odds of getting a straight with 4 connecters in omaha

[jay_shark]

If you read it carefully you will notice that he wants the probability of having a straight by the river NOT the flop .

You have to be careful not to over-count which is why I've broken it into cases . The question is not too difficult but it's just a pain in the ass to work out all the frivolous details .

[85chev]

I know the rules of Omaha, but like Jay said this gets a lot more complicated when you have until the river to hit the straight.

[findingneema]

I've been working on the river scenario; it just takes a lot longer. And the OP did ask what the odds were of flopping the straight, so I provided that answer.

[jay_shark]

I've added some more straights that I've omitted earlier .

458,569,579,589,6710,6810,6910,710j

Luckily for me I did all the dirty work already .

204*6=1224
364*2=728
+____
1952 +3952 =5904 . Don't worry I'll explain all this later .

[jay_shark]

For this installment , I'm going to calculate the number of straights of the following form : (note that x can only be used once with one of the other three cards repeated )

straights :

7 high - 345x : (4c2*4c1*4c1*36)*3 =10368
8 high- 456x: 4c2*4c1*3c1*37*2 + 3c2*4c1*4c1*37 =7104
9 high - 567x: 4c2*3c1*3c1*38 + 4c1*3c2*3c1*38*2 = 4788
10 high- 6710x : 4788 same as before
j high - 710j :7104 same as 8 high
q high - 10jq : 10 368 same as 7 high

Now just add them . 10368*2 + 7104*6 + 4788*12 = 120816

Here is where you have to be careful . There is some intersections of events that you have to exclude .

To be continued ....

[jay_shark]

There is actually a much better way of doing this . First you should recognize all of the possible straights .

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

Now we calculate the number of straights of the form
345xy where x,y =/ 6,7,8 but everything else
456xy where x,y =/ 7,8,9
567xy where x,y =/ 10
6710xy where x,y =/ j
710jxy where x,y=/ q
10jqxy where x,y can be anything .

first case :
345xy : 4c3*4c1*4c1*3 + 4c2*4c2*4c1*3 + 4c2*4c1*4c1*27*3 +4c1*4c1*4c1*27c2 = 30864

456xy : 4c3*4c1*3c1*2 + 4c1*4c1*3c3 + 4c2*4c2*3c1 + 4c2*4c1*3c2*2 + 4c1*4c1*3c2*27 + 4c2*4c1*3c2*27*2 4c1*4c1*3c1*27c2 = 22396

567xy : 4c3*3c1*3c1+ 4c1*3c3*3c1*2+ 4c2*3c2*3c1*2 +4c1*3c2*3c2 +4c2*3c1*3c1*34 + 4c1*3c2*3c1*34*2 + 4c1*3c1*3c1*34c2 = 24684

6710xy : same as above

710jxy : 3c3*4c1*4c1 +3c1*4c3*4c1*2 + 3c2*4c2*4c1*2 +3c1*4c2*4c2 + 3c2*4c1*4c1*33 +3c1*4c2*4c1*33*2 + 3c1*4c1*4c1*33c2 = 32044

10jqxy : 4c3*4c1*4c1 + 4c2*4c2*4c1*3 +4c2*4c1*4c1*36*3 + 4c1*4c1*4c1*36c2 = 51184

We have all the information we need to solve this . It is much easier to break it into cases where you compute the number of boards with best possible straight 7 high , 8 high , 9 high etc .

The total number of boards is just 30864*1 +22396*3 +24684*12 +32044*3 +51184*1 = 541576

Note that 1+3+12+3+1=20 so the total probability is just 541576/48c5 = 31.6 %