Omaha Probability Question

[Baxter]

Lets say a player is dealt AAA4.

1) What is the probability that any other player at a 10 player table is dealt the last ace.

2) What is the probability that a person at the 10 player table is dealt A2xx.

Thanks.

[BruceZ]

[ QUOTE ]
Lets say a player is dealt AAA4.

1) What is the probability that any other player at a 10 player table is dealt the last ace.

[/ QUOTE ]

9*1*C(47,3) / C(48,4) = 75%

or

1 - C(47,36) / C(48,36) = 75%

The first method is 9 times the probability of a single player having the last ace, since only one player can have it. The second method is 1 minus the probability that nobody has the last ace.


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2) What is the probability that a person at the 10 player table is dealt A2xx.

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9*[1*4*C(43,2) + 1*C(4,2)*43 + 4] / C(48,4) =~ 17.9%.

This assumes that some player still has AAA4. This is 9 times the probability that a single player has A2xx, since only one player can have it. 1*4*C(43,2) is the number of ways to have A2xx with x not a 2. 1*C(4,2)*43 is the number of ways to have A22x, with x not a 2, and 4 is the number of ways to have A222. C(48,4) is the total number of possible hands.