Re: two pairs

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1 total messages Started by Jeff Boyd Tue, 15 Sep 1992 22:24

#349

Author: Jeff Boyd
Date: Tue, 15 Sep 1992 22:24

34 lines
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Does anyone else here feel that counting is akin to witchcraft?

I took a slightly different approach to this one, treating each pair
-AND- the odd card as selections from sets of 4.  The result is the
same, but I think it's a bit more rigid in its logic.


   1)  select 3 of 13 sets of 4 (two pair slots, one singleton slot)

               13c3

   2)  select 1 pair from one of the 3 sets of 4

                4c2

   3)  select 1 pair from another of the 3 sets of 4

                4c2

   4)  select 1 singleton from the last of the 3 sets of 4

                4c1

   5)  select 2 of the 3 sets of 4 for the pair slots, -OR-
       select 1 of the 3 sets of 4 for the singleton slot

                3c2  or  3c1

Then

    P[2 pair]  =  (13c3)(4c2)(4c2)(4c1)(3c1) / (52c5)

               =  0.047539

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