Question concerning an example in SOR
 

I was rereading Sklansky on Razz and don't understand the rational for the answer to one of the examples. Any help would be appreciated.

In the example the hands are as follows:
Opponent: XX783
You hold: A2394
Q: What do you do if your opponent bets?
A: Raise because you are either a monster or a small underdog.

I understand if xx are 2 random cards you are a 78/22 favorite. Certainly you are ahead of a bluff. But how often do you see a 7 door card used as a bluff against a 3 when the pot is opened? I would put the odds of this being the situation, and you holding a "monster" as being pretty remote. If you always assume that your opponent has 2 random cards and raise based on this assumption then you will soon be broke.

If your opponent has 2 cards ranked 6 or below, which could be a reasonable holding, then you are a 55/45 dog and you are raising hoping to catch a 5,6 or 7 in the next 2 cards. You also hope that your opponent doesn't catch to improve. I don't understand the reason that you would grow the pot in a situation where you need to catch to possibly get ahead.

In holdem I may raise with a flush or straight draw but this is a semi-bluff and a big part of my potential profit is gained from where I can get my opponent to fold the currently best hand. But in Razz where your board is open to view by all players your 9 upcard defines your hand and your opponent will not fold to a raise, if I was the opponent I would probably reraise.

As a side note: One problem I have with the book is that the examples do not give any information concerning dead cards or previous betting. This information would make the problems much more usefull.

Re: Question concerning an example in SOR
 

[ QUOTE ]

Opponent: XX783
You hold: A2394
Q: What do you do if your opponent bets?
A: Raise because you are either a monster or a small underdog.

[/ QUOTE ]

I don't know why Sklansky would do it. I do this just to say: <font color="blue">"Hey, even with a 9, I have four-to-a-wheel and unless you have a better draw, do you REALLY want to invest more in this hand?" </font> About half the time they fold, because they have junk in the hole, I presume. Sometimes they don't fold and I check/call and then bet the river no matter what I hit and then they fold. This happens about a third of the time. So, I win more then half the pots this way.

But I don't know nuthin' 'bout no math - I just like pushing.

But if I had one card to come instead of two when I hit the fourth wheel, then I just call.

Re: Question concerning an example in SOR
 

The key is that villain can easily be betting his board here. Even presuming he started with a legitimate hand, he still has a pair here a lot; if he has exactly one pair you make a load of money on your raise, because he's way behind but he still has to call and try to suck out. Whereas if he does have a made 87, you're losing very little (and may actually be ahead depending on the dead card situation).

Against some very aggro players (like me sometimes) it can be better to call down, because they'll put bets in dead bluffing if you let them keep the betting lead.

Re: Question concerning an example in SOR
 

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Even presuming he started with a legitimate hand, he still has a pair here a lot;

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Is there a statistic on how often you can expect to pair in five cards?

Re: Question concerning an example in SOR
 

This question doesn't really have a clear answer in a vacuum IMO. In order to offer something really useful you'd need to know the action on previous streets and the other player's observed tendencies. That said, against lost opponents this is a raise b/c you will gain more in pot equity when he has a pair or some pudwhack hand than you will lose when he 3bets and turns his hand over. Also 8's are at least partial outs for you too, not just 5's, 6's and 7's. But, as you know, reads based on action will be the biggest factor.

As for how often he pairs, in a vacuum that's just a simple combinatorics problem and in this case, assuming he started with a 3-card 8 or better, there are 32 cards 8 or lower in the deck. We have A234 so there are 3 8's and 2 3's he could have in the hole (ignore the 7's b/c that's his door card). So 5 cards pair him out of 22 unseen giving him about a 23% chance of being paired. It's 22 b/c we have A234 and we know he has no 7's down and he already has 3 cards up so that's 10 already seen out of 32.

However, the actual probability will vary because we may be able to weight his hole cards toward lower or higher values using hand reading and there will be dead cards to account for. Those are things that you will have to determine and account for yourself at the table.

Re: Question concerning an example in SOR
 

Johnny, it was very generous of you to take the time to answer so completely.

I have a grown child and have never balanced a checkbook. Can't. Some sort of bizarre brain damage at birth, I imagine.

In Hold 'Em literature, I read there is a 50% chance the board will pair. (I cannot remember where I read it, it just stuck with me.) So, in 7 cards, you'd think there'd be a better an even chance, right? If I knew this, it might (along with all the things about reading the action you mentioned) go a long way toward whether or not I bet or call a marginal hand on the river.

You seem like a bright math guy, any simple percentages available for the simple-minded?

Re: Question concerning an example in SOR
 

I'm not much of a razz player; be warned.

Actually, the chance that villain has paired is much less than dealing out random low cards would predict. If villain doesn't suck, then he doesn't have an 8 in the hole; it's not healthy to draw to an 8-7 low. That means you are only ahead if the three paired him, and you already have a three.

This would be less true if he flat called in a late position before anyone entered, for instance.

One benefit of raising is that you can fold unimproved on the river if he 3 bets; you know you are beated, unless villain has steel (or perhaps just iron) balls. There's also an off chance that he's very loose-aggressive (which I suppose also means that he has steel balls), but you should have noticed that by now.

I don't have a recommendation here, but if he sucks then just raise.

Re: Question concerning an example in SOR
 

[ QUOTE ]
I'm not much of a razz player; be warned.

Actually, the chance that villain has paired is much less than dealing out random low cards would predict.

[/ QUOTE ]

I've only been playing a few weeks myself. I was asking a more general question - should probably go find the right forum for it. I was asking generally because anytime the player pairs anything, me or my opponent, it strongly affects the probabilities of making a hand at all. I'm trying to figure out how common it is, so that if I don't see a pair, what is the chance he is paired?

But I'll go find a better forum for this question and a better way to frame the question.

Re: Question concerning an example in SOR
 

I think your approach to the problem needs to be adjusted a little. Drawing an analogy from hold'em here isn't that useful IMO because your opponent's hand is not a random hand. He's not gonna have KJ or Q7 or 44 unless he's retarded so a percentage based on randomization is rather useless in razz.

What's important isn't how many cards you have seen, it's which cards you've seen that is important and how they relate to your opponent's range. Each situation is unique and a product of the exposed cards, which makes it more complicated because you don't share board cards like HE. If tracking exposed cards and continuous odds calculation are overly daunting then maybe stud games aren't for you. In HE you can get by with memorizing relatively few fairly standard situations (pair vs. overcards, AK vs. AJ, TP vs. NFD, bottom 2 vs. TPTK, overpair vs. OESD, etc.) You can do that in stud as well but to a much lesser extent.

To figure out how often he's paired, regardless of the situation, you need to:

1) determine the range of hole cards you think he has
2) count how many of these cards are exposed or have been exposed
3) count how many unseen cards pair him
4) divide by total number of unseen cards

Each situation will be different. He may give away that he's paired by giving up on his hand. He may give away his down cards on earlier streets. If he reraises your open with a 3 up with his 5 up and he catches 46 you're probably in deep [censored]. If he reraises your open with 3 up with his 5 up and he catches A2 you probably have a better chance even though this board is scarier. See the difference?

Re: Question concerning an example in SOR
 

[ QUOTE ]
Each situation is unique and a product of the exposed cards, which makes it more complicated because you don't share board cards like HE. If tracking exposed cards and continuous odds calculation are overly daunting then maybe stud games aren't for you. In HE you can get by with memorizing relatively few fairly standard situations (pair vs. overcards, AK vs. AJ, TP vs. NFD, bottom 2 vs. TPTK, overpair vs. OESD, etc.) You can do that in stud as well but to a much lesser extent.

[/ QUOTE ]
So, in other words, razz is more complicated than HE?

Re: Question concerning an example in SOR
 

Meh, not the best choice of words but you know what I meant. If HE was played lowball style then it would be simpler than razz. Stud high is more mathematically complicated than HE.

Regardless, that isn't the real point. I was trying to emphasize dynamic odds calculations rather than generic standard HE situations, even if the calculations themselves are simpler in razz. There won't be a single number for him to refer to. I guess he could form himself a table of scenarios and pick his odds out from that but I don't think that was what he was looking for.

Re: Question concerning an example in SOR
 

I got a 760 on my math SAT and I can't balance my checkbook either.

Re: Question concerning an example in SOR
 

[ QUOTE ]

any simple percentages available for the simple-minded?

[/ QUOTE ]

I hope this is what you're looking for. I don't think these calculations are very practical, but they will give you an idea of very basic odds in razz and some guideline of how to work out similar problems.

Assumptions: any baby up, opponent's range is a 3-card eight, with a random distribution, and no known dead cards. Keep in mind that dead cards probably have more effect on razz than stud or stud 8. The knowledge of the dead cards can drastically change the probabilities, and thus equity considerations. For instance, in the most extreme example, a 654 is a 65% favorite over A23 if you kill five cards of the 654.

4th street:

The easy way.
P(unpaired 8 after 4) = P(unpaired 8 after 3) x P(4th = unpaired 8OB card) = 1 x 20/49 = 40.8%

The long way.
P(pairing door card) = 3/49 = 6.1%
P(pairing hole cards) = 6/49 = 12.2%
P(pairing on 4th) = 18.4%
P(unpaired hand after 4) = 1 - P(paired hand after 4) = 1 - (3/49 + 6/49) = 1 - 18.4% = 81.6%
P(4th &gt; 8) = 20/49 = 40.8%
P(unpaired 8 after 4) = P(unpaired hand after 4) - P(4th &gt; 8) = 81.6% - 40.8% = 40.8%

5th street:

The easy way.
P(unpaired 8 after 5) = P(unpaired 8 after 4) x P(5th = unpaired 8OB) = 40.8% x 16/48 = 13.6%

The long way.
P(pairing door cards) = 6/48 = 12.5% (assuming unpaired after 4)
P(pairing hole cards) = 6/48 = 12.5% (assuming unpaired after 4)
P(pairing on 5th) = 25.0% (assuming unpaired after 4)
P(unpaired hand after 5) = P(unpaired hand after 4) - P(pairing on 5th) = 81.6% - 25.0% = 56.6%
...

I'm too tired to do the rest, but you get the idea. This should give you a decent guideline in case you want to play with 3-card 7's or 9's.

Re: Question concerning an example in SOR
 

I'm not sure what you're attempting to demonstrate with this calculation. In your 4th street problem you calculated the probabllity that he'd make a 4-card 8 if he started with a 3-card 8. In your 5th street problem you calculated the probability that he'd have an unpaired hand on 5th if he started with an unpaired hand on 3rd (without the 8 or better criterion you imposed on 4th). How are those numbers useful? Where am I going to apply them when I make a decision at the table? What I really need to know is how my hand stacks up relative to the other guy's, not how it ranks on the scale of all hands possible. And in order to determine that we must take both players' cards into account.

Re: Question concerning an example in SOR
 

I would suggest you reread my post again, with emphasis on the preface, as well as Prax's questions in this thread. I could get into more details if you'd like.

Let's just tackle one: Where am I going to apply these numbers at the table? Um, how about against an aggressive player who's autobetting his board just because he's low? It's been awhile since I've played razz, but I'm pretty sure there are still players out who do that.

Re: Question concerning an example in SOR
 

You are a slight dog against two wheel cards others than 3s. You are a huge favorite if villain paired his 3.

Re: Question concerning an example in SOR
 

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If your opponent has 2 cards ranked 6 or below, which could be a reasonable holding, then you are a 55/45 dog.


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I'm not convinced that this is correct: ProPokerTools Razz simulation gives 55/45 only in the worst possible case: opponent started with 567. In any other case, our winning probabilities are much better (e.g. A47: 50,3/49,7).

In general (ProPokerTools calculations)

opponent started with 7 low without a 3: 52/48 (we are a 2%-underdog)
opponent started with 7 low including a 3: 26/74 (we are a 24%-favorite)

A simplified calculation:

Since our opponent will bet in both cases, we cannot derive any conclusion about his hand from the fact that he has bet.

There are 19 unknown cards below 7 including two 3s. The probablitiy that opponnt has a 3 in the hole is approximately 20%. In 2 cases out of 10 we are a 24%-favorite, in 8 cases out of 10 we are a 2%-underdog, overall we are a 3,6%-favorite.

(ProPokerTools simulation gives 53,7% for A2349 against (6- 6-)783.)

Re: Question concerning an example in SOR
 

Ok, here's two answers, assuming your opponent started unpaired and 8 or better:

xx783
a2349

If villain cannot have an 8 in the hole (aka good player):
Two 3's in the deck could pair
32-8 seen= 24 low cards left
Deal out two low cards to represent hole cards
Odds of not pairing are 22/24*21/23 = 84%

If villain can have an 8 in the hole (aka dubious player):
Two 3's and three 8's in the deck could pair
32-8 seen= 24 low cards left
Deal out two low cards to represent hole cards
Odds of not pairing are 19/24*18/23 = 62%