Omaha H/L probability question

[J.Copperthite]

A hand came up in my Omaha H/L table that I found interesting. The final two players when the reached the showdown, both showed A-Q-2-3. I wanted to know how to calculate the probability in Omaha of being dealt the exact 4 ranks of cards like this, regardless of suitedness?

[gaming_mouse]

What is the chance that there are two people at the table in a full ring game who have the same hand?

We consider all possible (10 choose 2) pairs of players whose hands might coincide. For any given pair of players, the chance is:

(3^4)/(48 choose 4)

Our first order approximation (using inclusion-exclusion) is therefore:

(10 choose 2)*(3^4)/(48 choose 4)=0.0187326549

This is an upper-bound on the true probability. To refine our appoximation, we now consider the 2nd term.

Finding the second term (which we must subtract off) is more difficult, as there are ((10 choose 2) choose 2) = 990 pairs of pairs of players. We can break these pairs of pairs down into two types:

1. Those that overlap (eg, {1,3} & {1,7})
2. those that do not (eg, {1,3} & {2,4})

By multinomial, there are (10!/(2!*2!*6!))/2=630 non-overlapping pairs of pairs. There are therefore 360 pairs of pairs of type 1 (overlapping).

Note that an overlapping pair of a pairs corresponds to 3 people all having the same hand, which we calculate as:

((3^4)/(48 choose 4))*((2^4)/(44 choose 4))

Call this P1.

Whereas the chance of a non-overlapping pair of pairs is simply:

((3^4)/(48 choose 4))^2

Call this P2

The second term of inclusion-exclusion is: P1*360 + P2*630, which works out to:

0.000126835808

Refining our approximation now (which is guaranteed to be as good as the above number)

0.0187326549-0.000126835808=0.0186058191

About 1 in 54

[callydrias]

[ QUOTE ]
We consider all possible (10 choose 2) pairs of players whose hands might coincide. For any given pair of players, the chance is:


(3^4)/(48 choose 4)

[/ QUOTE ]

It's been a while since I've taken a combinatorics class. Can you explain this part of your answer? I understand this doesn't take into account 3 or more or multiple pairs of people holding the same hand. A couple things are bugging me (which may be easily resolved once I figure this part out):

1. Does this answer taken suitedness into account? e.g Is A [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/spade.gif[/img] equal to A [img]/images/graemlins/diamond.gif[/img]2 [img]/images/graemlins/diamond.gif[/img]3 [img]/images/graemlins/diamond.gif[/img]4 [img]/images/graemlins/club.gif[/img] or not?

2. Does this answer take pairs/trips/quads in a single hand? I suspect not because the numerator term seems to describe a choice for each card which has exactly 3 options.

TIA

[gaming_mouse]

[ QUOTE ]
1. Does this answer taken suitedness into account? e.g Is A [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/spade.gif[/img] equal to A [img]/images/graemlins/diamond.gif[/img]2 [img]/images/graemlins/diamond.gif[/img]3 [img]/images/graemlins/diamond.gif[/img]4 [img]/images/graemlins/club.gif[/img] or not?

[/ QUOTE ]

No. The OP specified suits are to be ignored. So the hands above are equal.

[ QUOTE ]

2. Does this answer take pairs/trips/quads in a single hand? I suspect not because the numerator term seems to describe a choice for each card which has exactly 3 options.


[/ QUOTE ]

You are right. I assumed cards of 4 different ranks. I should have made that assumption explicit. Although we can ignore trips obviously, but not pairs....