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  #1  
Old 03-21-2007, 06:01 PM
myxomatosized myxomatosized is offline
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Default odds of getting a straight with 4 connecters in omaha

what are the chances of having a straight by the river with a hand like 6789 in omaha? i'm not sure how to do the math but it seems pretty good. maybe like 50%. that's just a pure guess.

also what are the chances of flopping a straight with the same hand?

thanks
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  #2  
Old 03-21-2007, 07:12 PM
jay_shark jay_shark is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

Any board that contains a 10 or a 5 gives you a straight . Lets also exclude all boards that give you ties .

The prob. that a 10 is excluded is 44c5/48c5=0.634 . The prob. it is included is 1-0.634 =0.3657 .The probability a 5 is included is the same number . So you add the two probabilities and subtract the times when they appear together .

The probability that a board contains a 5 and a 10 is [4*4*40c3 + 4c2*4c1*40c2+4c1*4c2*40c2 + 4c3*4c1*40c1+4c1*4c3*40c1 + 4c4*4c1+4c1*4c4 ]/48c5

=0.114

The overall probability is 0.3657*2 - 0.114=0.616 or 61.6%
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  #3  
Old 03-21-2007, 07:25 PM
85chev 85chev is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

Are you talking pre-flop hand of 6789 (ie in the pocket)?

In that case you need to catch a 3,4,5, OR 10,J,Q OR 5,6,7 OR 8,9,10 on the board.

jay_sharks calculations are wrong because you can only use two cards from your pocket and I am pretty sure his calculations are based on using all of your cards.

I will figure out the probability and post it later.
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  #4  
Old 03-21-2007, 07:27 PM
ScottHoward3.0 ScottHoward3.0 is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

[ QUOTE ]
Any board that contains a 10 or a 5 gives you a straight .

[/ QUOTE ]
This sounds wrong to me, but im not a math guy.
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  #5  
Old 03-21-2007, 07:32 PM
jay_shark jay_shark is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

I have too much math in my head that I forgot about the rules to Omaha .

I've done this in the past somewhere but I'll do it again .
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  #6  
Old 03-21-2007, 07:37 PM
eljizzle eljizzle is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

[ QUOTE ]
Are you talking pre-flop hand of 6789 (ie in the pocket)?

In that case you need to catch a 3,4,5, OR 10,J,Q OR 5,6,7 OR 8,9,10 on the board.

jay_sharks calculations are wrong because you can only use two cards from your pocket and I am pretty sure his calculations are based on using all of your cards.

I will figure out the probability and post it later.

[/ QUOTE ]

I'm guessing that you would take the total number of boards that could exist (not taking suits or order into consideration) and look at that number compared to how many contain these card combinations....granted this just means you'll make a straight, not the nuts by any means.

3,4,5
4,5,6
4,5,7
4,5,8
4,6,7
4,6,8
4,7,8
5,6,7
5,6,8
5,6,9
5,7,8
5,7,9
6,7,T
6,8,T
7,6,T
8,9,T
8,T,J
T,J,Q

I may have forgot a couple, but you get the point...
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  #7  
Old 03-22-2007, 01:38 AM
85chev 85chev is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

No worries Jay. I tried doing this earlier, but I did'nt have time. Essentially though, you have two of the outs to any of your straights so that will hurt your chances.

eljizzle you're on the right track, but I am just too tired to do the math right now.
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  #8  
Old 03-22-2007, 03:11 PM
jay_shark jay_shark is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

The math is horrendous , but I'll break this up into parts . Clearly the denominator should be 48C5 . The tricky part is the numerator because there are so many cases to consider .

If you use two of your hole cards then the number of straights is :

345,8910
457,7910
578,7810
456,910j
568,810j
567,10jq

For each of these it can be tricky because there is some overlap in events . Lets start with the number of boards in which exactly three of the cards are used (multiplicity).

Start with 345 : 4c2*4c2*4c1*3c2 + 4c3*4c1*4c1*3c2=624
8910: 3c2*3c2*4c1 + 4c2*3c2*3c1*2 + 3c3*3c1*4c1*2+ 4c3*3c1*3c1 = 204

457 : 4c2*4c2*3c1 + 4c2*4c1*3c2*2 + 4c3*4c1*3c1*2+ 3c3*4c1*4c1 =364

If you add everything you should get 624*2+204*6+364*4=3928

Ok I have 3928 number of flops that contains exactly three of the aforementioned numbers on the list .

Stay tuned and I'll finish the rest ....
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  #9  
Old 03-22-2007, 06:17 PM
85chev 85chev is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

[ QUOTE ]
The math is horrendous

[/ QUOTE ]

You got that right. I tried doing this last night, but after about 4 pages of work I figured I will go and work this over with one of the TA's in my Combinatorics class. I will do that on Monday and then post the answer.
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  #10  
Old 03-22-2007, 08:47 PM
findingneema findingneema is offline
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Default Re: odds of getting a straight with 4 connecters in omaha

Do you folks know the rules of Omaha? You have to use EXACTLY two of your cards! Here's the right answer for the flop:

straights:
7-high: 345
8-high: 456,457,458
9-high: 567,568,569,578,579,589
T-high: 67T,68T,69T,78T,79T,89T
J-high: 7TJ,8TJ,9TJ
Q-high: TJQ

Total flops: 48c3 = 17296

7-high: (4c1)^3 = 64
8-high: (4c1)^2 * (3c1) * 3 = 144
9-high: (4c1) * (3c1)^2 * 6 = 216
T-high: (4c1) * (3c1)^2 * 6 = 216
J-high: (4c1)^2 * (3c1) * 3 = 144
Q-high: (4c1)^3 = 64

You flop a straight on 848 flops, or 4.90% of the time.
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