a razz odds question

[RustyBrooks]

Of course I factored them into the equation. If you have a 57% chance of drawing 2 3 4 5 6 7, then you have a 43% chance of catching all the other cards (A 8 9 T J K Q). I'm just interested in the cases where you catch good and he catches bad. There are 4 branches to this probability tree and we're only interested in how often 1 particular branch happens.

And yeah, technically if he catches a T and you catch a 9 your hand looks better. But I don't know how often he'll fold there and you want him to, because his hand IS better.

But this is kind of balanced out by the fact that there are a couple cards each of you can catch that look good but aren't. If you want really comprehensive results, you'd probably need to enumerate all the cases and consider how often you'd get cases you like. The math I did above is kind of shorthand... like you're not going to be sad at all if you get a 9 and he gets a K even though he's still ahead, because he might fold.

The cases are enumerable but I don't think you'd want to do it by hand. There are 160-ish distinct 3rd streets (that's kind of just a ballpark guess, actually probably a bit fewer)

And yeah of course I am not factoring in known up cards. These can considerably sway the decision one way or the other.

[Praxising]

[ QUOTE ]
I suppose part of the problem is that I'm not entirely sure what the odds need to be on 4th to continue if I catch a 9 or a T and my opponent catches a baby card. If you can enlighten me on this subject also, I would be much obliged.


[/ QUOTE ]

Presuming you have a nice starting hand and your cards are live, you raise third so you can call a brick on 4th. (Also presuming one opponent.) Then you know you can call because you have the pot odds. Read the Razz book included in Sklansky on Poker. He explains it all.

If everyone just called on 3rd, fold your brick. That's the standard line.

[cjk73]

[ QUOTE ]
Say you have AT8 and your opponent has A68. You know your 3 cards, your opponent's up-card and 6 other up cards, for a total of 10, so there are 42 unknown cards.

You have 24 cards to improve - so you'll improve 57% of the time.
Your opponent needs to catch J K Q or 8 - we'll ignore his pair cards for a moment. So there are 14 bad cards for him, he'll catch bad 33% of the time.

[/ QUOTE ]

I am not trying to be nitty, really just trying to learn razz math....if you know villains hand to the point of being able to determine what is good for you bad for him, dont you have to take into account his 6 and only count 23 good cards for you?

[appears]

[ QUOTE ]
I suppose part of the problem is that I'm not entirely sure what the odds need to be on 4th to continue if I catch a 9 or a T and my opponent catches a baby card. If you can enlighten me on this subject also, I would be much obliged.


[/ QUOTE ]

I've read (and re-read) SOP. But what constitutes a brick, or at least an auto-fold brick, isn't so clear, and the ante structures of the games that Sklansky describes are different. Sklansky makes it clear that in a pot that was double-raised on 3rd st., 4th st. is an automatic call as long as you started with a three card hand. In the 15-30 game Sklansky describes, that means 88-15 or about 6-1 gives you that automatic call. Well, I've been playing the 2-4 on stars, where in a single raised pot, I can be getting 11-2 or 5.5-1 on 4th st., so I suppose I still need to call if I brick.

Actually, I was hoping to pry the odds from someone so I wouldn't have to sit down and do the math myself, but I just did, and maybe I'm better off for it.

[appears]

[ QUOTE ]
Of course I factored them into the equation. If you have a 57% chance of drawing 2 3 4 5 6 7, then you have a 43% chance of catching all the other cards (A 8 9 T J K Q). I'm just interested in the cases where you catch good and he catches bad. There are 4 branches to this probability tree and we're only interested in how often 1 particular branch happens.

And yeah, technically if he catches a T and you catch a 9 your hand looks better. But I don't know how often he'll fold there and you want him to, because his hand IS better.

But this is kind of balanced out by the fact that there are a couple cards each of you can catch that look good but aren't. If you want really comprehensive results, you'd probably need to enumerate all the cases and consider how often you'd get cases you like. The math I did above is kind of shorthand... like you're not going to be sad at all if you get a 9 and he gets a K even though he's still ahead, because he might fold.

The cases are enumerable but I don't think you'd want to do it by hand. There are 160-ish distinct 3rd streets (that's kind of just a ballpark guess, actually probably a bit fewer)

And yeah of course I am not factoring in known up cards. These can considerably sway the decision one way or the other.

[/ QUOTE ]

Here's the problem the way I see it. Let's say I have T2/A v xx/A. For me to at least appear to outdraw my opponent, I must catch a 2, 3, 4, 5, 6, 7, or 8 while my opponent catches an A, 9, T, J, Q, or K. So that's 27 cards for me and 21 for my opponent. Assume it's an eight-handed game, so there are six other cards out, but they're all random, so I'll just consider them unknowns. So that's: (27/48)*(21/47) = .56*.45 = .25 = 3-1. If I assume my opponent will still call me if he catches a 9 and I catch an 8 or better, then the odds drop to 4-1. But is this a reasonable assumption?

I guess this kind of goes back to what I was trying to say before but might not have said very well. There are few instances where you catch "good" and your opponent catches "bad" but not "bad" enough to warrant a fold. Those instances might be: xx/A8 v xx/A9, xx/A8 v xx/AT, (maybe) xx/A7 v xx/A9 . But in all other instances, you'll have xx/A[insert 2,3,4,5,6,7 here] v xx/A[9,T]. So far more often that not, 9s and Ts are bad cards for your opponent. But again, I'm not sure how to put an exact figure on that.

In any event, I think I learned something here. Thanks for the responses.

[RustyBrooks]

It's quite difficult to put an exact number on the sitation without enumerating the possibilities, for the reason you've noted. The brickness of a card for your opponent is dependent on what you draw, for the middle cards like 9 and T.

As for what cjk73 said, we know what his hand is just for the purposes of the example. I didn't include 6 and 8 in the list of bricks because you won't know that they're bricks - you're better off assuming that any low card that falls didn't pair him unless you have good reason for the contrary.

There's an awful lot of hand-waving in the math here, I admit, but I think it's pretty close/reasonable.

And I'm glad you went through the effort to do the math, appears, it's far better than going by rules of thumb, which may or may not apply to the situation at hand. At the table I do math like this in only the vaguest of terms, I think of 1/3 1/4 1/2 etc and not "57% this" and "33%" that.

Something I am struggling to come up with is a short/fast enough way to discount my outs on later streets, when I am drawing to beat my opponent's current hand, but not his best *possible* hand - so sometimes I will make my hand and lose. It's easy to do with pencil and paper, but I haven't found sort of any good short way to do it in 5 seconds. With holdem you have the same problem but you sort of just get used to discounting outs roughly. But holdem is simpler because you only have to consider which of your outs could be good for your opponent. In stud games your opponent's final card is wholly independent of yours.

[runninice]

in a razz thread you posted

I'm at the tables at the moment but I'll post some graphs later about this.

what graphs? id like to graph my razz play

[RustyBrooks]

Ah, check this out:
http://www.rustybrooks.com/poker/new_analysis/

It's an analysis method I'm working on. It lets you look at your equity against a range of hands on this street, vs your equity against the same range on the next card.

It's quite time intensive at the moment and fairly fragile. I might put an interface to it on the web in the near future.