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Topic: Another combinatorial probability question
Replies: 3   Last Post: Jan 10, 2011 8:00 PM

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 Alexander Solla Posts: 9 Registered: 12/17/10
Another combinatorial probability question
Posted: Jan 10, 2011 7:12 PM

Hi everybody,

I'm working through "Schaum's" notes on probability in preparation for
drawing 5 cards from a standard deck will result in a full house. My
answer disagrees with theirs by a factor of 6, which makes me think
I'm double counting based on order (or something like that).

The strategy: count the number of admissible hands, divide by the
number of possible hands (52-choose-5). A hand is "admissible" if it
is a full house: if it is made up of 3 cards from one "rank" and 2
cards from another "rank".

I guess I am messing up the counting:

For each of the 13 ranks, there are (4-choose-3) ways to make a
triple.
For each of the remaining 12 ranks, there are (4-choose-2) ways to
make a pair.

Any triple can be joined with any of these pairs, so that these
quantities multiply:

13 * (4-choose-3) * 12 * (4-choose-2) = 3744 hands are admissible

"Schaum's" answer key says there are 13 * 2 * 4 * 6 = 624 admissible
hands. My counting argument is pretty simple. I don't see where I am
going wrong. Can anybody help? Thanks.

Date Subject Author
1/10/11 Alexander Solla
1/10/11 Nmacgre
1/10/11 Henry
1/10/11 Robert Israel