I'm trying to work out the number of combinations of certain categories of hands in plo.

The number of starting hands in plo is 52C4 = 270,725.

The probability of getting dealt aces is C(4,2)*C(48,2) / C(52,4) = (6*1128)/270725 = 2.7% ie there are 6,768 combos

• How many combos of broadway kings ie KKJT?
  
• How many combos of high card wraps? Defined as 4 cards in a row from JT98 to AKQJ?
  
• How many combos of high gappers from QJ98 to AKQT?

The reason that I'd like to know this is so that I can get a feel for how often someone has certain types of hands when they 3bet. We obviously need to put consideration into what hands certain players like to 3bet ie often people don't 3bet 'bad' AAxx pre-flop or even good AAxx hands oop pf. However, i feel like i can use this analysis as a starting block to build on.

Any help would be much appreciated.

Just do what you did for the AA. For example the broadway kings:

Considering T+ as broadway cards:
COMBIN(4;2)*COMBIN(16;2)
if KKK is included then add this:
COMBIN(4;3)*COMBIN(16;1)

And so on.

as above for broadway kings.

For str8s JT98 - AKQJ....

There are only 4 different str8s, A high, K high, Q hgh and J high, in this range.

for each card there are 4 different choices each time

so number of combinations is

4 * 4^4 = 4^5 = 1024 combinations.


One card gappers is a bit more annoying, but still the quickest plan would be to count them all.... off the top of my headi think theres 3 for each high card, and 3 high cards, so theres 9.

9 *4^4 = 2304 combinations.

Any hand with no pairs in you can count like this. As soon as you get a pair youll need to use C(x,y) as youve done.

Thanks for the help. I've drafted the following spreadsheet. Any glaring mistakes?



Also, im thinking of adding double-premium pairs (TT+) ie QQJJ. How many combos of these are there? I'd also need to be careful of not double-counting the AA or KK hands that i've not already included.

The double-suited hands have a mistake. In the first possibility the numer is not (26/50) but (39/50) since there are only cards from one suit in the hand already, leaving 3 possible suits to double with. Also there's a simpler way to do it:

=(12/51)*(39/50)*(12/49)+(39/51)*(24/50)*(12/49)

Wtf is with 4*4*4 on the high wraps and rundowns math? I mean there are 4 wraps possible and each wrap has 4*4*4*4 possibilities so it's 4*4*4*4*4. Anyway the result is the same

Ashley if you get this spread sheet worked out plz PM me and let me know.

Thanks for the help guys. Please could you check the calcs for these further categories that I have done.

I've largely followed the logic from the above help that you provided (thanks alot!).

Double pairs (from 22-99) I'm including pairs TT+ elsewhere in the spreadsheet and so don't want to double count!? 4C2*4C2 *8 pairs (2 thru to 9's)= 288?

Medium pair wraps (where medium pair is 55-99 and medium wrap is consecutive cards 5-T?

(think ive listed them all here! ) Can you check the C calcs as I don't have a calculator at home so am doing in excel (8C2 = (8*7)/(2*1) right)?

5567 - 4C2*8C2 = 6*28 = 168?
6678 - as above
5667 - as above
6678 - as above
5677 - as above
6778 - as above
7789 - as above
7889 - as above
889T - as above
7899 - as above
899T - as above

Total = 11*168 = 1848 combos

Medium pair gappers:

J998 = 4C2*8C2 = 6*28 = 168?
T997 = as above
T887= as above
9986= as above
9886= as above
9776= as above
8875= as above
8775= as above
7764= as above

Total = 9*168 = 1512 combos

Mid gappers (JT97 - 9865) - 6 straights so 6*4^4 = 1536 combos?

High double-gappers (AKJ9 - KJ98) - 6 types so 6*4^4 = 1536?

Mid top gappers (J987 - 5679) - 4 types so 4*4^4 = 1024?

Mid two gappers (QTJ7 - J876) - 6 types so 6*4^4 = 1536?

Mid- double gappers (QJ97 - J967) - 6 types so 6*4^4 = 1536?

Low-wraps (7654 - 6543) - two straights so 2* 4^4 = 512?

Low gappers (7653 - 6532) - 4 straights so 4*4^4 = 1024 combos?

Low top-gappers (7543 - 6432) - 2 types so 2*4^4 = 512?

High-wrap danglers (AKQ8 - JT92)

AKQ8-AKQ2 - 7 types > 7*4^4 = 1792
KQJ7 - KQJ2 - 6 types > 6*4^4 = 1536
QJT6-QJT2 - 5 types > 5*4^4 = 1280
JT95-JT92 - 4 types > 4*4^4 = 1024

Total = 5632

I'll take a look after coffee.

In the meantime, you have the C function in excel. Just click your heels three times and type in the following.

=combin(4,2)

J998 = 4C2*8C2 = 6*28 = 168?

Your calc includes JJ99 and 9988, which i assume you don't want since you count those elsewhere, right?, so make that

C(4,2) *4 *4 = 96

Double pairs (from 22-99) I'm including pairs TT+ elsewhere in the spreadsheet and so don't want to double count!? 4C2*4C2 *8 pairs (2 thru to 9's)= 288?

c(4,2) * c(4,2) * c(8,2)= 1008

ie 9988, 9977, down to 9922 and including 3322, right? There are c(8,2) such combinations, and four cards for each rank to get to that 1008 number.

I don't believe your calculation for aces is correct. It should be

1*C(4,2)*C(12,2)*C(4,1)^2 = 6,336 combos. (2.34%)

That is, we are choosing 1 rank (ace), choosing 2 of 4 suits, then choosing 2 ranks from the remaining 12 ranks, then choosing 1 suit from 4 twice.

I was playing around with making a similar spreadsheet to yours trying to figure out what % of hands certain categories make up. It gets quite confusing and you have to be careful about overlapping categories.

Anyone made any progress on such a spreadsheet and want to share?

i have done this thing before, but never finish it. it just has some different breakdown of hand types.

with AAxx calc i has the same result as OP, 6768 and about 2.5% of total hands.
COMBIN(4;2) = four aces, choose two of them
COMBIN(48;2) = other remaining 48 cards, choose two of them
so =COMBIN(4;2)*COMBIN(48;2)

I was about to post then I saw that this might save you some work:
http://en.wikipedia.org/wiki/Poker_probability_(Omaha)

could you explain the equation you have in the middle please i don't see how you get it?

I think your calculation for aces ignores the possibility that the other 2 cards could also be aces (you could trips or quads, and while those are bad hands they still are technically AAxx hands). You select from only the remaining 12 ranks, so while your calculation is probably a little more useful since it more accurately represents a villian's range, I don't think it's fair to say OP's calculation is wrong.