#1
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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 |
#2
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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% |
#3
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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. |
#4
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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. |
#5
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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 . |
#6
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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... |
#7
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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. |
#8
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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 .... |
#9
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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. |
#10
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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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