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View Full Version : Omaha Preflop Probabilities (Can someone check my math please)

sthief09
08-08-2005, 03:46 PM
<font class="small">Code:</font><hr /><pre>
=4*4*COMBIN(50-8,2)/COMBIN(52,4) getting dealt A2xx
=1-(1-[above])^10 anyone having A2xx
=3*3*COMBIN(50-8,2)/COMBIN(52,4) A2xx if I have A2xx
=1-(1-[above])^9 chances of anyone else having A2xx
=4*3*COMBIN(52-4-3-4,2)/COMBIN(52,4) someone having A2xx if I have A3xx
=1-(1-[above])^9 anyone having A2xx if I have A3xx

</pre><hr />
and results:
5.1% getting dealt A2xx
40.7% Anyone having A2xx
2.9% A2xx if I have A2xx
23.0% chances of anyone else having A2xx
3.6% someone having A2xx if I have A3xx
28.3% anyone having A2xx if I have A3xx

LetYouDown
08-08-2005, 04:15 PM
On your first one...the probability of having A-2-X-Y where X and Y aren't A or 2. You have 50-8 but you need 52-8 or 50-6 instead. This brings the probability slightly higher and looks right.

I haven't looked hard at the other numbers, but my guess is that they're at least slightly off. I believe you overlooked the Inclusion/Exclusion principle (sticky, for the love of GOD, lol). So you're double/triple counting times when more than one person has the hand in question.

BruceZ
08-08-2005, 04:40 PM
[ QUOTE ]

=4*4*COMBIN(50-8,2)/COMBIN(52,4) getting dealt A2xx

[/ QUOTE ]

If you want x to not be A or 2, then you want COMBIN(52-8,2). This will not count cases where x can be another A or 2.

[ QUOTE ]
=1-(1-[above])^10 anyone having A2xx

[/ QUOTE ]

This is an approximation since the hands are not independent. For the exact answer, use inclusion-exclusion (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=&amp;Number=417383&amp;page=&amp;v iew=&amp;sb=5&amp;o=&amp;fpart=). This would be complicated if you also want to count cases where x can be A or 2.

[ QUOTE ]
=3*3*COMBIN(50-8,2)/COMBIN(52,4) A2xx if I have A2xx

[/ QUOTE ]

Should be COMBIN(52-8,2). Does not count cases where x can be another A or 2.

<font color="red"> COMBIN(50-8,2) was right since you have to subtract the two x. However, the denominator should be C(48,4).</font>

[ QUOTE ]
=1-(1-[above])^9 chances of anyone else having A2xx

[/ QUOTE ]

Same comments as above about inclusion-exclusion.

[ QUOTE ]
=4*3*COMBIN(52-4-3-4,2)/COMBIN(52,4) someone having A2xx if I have A3xx

[/ QUOTE ]

If x is not A,2,3 then this should be COMBIN(52-<font color="red">14</font>). If your opponent can have a 3, then it should be COMBIN(52-<font color="red">11</font>). Either way, this does not count x can be another A or 2.
<font color="red">The denominator needs to be C(48,4).</font>

[ QUOTE ]
=1-(1-[above])^9 anyone having A2xx if I have A3xx

[/ QUOTE ]

Same comments as above about inclusion-exclusion.

aloiz
08-08-2005, 04:43 PM
Need to assume that xx does not contain an A,2, and in the last two cases a 3 for the following to be correct.

getting A2xx =
4 * 4 * C(44,2)/C(52,4) =~ .056

anyone at a 10 person table getting A2xx (need inclusion exclusion for this)=
10*4*4*C(44,2)/C(52,4)
- C(10,2) *4*4*C(44,2) * 3*3*C(42,2)/C(52,4)/C(48,4)
+ C(10,3) *4*4*C(44,2) * 3*3*C(42,2) * 2*2*C(40,2)/C(52,4)/C(48,4)/C(44,4)
- C(10,4) * 16 * C(44,2) * 9 * C(42,2) * 4 * C(40,2) * C(38,2)/C(52,4)/C(48,4)/C(44,4)/C(40,4) =~ .467

A specific person having A2xx given that I have A2xx
Your calculation is correct

Someone else having A2xx given that I have A2xx (again inclusion/exclusion)=
9*3*3*C(42,2)/C(52,4)
- C(9,2) *3*3*C(42,2) * 2*2*C(40,2)/ C(48,4)/C(44,4)
+ C(9,3) *9*C(42,2) * 4*C(40,2) * C(38,2)/ C(48,4)/C(44,4)/C(40,4) =~ .226

A specific person having A2xx given that I have A3xx:
Your's looks right

Someone else having A2xx given that I have A3xx =
9*4*3*C(41,2)/C(48,4)
- C(9,2) *4*3*C(41,2) * 3*2*C(39,2)/C(48,4)/C(44,4)
+ C(9,3) *4*3*C(41,2) * 3*2*C(39,2) * 2*C(37,2) / C(48,4)/C(44,4)/C(40,4) =~ .398

aloiz

BruceZ
08-08-2005, 04:48 PM
[ QUOTE ]
I haven't looked hard at the other numbers, but my guess is that they're at least slightly off. I believe you overlooked the Inclusion/Exclusion principle (sticky, for the love of GOD, lol). So you're double/triple counting times when more than one person has the hand in question.

[/ QUOTE ]

Double/triple counting would happen if he just multiplied by the number of opponents, assuming mutual exclusivity. He's not doing that. He is using the approximation of independence, which should actually give an answer that is less than the actual value. This is because each opponent who does not have it makes it more likely for the next opponent to have it. We are approximating all of the probabilities of not having it as equal, which gives a result that is too high, and then we subtract from 1, which makes it too low. This should give a lower bound, and mutual exclusivity should give an upper bound, so we can see how accurate it is, and whether we need to use inclusion-exclusion.

LetYouDown
08-08-2005, 04:58 PM
You're right (shocker). I should have looked harder before I posted. Time for me to go read the inclusion/exclusion post again =).

sthief09
08-08-2005, 05:09 PM
[ QUOTE ]

I haven't looked hard at the other numbers, but my guess is that they're at least slightly off. I believe you overlooked the Inclusion/Exclusion principle (sticky, for the love of GOD, lol). So you're double/triple counting times when more than one person has the hand in question.

[/ QUOTE ]

yeah I know about that but I knowin hold'em it's fairly clsoe to just any one person ahving it. but that's with something like what are the chances that someone has AA if you have KK. the chances of 2 people having AA are almost 0 so it's not worth considering

BruceZ
08-08-2005, 05:21 PM
[ QUOTE ]
[ QUOTE ]

I haven't looked hard at the other numbers, but my guess is that they're at least slightly off. I believe you overlooked the Inclusion/Exclusion principle (sticky, for the love of GOD, lol). So you're double/triple counting times when more than one person has the hand in question.

[/ QUOTE ]

yeah I know about that but I knowin hold'em it's fairly clsoe to just any one person ahving it. but that's with something like what are the chances that someone has AA if you have KK. the chances of 2 people having AA are almost 0 so it's not worth considering

[/ QUOTE ]

That's irrelevant for your calculation. The fact that the probability of 2 people having A2xx might be small would mean that you could assume mutual exclusivity and just multiply by the number of opponents, which is the same as using just 1 term in inclusion-exclusion. You are using the approximation of independence which requires a different assumption, namely that the events of each player having A2xx are independent of each other, meaning that one player's results do not affect the probabilities for other players, so you can multiply the probabilities by raising to the 10th power. I don't think either one of these are very good approximations for omaha, and they appear to give answers that differ by over 10%.

See aloiz's post, and my response to his post, for the exact answers by the inclusion-exclusion computation. These numbers are higher than yours as predicted, and one is almost 12% higher. It looks like the inclusion-exclusion principle has really found its use in Omaha.

BruceZ
08-08-2005, 06:30 PM
This is what we need. I only found one thing that needs to be recomputed, and one comment.

[ QUOTE ]
Need to assume that xx does not contain an A,2, and in the last two cases a 3 for the following to be correct.

getting A2xx =
4 * 4 * C(44,2)/C(52,4) =~ .056

anyone at a 10 person table getting A2xx (need inclusion exclusion for this)=
10*4*4*C(44,2)/C(52,4)
- C(10,2) *4*4*C(44,2) * 3*3*C(42,2)/C(52,4)/C(48,4)
+ C(10,3) *4*4*C(44,2) * 3*3*C(42,2) * 2*2*C(40,2)/C(52,4)/C(48,4)/C(44,4)
- C(10,4) * 16 * C(44,2) * 9 * C(42,2) * 4 * C(40,2) * C(38,2)/C(52,4)/C(48,4)/C(44,4)/C(40,4) =~ .467

[/ QUOTE ]

OK

[ QUOTE ]
A specific person having A2xx given that I have A2xx
Your calculation is correct

[/ QUOTE ]

His denominator needs to be C(48,4) instead of C(52,4).

[ QUOTE ]
Someone else having A2xx given that I have A2xx (again inclusion/exclusion)=
9*3*3*C(42,2)/C(52,4)
- C(9,2) *3*3*C(42,2) * 2*2*C(40,2)/ C(48,4)/C(44,4)
+ C(9,3) *9*C(42,2) * 4*C(40,2) * C(38,2)/ C(48,4)/C(44,4)/C(40,4) =~ .226

[/ QUOTE ]

The denominator in the first term should be C(48,4). That would change this to ~.327 if everything else is right.

[ QUOTE ]
A specific person having A2xx given that I have A3xx:
Your's looks right

[/ QUOTE ]

His denominator needs to be C(48,4) instead of C(52,4).

[ QUOTE ]
Someone else having A2xx given that I have A3xx =
9*4*3*C(41,2)/C(48,4)
- C(9,2) *4*3*C(41,2) * 3*2*C(39,2)/C(48,4)/C(44,4)
+ C(9,3) *4*3*C(41,2) * 3*2*C(39,2) * 2*C(37,2) / C(48,4)/C(44,4)/C(40,4) =~ .398

[/ QUOTE ]

You said this would not allow a 3, but the C(41,2) in each term allows 3s. 52 - 4 aces - 4 deuces - 1 three - 2 x = 41. To exclude 3s we would make it C(38,2). Allowing 3s may be preferable though.

BruceZ
08-08-2005, 06:42 PM
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