Question concerning an example in SOR

[Johnny#5]

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.

[Andy B]

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

[templar999]

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any simple percentages available for the simple-minded?

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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 > 8) = 20/49 = 40.8%
P(unpaired 8 after 4) = P(unpaired hand after 4) - P(4th > 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.

[Johnny#5]

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.

[templar999]

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.

[betgo]

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

[NBourbaki]

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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.)

[Tha Stunna]

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%