Sklanskly come up with 49.8?

His next example deals with a razz game. I have a 4,3,2,A and the number of seen cards as 10. He sayes you have a 81.8% chance to hit a 8 or lower, if none of the seen cards are a 8,7,6,5. Once again if I count the seen cards (14) and unseen cards (38). I come up with 16 cards to make the 8 lower and 22 won't make it. At least this is closer with a difference of Sklansky's 81 to my 72 (once again I know I'm probably doing the math wrong, thats why I'm ask this question).

So I'm all screwed up, could someone help me here on what I'm doing wrong.

Oldvarmint

[Alex/Mugaaz]

[ QUOTE ]
In chapter 5 (Pot Odds)of Sklanskys "The Theory of Poker". He gives an example of counting all the "seen" cards. He starts out with a 7 card stud open ended straight draw (6,7,8,9)after 4 cards have been dealt. The number of "other" seen cards are 8. He then gives the odds if none of the seen cards are a 5 or a 10; to hit the straight as 49.8 percent. Could someone tell me where I'm missing this. I see a total of seen cards (counting my 4) as 12, total cards left unseen 40. So I see it 8 cards can help me hit the straight and 32 won't. Which means to me you have a 25% chance of hitting it (I know I'm probably doint this all wrong). How does Sklanskly come up with 49.8?

His next example deals with a razz game. I have a 4,3,2,A and the number of seen cards as 10. He sayes you have a 81.8% chance to hit a 8 or lower, if none of the seen cards are a 8,7,6,5. Once again if I count the seen cards (14) and unseen cards (38). I come up with 16 cards to make the 8 lower and 22 won't make it. At least this is closer with a difference of Sklansky's 81 to my 72 (once again I know I'm probably doing the math wrong, thats why I'm ask this question).

So I'm all screwed up, could someone help me here on what I'm doing wrong.

Oldvarmint

[/ QUOTE ]

You have 3 more cards coming, not 1.

[oldvarmint]

Could you show me the math that because it doesn't add up to me?

[senjitsu]

You have 8 outs three times -- the technically accurate way is to multiply the odds of not hitting three times.

so 32 cards out of 40 unseen will not help you on the first card. On the second card, 31/39 (since you have seen a card) and on the third card, 30/38

32/40*31/39*30/38= .5020 those are your odds of catching 3 cards that dont make your straight. Subtract from 1 to get .498, or 49.8 percent.


jc


[ QUOTE ]
Could you show me the math that because it doesn't add up to me?

[/ QUOTE ]

[senjitsu]

[ QUOTE ]

So I'm all screwed up, could someone help me here on what I'm doing wrong.

Oldvarmint

[/ QUOTE ]

You're dividing the # of outs by the number of non-outs... your chances of making your hand _for each individual card_ are the number of outs devided by the total number of unseen cards. For example, in the razz problem, your chances of making your hand on 5th st are 16/38. If you do not make it on 5th st, your chances on 6th are 16/37 (because you have now seen an additional card), and if you dont make it on 6th st, your chances are 16/36 on seventh.

You can figure out your odds through multiple streets by using the method i describe in my previous post.

[oldvarmint]

Thank you so much, thats what I needed. I'm not to bright, but I still try.

OV