Basic Omaha 8 Probability Calculations

[franknagaijr]

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 0.63829

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

[franknagaijr]

Buzz - This is the stuff I didn't quite get six months ago. Please correct any errors and put it in the stickies.

[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.


[/ QUOTE ]
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.


[/ QUOTE ]
From formula given in A1

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

[Buzz]

Thanks Frank - Good idea.

It will take me a while to wade through it.

Ideally, I think you explain at a level the potential learner can understand. But there's a whole spectrum of learning readiness levels in math.

A general problem with explaining probability, or anything related to math, is that people are at different levels of understanding and learning readiness.

Good idea though. I'll see what I can do.

Buzz

[franknagaijr]

[ QUOTE ]
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.


[/ QUOTE ]
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.


[/ QUOTE ]
From formula given in A1

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

[/ QUOTE ]

I cannot seem to edit my first post. I like your additions, and would suggest dumbing them down as follows.

1) The bang (!) symbol works as follows:
4! = 4 * 3 * 2 *1
5! = 5 * 4 *3 * 2 * 1
2) Your addition to A3 could use a verbal explanation of what you are up to. I understand it, but I would not have a year ago at this time. (It nicely drives home the point of backing out any permutations, also)

[RobNottsUk]

Good idea to explain factorial then in a preliminary section.

Guess if you can't edit, then this'll have to be a rought draft and copied and pasted into fair version.

[franknagaijr]

[ QUOTE ]
Good idea to explain factorial then in a preliminary section.

Guess if you can't edit, then this'll have to be a rought draft and copied and pasted into fair version.

[/ QUOTE ]

Are you (Rob) game to repost the edits as a new thread for stickying? If not, I can probably get to it within the next week?

[THEjDonk]

I had a brief look at this post. Your calculations on Q7 seem to count hands like 666, 778, 88A and so on. Once you have determined the rank of a low card that doesn't pair you, there are four less cards in the deck that will allow you to flop a low hand (not one card less).

I'm not going to look through all the other calculations you've made right now, but I think your post could need a little editing.

[franknagaijr]

[ QUOTE ]
I had a brief look at this post. Your calculations on Q7 seem to count hands like 666, 778, 88A and so on. Once you have determined the rank of a low card that doesn't pair you, there are four less cards in the deck that will allow you to flop a low hand (not one card less).

I'm not going to look through all the other calculations you've made right now, but I think your post could need a little editing.

[/ QUOTE ]
D'oh! Thanks. Indeed I can see that 7 and 8 are wrong, and I need to back out the duplicates. I'll plan to rewrite this and post this in the new forum software.