Probabilities of card combinations

[Bertil]

Is this forum the place to ask about certain card cominations?
Specifically, I would like to know the probabilities for holding these
two hands in bridge:

AQJx-QJx-QJx-QJx and QJxx-QJx-QJx-AQJ.
When searching a bridge computer software none was found in 11million
random deals.
Could the odds of being dealt one of these hands be that low?

Bertil

[Jim Heckman]

On 10-Feb-2012, Bertil <stigf...@hotmail.com>
wrote in message
<f3b7b097-e27a-48a4...@n12g2000yqb.googlegroups.com>:

Depending on exactly what you mean, yes they could be that low.
First, I'll assume you mean the probability that a particular
player, say South, holds the hand in question, as opposed to any of
the four players at the table holding the hand. I'll also assume
you've listed the suits in order, so that AQJx-QJx-QJx-QJx means
specifically spades-hearts-diamonds-clubs.

With these assumptions, for the first of your desired hands there
are (13-4) = 9 non-face-card possibilities for each of the x's, so
9^4 = 6561 such hands. For the second desired hand there are
(9 choose 2) = 9! / (2! * (9-2)!) = 36 possibilities for the first
two x's, and 9 possibilities for each of the other two x's, so
36*(9^2) = 2916 such hands. But there are (52 choose 13) =
52! / (13! * (52-13)!) = 635,013,559,600 possible bridge hands, so
the probability of holding the first hand is approximately 1 in
96,786,094 and of the second approximately 1 in 217,768,710.

--
Jim Heckman