Pinochle Question

[The Dude]

Last night while playing pinochle, double aces around was dealt two hands in a row. So here's my question.

What is the probability that on any given hand, all eight aces will be dealt to one team? (There are 48 cards, two of every suit and rank from A to 9. Each player is dealt 12 cards, and there are two teams of two players.)

Also, of the times all eight aces are dealt to one team, what % of the time will they be split 8-0, 7-1, 6-2, 5-3, and 4-4?

[Tom1975]

One team should get all the Aces around 20.3% of the time. The easiest way to figure this out to is to calculate the odds of one team being dealt no Aces:

(40/48)*(39/47)*(38/46)*(37/45)*(36/44)*(35/43)*(34/42)*(33/41)

[BruceZ]

[ QUOTE ]
One team should get all the Aces around 20.3% of the time. The easiest way to figure this out to is to calculate the odds of one team being dealt no Aces:

(40/48)*(39/47)*(38/46)*(37/45)*(36/44)*(35/43)*(34/42)*(33/41)

[/ QUOTE ]

That expression is the proabability of no aces in 8 cards. We want the probability that one team gets no aces in 24 cards times 2, since it can happen to either team.

2*C(40,24)/C(48,24) =~ 0.40% or 1 in 256.5.

[BruceZ]

[ QUOTE ]
Last night while playing pinochle, double aces around was dealt two hands in a row. So here's my question.

What is the probability that on any given hand, all eight aces will be dealt to one team? (There are 48 cards, two of every suit and rank from A to 9. Each player is dealt 12 cards, and there are two teams of two players.)

[/ QUOTE ]

2*C(40,24) / C(48,24) =~0.4% =~ 1 in 256.5.


[ QUOTE ]
Also, of the times all eight aces are dealt to one team, what % of the time will they be split 8-0, 7-1, 6-2, 5-3, and 4-4?

[/ QUOTE ]

8-0: 2*C(8,8)*C(16,4) / C(24,12) =~ 0.13%

7-1: 2*C(8,7)*C(16,5) / C(24,12) =~ 2.58%

6-2: 2*C(8,6)*C(16,6) / C(24,12) =~ 16.58%

5-3: 2*C(8,5)*C(16,7) / C(24,12) =~ 47.38%

4-4: C(8,4)*C(16,8) / C(24,12) =~ 33.32%

Note: These sum to exactly 1 (a good thing).

[The Dude]

[ QUOTE ]
=~ 1 in 256.5.

[/ QUOTE ]
Hmm. Either I've been on the very low end of the distribution in how many times this has happend while I've been playing, or I've grossly overstimated the number of hands of pinochle I've played in my lifetime.

Thanks Bruce.