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




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 a (nonacademic) exam. I was asked to calculate the probability that 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 (52choose5). 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 (4choose3) ways to make a triple. For each of the remaining 12 ranks, there are (4choose2) ways to make a pair.
Any triple can be joined with any of these pairs, so that these quantities multiply:
13 * (4choose3) * 12 * (4choose2) = 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.



