Q1 - What is this notation I keep seeing that looks like this:
c(4,2)

A1 - This is the number of combinations in which four cards could go into two spaces. For example, if you are dealt two of the four aces, you could have (Ah Ac), (Ah Ad), (Ah As), (Ac Ad), (Ac As), (Ad As) for six possibile combinations.

Q2 - What is the difference between a combination and permutation?

A2 - Permutations counts different sequences of the same cards, while combinations do not. In the example in Q1, (Ah Ac) and (Ac Ah) are both counted as two distinct permutations, but as one combination. In poker probability, we are generally not interested in permutations. Sometimes when your number appears to be wrong, it may be because you let a permutation slip into your figures.

Q3 - How can I figure out C(45,4)

A3 - If you have microsoft excel, type in '=Combin(45,4)' and you will get the answer.

Q4 - I want to figure out the probability of being dealt A2xx, with x being any card.

A4 - If you have a fresh deck, your denominator will be C(52,4). The tricky part will be to figure out the number of combinations you have that meet the combination, avoiding any double counting. perhaps the best way to avoid double counting is to isolate the individual elements of your problem, in this case, Aces, Deuces, and any other card. There are four Aces, four Deuces, and 44 'x' cards, and our valid combinations are as follows:

c(4,1) * c(4,1) * c(44,2)
In other words, four aces in one spot, four deuces in the second spot, and four x cards in the third spot.
Divide this by c(52,4) and you should arrive at 0.055909

Q5 - Doesn't the answer in A4 avoid possibilities such as AA2x and A22x?

A5 - Yes! and this is where you need to be careful to avoid double counts and overlaps as you figure out this problem. To get to the correct answer, you need to first figure out the different combinatory ways you can meet all of the conditions, add up those combinations, and get the right answer.

Observe that the possibilities below are not overlapping.
A2xx = c(4,1) * c(4,1) * c(44,2)
AA2x = c(4,2) * c(4,1) * c(44,1)
AAA2 = c(4,3) * c(4,1)
A22x = c(4,1) * c(4,2) * c(44,1)
A222 = c(4,1) * c(4,3)

If you add all of these, and divide by c(52,4), you should arrive at 0.063829

Q6 - Is there an alternate method to get to the answer in Q5?

Q6 - You could figure out all of the combinations that do not meet the condition, and subtract the possibility of this not happening from 1 as follows

xxxx
xAAA
xxAA
xxxA
x222
xx22
xxx2

In this case, the alternate method is equally tedious, but in some cases, the alternate method will save time in getting to the correct answer.

Q7 - Let's say I have been dealt AA23 What are the chances of my flopping a made low?

A7 - There are now 48 unknown cards, and our denominator will be c(48,3). In the 48 unknown cards, we need to break them down as follows to arrive at the correct answer

A = 2 aces
2 = 3 deuces
3 = 3 treys
L = 20 other low cards that do not duplicate those in our hands

And then we can figure out the possibilities as follows:
LLA = c(20,2) * c(2,1)
LL2 = c(20,2) * c(3,1)
LL3 = c(20,2) * c(3,1)
LLL = c(20,3)


You should arrive at 0.153793 as the chance of flopping any made low.

Q8 - What if I want to also include the chance of flopping any two-card low draw

A8 - We need to add in the possibilities that were not included in A7.

H = Any of the 20 unknown non-low cards

LLH = c(20,2) *c(20,1)
LAH = c(20,1) * c(2,1) * c(20,1)
L2H = c(20,1) * c(3,1) * c(20,1)
L3H = c(20,1) * c(3,1) * c(20,1)

If you add this to the A7 answer, your total chance should come out to 0.558511


From RobNottsUK

I like it, here's a suggested enhancement for non Excel users, or ppl who like pencil & paper


Quote:


Q1 - What is this notation I keep seeing that looks like this:
c(4,2)

A1 - This is the number of combinations in which four cards could go into two spaces. For example, if you are dealt two of the four aces, you could have (Ah Ac), (Ah Ad), (Ah As), (Ac Ad), (Ac As), (Ad As) for six possibile combinations.


The formula for combinations of n cards in r spaces is :

c(n,r) = n! / ((n-r)! x r!)


Quote:

Q3 - How can I figure out C(45,4)

A3 - If you have microsoft excel, type in '=Combin(45,4)' and you will get the answer.


From formula given in A1

c(45,4) = 45! / (45-4)! * 4! = 45! / 41! * 4! = 45*44*43*42 / 24

Buzz - Kraada was kind enough to point out a typo, which I cannot edit:

If you add all of these, and divide by c(52,4), you should arrive at 0.63829

Should be 0.063829

Thanks
FNJ

Also, Q7 and Q8 give an incorrect answer as they don't factor in duplicates. When I'm feeling it, I'll rewrite those.

Done.

very cool stuff... i was aware of combinations/permutations, but i've never seen a way to use it this involved in a 52-card deck.
this would be excellent for stuff like stud, stud/8, and razz for me.
are there any other threads here that expound on such goodies?
thx
$pike

Hi Mods

Maybe you can link this in the main faq and unsticky it?

Also another topic for FAQ: Where can I find live PLO8?

thx
fnj

I'll do that. I keep thinking I can make the probability section more meaningful, but I never seem to find the time to do that. There are some genuine probability wizards on the probability forum, geniuses who know much more about probability than most or all of the individuals who respond to probability questions on the Omaha-8 forum. However, people may find more interest in answering probability questions related to Omaha-8 and thus get better answers if specific questions are posted directly on the Omaha-8 forum. If in doubt, cross posting seems fine.
There are lots of casinos, here and there, that have fixed-limit-Omaha-8, but I don't know of any that have pot-limit-Omaha-8 on a sustained basis.

Buzz

Exactly. Maybe just a link to a generic 'hey, where's the live plo8?' thread with the generic responses, like 'Dave's not here Man'.

There currently is no B&M sustained PLO-8 anywhere, to the best of my knowledge. I hope that changes. I love that old Cheech and Chong dialogue skit. I think I have the CD somewhere.

Buzz

for a hand such as AA23 for example...

against a random hand, what is the percentage of times that it:

scoops
loses
chops
quarter
3/4

How do you calculate this?


http://forumserver.twoplustwo.com/42...rticle-290870/

I really would like to know the answer but the link is broken...

http://forumserver.twoplustwo.com/42...rticle-290870/

Methodology for Tournament Hand Mathematics in a 'prize-steps' situation when heads-up

1) Determine the respective prizes for all remaining places
2) Place this information into a tournament equity model to determine your preflop prize-pool equity using an ICM calculator to determine your prize EV if you fold the hand.
3) Calculate the chip distribution and determine your prize pool equity (ICM value) if you flat call and fold on the flop.
4) Calculate your prize pool equity if you are all-in and scoop, win 3/4, chop, win 1/4 and lose
5) Calculate your prize pool equity if you flat call, push the flop, and your opponent folds.
5) Determine what percentage of flops will cause you to put all of your chips in the middle, and what percentage will cause you to check-fold
6) For the flops where you intend to put it in the middle, assign rough values to the chances that your opponent will fold
7) Create a matrix-tree of all of the possibilities involved in seeing a flop with chips behind and determine the respective chances of you folding, villain folding, and all of the possible outcomes when your chips go in the middle. (This should add up to 100%)
8) Multiply each of the matrix tree percentages times the appropriate prize pool equity number
9) Sum up the prize pool equity numbers from step 8 as your PPE for a stop and go
10) Create a matrix tree for reraising pre-flop and all of the possible outcomes. (Again, should be 100%)
11) Multiple each percentage times its corresponding PPE number.
12) Sum up the PPE for reraising preflop
13) Compare your stop-n-go with reraising preflop and also with folding preflop

Bump.

Threads older than February 1 are to be automatically archived next month in preparation for an upgrade. I'm bumping all 2+2 threads cited in our two stickies that are not already archived so that they will not be lost to us.

This includes all the wells.

Buzz