a razz odds question
 

I've read in more than one place that the odds of outdrawing a single opponent on 4th st. is 4-1 . Can someone show me the math? I can't seem to come up with that number. Also, if this number is correct, does this mean that on any given street, if you are getting 4-1 pot odds, you need to be calling if you feel you are at most only one card behind? Two examples : 1) defending the bring-in against a single opponent showing a baby upcard when you have a hand like A2/K ; 2) you have caught bad on fourth while your opponent appears to have caught good ... say A2/72 v xx/A2 . (Assume in both examples that the cards that are out don't have any particular sway.)

I also guess it's worth mentioning that I read through this whole lengthy post on defending the bring-in but wasn't exactly sure what to draw from its conclusions. I assume that the odds of outdrawing someone on 4th st. are pretty relevant to answering that question. But I'll hold off on any follow-up questions until someone responds and sets me on the right path.

Re: a razz odds question
 

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.

So for him to catch bad, and you to catch good is 33% * 57% = 20% or in this case 4:1.

This ignore subtleties like, is a T a bad card for him if you catch good? Your board will be better but your hands could be about equal so that's a slight advantage. Also, there are more cards for him that are bad but you don't know what they are (his pair cards) and he doesn't know what cards pair you, but I tend to think that sort of evens out.

As the streets progess the numbers change a bit because a) there are fewer cards to come and b) there are more known cards. Being "one card ahead" is worth more the later the street it is, and also, how low your current hand is matters a lot. On earlier streets a primary concern is not only what you and your opponent's current hand is, but what the next best hand you're drawing to is. So there's a big difference between A2T9 and A2T6 and a tremendous difference between A2T69 and A2T65 etc. The more likely that you can not only outdraw your opponent's current hand, but ALSO outdraw what he's drawing to, the better.

Re: a razz odds question
 

Note that on 3rd your chances are worse if he has a much lower 3 card hand than your 2nd-lowest card. This is a close example because he has a 3 card 8 and you're drawing to a 3 card 8. If he has a 3 card 6, like A56 and you have the same AT8 then you will still "outdraw" him 1 in 5 times but a lot of those times he'll call anyway, like if you have AT8 and he has A56 and he draws a T and you draw a 7, he'll have A56T vs your xx87, not only are you not *actually* ahead because of your T, but he might call you.

Realistically in the 3rd st case you are making a call as an underdog with the hopes of picking it up outright or becoming a favorite, in this example you have fewer chances of becoming an outright favorite and fewer chances of winning the pot outright. The better his 3 card hand appears to be, and the worse your 2nd low card is, the trickier it's going to be from 4th on.

Re: a razz odds question
 

[ QUOTE ]
So for him to catch bad, and you to catch good is 33% * 57% = 20% or in this case 4:1.

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Rusty, isn't 80/20 4 to 1? Worst case here, rounding the 33 down to 30 and the 57 up to 60, is 60/30 or 2 to 1.

Re: a razz odds question
 

When there are two independent events, you multiply the probabilities together. You need to catch good, .57, and he needs to catch bad, .33, so BOTH of those happening happens .33*.57=.20

so 20% of the time it does happen and 80% of the time it doesn't, so 80:20 or 4:1

Re: a razz odds question
 

[ QUOTE ]
When there are two independent events, you multiply the probabilities together. You need to catch good, .57, and he needs to catch bad, .33, so BOTH of those happening happens .33*.57=.20

[/ QUOTE ] To outdraw him, I only need him to catch worse, as in he can stay the same, get a brick, or a worse card than my draw. No? And, I'm not sure these are considered independent events, as one event affects the outcome of the other. (He gets my good card, so I have one less good card to get or vice versa depending on who goes first.)

Re: a razz odds question
 

The tendencies of your opponent are SUPER important here. If your opponent will rigorously fold if he bricks, then you are correct to call with ANY baby face up if you're getting better than 4:1 immediate odds from the pot.

However, also consider that the pot is small, and you are unlikely to be getting MUCH better than 4:1 from the pot, and if you are then he'll call more liberally on 4th. So this is realistically worth only a fraction of a small bet every time you try it.

Certainly it makes you look looser than you are though.

Re: a razz odds question
 

[ QUOTE ]
[ QUOTE ]
When there are two independent events, you multiply the probabilities together. You need to catch good, .57, and he needs to catch bad, .33, so BOTH of those happening happens .33*.57=.20

[/ QUOTE ] To outdraw him, I only need him to catch worse, as in he can stay the same, get a brick, or a worse card than my draw. No? And, I'm not sure these are considered independent events, as one event affects the outcome of the other. (He gets my good card, so I have one less good card to get or vice versa depending on who goes first.)

[/ QUOTE ]

Yes of course, if you catch a good one, he has one less good one to catch, the effect is minimal enough to consider independent (about 1/42 or a little more than 2%)

What cards is he going to draw to "stay the same" in your opinion?

In order for you to have a better hand than him on 4th, he has to get a T or worse, period, and you have to get a 9 or better - except that a 9 is not really that good so I'm not going to include it. If you draw a 9 and have T98A vs his A68T, you are not winning, and he is not folding. A 9 is only good for you if he catches really bad and maybe even then it's not great.

My math is correct - we're arguing really over what constitutes a situation good enough to warrant calling 3rd, and it's well established that we disagree on that.

Re: a razz odds question
 

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

Re: a razz odds question
 

1.) Why didn't you count 9s and Ts as bad cards? On 4th st. if you have 26/A8 and your opponent shows xx/A9, you've technically outdrawn your opponent but he probably isn't going anywhere in the hand ; but if you omit 9s and Ts as bad cards, then doesn't that also completely omit from the equation those instances where you might connect with xx/A2 v xx/AT and induce a fold there?

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.

2.)The way you did the math, all six upcards must be 9s or Ts because you didn't factor them into the equation. But how should six random upcards be factored in?

Re: a razz odds question
 

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.

Re: a razz odds question
 

[ 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.


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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.

Re: a razz odds question
 

[ 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?

Re: a razz odds question
 

[ 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.

Re: a razz odds question
 

[ 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.

Re: a razz odds question
 

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.

Re: a razz odds question
 

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

Re: a razz odds question
 

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.