this is more like 2010 stud forum...bass3p constantly complaining about something. i swear 90% of your posts or pessimistic/whiny/bitching about something

takes me 30 minutes to an hour to get a sng going. regs stopped regging any of the sngs when im on. time to turn into razitup and play 8 tables at once of lowstakes poker.

way to prove him wrong, bro

OK, so, if you don't know anything about combinatorics at all, I recommend that you learn some, it's probably the most useful thing to know about poker ever. The 2nd most useful thing if you're interested in problems marginally more complicated than this is basic programming. I recommend something like python, ruby, etc. Razz is very easy to write programs for because there are no suits, it makes things remarkably less complicated.

OK, so, let's start with, how many combinations of hole cards ARE there? Well, there are 52 cards and 3 slots, which means that there are C(52,3) distinct combos. Note that we consider A23 and 32A to be the same here. This makes sense if you are considering combos prior to seeing door cards. Google can calculate this for you by the way, "search" for "52 choose 3" and you get 22,100

OK, so, let's consider a particular unpaired hand, such as A23. Any given unpaired hand is going to have the same number of combinations, i.e. there are just as many A23 as 25K or 379 or whatever. Well, you have 4 choices for the first slot, 4 choices for the 2nd and 4 choices for the 3rd. That means there are 4*4*4 combos of A23, which is 64.
So the chance of getting any particular combo is 64/22,100.

So you can count how many 3-card wheels there are. These would be A23 A24 A25 A34 A35A45 235 235 245 345. So total there would be 10*64, i.e. the chance is 640/22100 or about 2.89%

3-card sixes add the following hands: A26 A36 A46 A56 236 246 256 346 356 456, so 10 more sets for a total of 20 sets, which comes out to 1280 combos. About 5.79% of hands dealt.

You can continue on in this way. Counting which combos you're adding is tedious but there's a formula to it. If you want to know how many combos there are of 3 unpaired cards, that are "X or under" (such as 9 or under) then think of it this way:
The first card can be any card X or under, there are X*4 of these.
The second card can be any card X or under, except the first one you chose, there are (X-1)*4 of these
The third card can be any card X or under, except the 1st or 2nd you chose, there are (X-2)*4 of these.
You multiple these 3 numbers together and then divide by 6. Why divide by 6? Because the method above will produce both hands like A23 and 32A and we have already noted these are the same hand. Each logical combination will be produced 6 times, i.e.
A23
A32
23A
2A3
32A
3A2

So for our 3-card-6 example this would be (6*4)*(5*4)*(4*4)/6 = 1280

You can re-order this as (6*5*4)*(4*4*4) and then you can easily see the relationship to finding out how many combos under, say, 7 there are compared to 6. Consider
(7*6*5)*(4*4*4) = (6*5*4)*(4*4*4)*N
where N is the factor of how much larger 3 card 7s or better are than 3 card 6s or better. So, clearly this is the same as
(7*6*5) = (6*5*4)*N
and further
7 = 4N
N = 7/4 = 1.75

Whew, so there's some tl;dr for you. Now a little chart. Divide the combos by 22,100 to get the probability of getting dealt one.

N	combos
5	640
6	1280
7	2240
8	3584
9	5376
T	7680
J	10560
Q	14080
K	18304

Posts like Rusty's are why I love 2+2. Thank you so much.

implying that i care, bro. i was winning at about ~60-65% against most of the hu razz regs between $15-7, but can't seem to get an edge over genrih v. the games are dead because waiting for a 30$ game takes as long it did for a $100-$500 hu game on fulltilt. you're pretty much only capable of winning pocket change unless you want to turn to cash games where theres a minimal edge since you pretty much have to hope that you run good and villain runs slightly less good than you that he calls to the river. you're much better off making the transition to omaha or stud because the skill gap is exponentially bigger.

Consider discussing this somewhere else.

I totally agree. Great post!

And maybe razz is a more complicated game than the appearances could dictate.

This is a really handy chart (note: didn't check the math yet), but hopefully no one will take away from it the idea that a villain who's got a 50% VPIP in razz is probably playing all 3-card J's or better and nothing with a Q, K, or pair.

Yeah my post #4 in this thread lays out the approximate vpip progression that I usually assume. It's not constant, some people order hands a bit different than others.

I have a chart around somewhere of hands ranked like pokerstove, i.e. hot/cold equity vs N random hands.

bass3p, sn? I will snap-sit you any amount of tables!

What, is your PM broken?

One of those questions you should be able to figure out pretty easily...

Thanks for the info. So from your explanation and math, does the following breakdown make sense?

N combos
5 640
6 1280 640 W-W-W + 640 W-W-6
7 2240 640 W-W-W + 640 W-W-6 + 640 W-W-7 + 320 W-6-7
8 3584 640 W-W-W + 640 W-W-6 + 640 W-W-7 + 320 W-6-7 + 640 W-W-8 + 320 W-6-8 + 320 W-7-8 + 64 6-7-8

Where W = 5 or less. I'm just trying get a table that makes more sense since someone might play any 2 wheels with a 9 but not all 3 card 9's.

If I'm starting to get this I'll give pairs a shot.

CC

This is very helpful. One question - is there a shortcut for counting combos with a particular door card? I raise this as it seems to me that it might be useful to be able to construct a range such as, for example, any 3-card-8 or better plus any unpaired 3-card-9 or worse with an A or 2 as the door card.

Looking quickly at your example, it seems that the formula for determining the chances of being dealt a particular X-high combo or better with a particular door card would be:

1*4 * (X-1)*4 * (X-2)*4 /6

I.e., we take the original formula of [X*4 * (X-1)*4 * (X-2)*4/6] and replace X with 1, since we are picking the door card first and there can only be one door card. So the formula will be (X-1)*(X-2)*64/6. If we have a range of door cards consisting of D possible door cards, the formula would be D*(X-1)*(X-2)*64/6

That means that the odds of getting an unpaired K-high with an A or 2 as a door card are 2816/22,100:

2*(13-1)*(13-2)*64/6 = 2816

The odds of getting getting an unpaired 8-high or better with an A or 2 as a door card are 896/22,100 under the same formula, so the odds of getting an unpaired 9-high or worse with an A as a door card is 1920/22,100. So the range consisting of any 3-card-8 or better any unpaired 3-card-9 or worse with A or 2 as the door card would appear to be ~25% range.

(3584 + 1920) / 22,100 = 24.9%

However, the foregoing is a pretty cumbersome way of calculating a range (in particular the step of subtracting out the hands in the first range with the relevant door cards). I was wondering if there are any established shortcuts for adding combos for razz. (I'm guessing no, but it never hurts to ask...)