Calculating probability for omaha starting hands to occur

[Vammakala]

Dunno, I fought with this for a time and all my math books etc. are not available at the moment. So...

How do I calculate the probability of being dealt A2xx where xx are any two cards. A friend suggested 1/13 * (probability of deuce coming with the last 3), but it gives ridiculously low percentage compared to what it appears to be.

Thanks,
-Vamm

[rjp]

Post deleted by rjp

[rjp]

You can get A2xx in a number of ways:

A2xx, Ax2x, Axx2, xxA2, 2Axx, 2xAx, 2xxA, xx2A (I think I got them all)

Basically you want to calculate the probability of getting an Ace or a Two on each of the four cards dealt:

8/XX: Probability of being dealt an Ace or Two at a specific card
4/XX-X: Porbability of being dealt the other Ace or Two (whichever is needed) at another specific card

So, you can then calculate this by calculating the probabilities of getting an A2 in each of the possible variations (C means a card you want to get):

CCxx: (8/52)*(4/51)*(44/50)*(44/49)
xCCx: (44/52)*(8/51)*(4/50)*(43/49)
xxCC: (44/52)*(43/51)*(8/50)*(4/49)
CxxC: (8/52)*(44/51)*(43/50)*(4/49)
CxCx: (8/52)*(44/51)*(4/50)*(43/49)

This calculates to 4.68%

I'm just getting into Omaha, so I think I have this right, but realize I might have mucked it up. [img]/images/graemlins/smile.gif[/img]

You just have to make sure you have the probability associated with each possible variation.

[BruceZ]

[ QUOTE ]
You can get A2xx in a number of ways:

A2xx, Ax2x, Axx2, xxA2, 2Axx, 2xAx, 2xxA, xx2A (I think I got them all)

[/ QUOTE ]

You are missing xA2x, x2Ax, xAx2, and x2xA. You should know that there have to be 12 all together since there are C(4,2) = 4*3/2 = 6 ways to choose the 2 positions out of 4 cards, times 2 ways to order the A and 2.


[ QUOTE ]
CCxx: (8/52)*(4/51)*(44/50)*(44/49)
xCCx: (44/52)*(8/51)*(4/50)*(43/49)
xxCC: (44/52)*(43/51)*(8/50)*(4/49)
CxxC: (8/52)*(44/51)*(43/50)*(4/49)
CxCx: (8/52)*(44/51)*(4/50)*(43/49)

This calculates to 4.68%

[/ QUOTE ]

Now you are just missing xCxC. There must be 6 of these. Each of these has the same probability (8/52)*(4/51)*(44/50)*(43/49), so just multiply that by 6 to get the final answer of ~5.59%. Note that this does not include cases where there are more than one A or 2.

Another way to compute this is just to note that there are 16 ways to make A2, so there are 16*44*43/2 = 15136 ways to make A2xx where x is not an A or 2. The total number of hands is C(52,4) = 52*51*50*49/(4*3*2*1) = 270725, so the probability of A2xx is 15136/270725 =~ 5.59%, same as before.

If you want to include cases where you have more than one A or 2, here is the number of each case, where x is not an A or 2, and these can come in ANY ORDER. Note that there are 6 ways to make a pair AA or 22, and 3 ways to make AAA or 222.

AA2x: 6*4*44 = 1056
22Ax: 6*4*44 = 1056
AAA2: 3*4 = 12
222A: 3*4 = 12
AA22: 6*6 = 36
A2xx: 16*44*43/2 = 15136 (from above)
======================
total = 17296

17296/270725 =~ 6.39%.


The problem with the first guy's method 1/13 * (probability of deuce coming with the last 3) is that it only counts cases where the first card is an A, so the correct answer is roughly 4 times that. That would be larger than 6.39% since it would over count the times that there are more than one A or 2.

[rjp]

Thanks for finding the mistakes I knew were in there. [img]/images/graemlins/smile.gif[/img]