Maths Question Omaha Hilo Starting Hands

[Big Dave D]

I am quite interested in working out what hands occur which % of time so I can work out how often I should be seeing the flop. For example, how often does an A2 occur, A3 suited etc. Of course this has to be refined somewhat so that it may be A3 without trips, or A3 with another wheel card. Has anyone any techniques for this?

Thanks

Dave

[bigpooch]

There are C(52,4) = 270725 possible O8 hands.

A) Any A2:

AAA2: 4x4 = 16
A222: 16
AA22: 6x6 = 36
AA2x: 6x4x44= 1056
A22x: 1056
A2 other: 4x4xC(44,2)= 15136

total A2: 17316
chances of getting A2: 0.0639616 or about 1/15.6

B) For A3 hands, here are a few cases:

B1) For A3 with another suited card without a deuce:

A3 of same suit and two other (non deuce):
4xC(46,2) = 4140

A and 3 but non-3 is of same suit (no deuce:
4x(C(11,2)x3+(11x(3+3x36))) = 5544

total: 9684

B2) Other A3 with AA and a 3

AAA3 (no suited ace): 4
AA33 (no suited ace): 6
AA3x (no suited ace, no deuce): 6x2x20 = 240

total: 253

[Buzz]

Bigpooch - I follow your reasoning (and agree) until you get to B.

For A3XX suited or double-suited to the ace, could be all in same suit, but X is not deuce, which seems to be what you calculated, I get a total of only 8634 hands as follows:
AAA3s---12---4*3
AA3s3t---6---6*1
AA3n3s---24---6*2*2
AA3sXs---120---6*2*10
AA3sXt---120---6*2*10
AA3sXn---240---6*2*20
AA3nXs---240---6*2*20
A3n3n3s---12---4*3
A3n3sXs---120---4*3*10
A3n3sXn---360---4*3*30
A3n3nXs---120---4*3*10
A3sXsXs---180---4*1*10*9/2
A3sXsXn---1200---4*1*10*30
A3sXnXn---1740---4*1*30*29/2
A3nXsXs---540---4*3*10*9/2
A3nXsXn---3600---4*3*10*30
total---8634

Of the above hands, I'd probably not play the 12 A333 hands, but I'd probably play the rest of them, at least some of the time.

I'm never sure my math is correct, and there are lots of places I could have made a mistake in the above table - so I’m hardly one to suggest your math is off, but perhaps you have some duplications in your figure of 9684.

Well, okay, I can check my math. Here goes.
A3subf---8634
AAA3n---4---4*1
AA3n3n---6---6*1
A3n3n3n---4---4*1
AA3nXn---240---6*2*20
A3n3nXn---360---4*3*30
A3nXnXn---5220---4*3*30*29/2
AAAA---1
AAAX---160---4*40
AAXX---4680---6*40*39/2
AXXX---39520---4*40*39*38/6
XXXX---91390---C(40,4) 40*39*38*37/24
3333---1
333X---160
33XX---4680
3XXX---39520
total---194580
should be C(48,4) = 194580. Checks. (no deuces)

Buzz

[bigpooch]

Thanks for the correction! Also, your post is quite clear
and easily checks.

[chaos]

The hardest part for me is not that calculations themselves but defining exactly what I want to calculate and then listing all the cases to calculate without omissions or duplications.

For example many people say they will play any four cards 9 or higher. At first glance this appears to be
C(24,4) = 10,626. But this number is too high because what you actually want to calculate is any four cards 9 or higher minus the obviously unplayable hands like quads and trips. So you would need to calculate the unplayable hands and subtract them from 10,626. Or you could try and enumerate all the hands you would play and calculate the number of combinations of each.