Poker Hands_solutions

# Poker Hands_solutions - STAT 230 Introduction to...

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STAT 230: Introduction to Probability and Random Variables Summer 2009 Problem set: Introduction to Poker Solutions PROCEDURE: We will first choose from the 13 kinds common to each suit, then we choose the suit for each group of cards that have the same kind. The total number of poker hands is 52 5 Pair : We choose the kind for the pair. There are 13 1       choices for that. Now we have 12 kinds from which to choose the remaining 3 which works out to be 12 3 combinations. So, the total number of kind choices for this case is 13 12 1 3    ×       . In choosing the suits, for the pair we have 4 2       choices and for the rest of the cards we have 4 1       choices each. So, the number of suit choices in this case is 4 4 4 4 1 1 1 2         × × ×                 The total number of hands with one pair is 4 4 4 4 13 12 1 1 1 2 1 3            × × × × ×                      

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Two Pairs : We have to choose 2 kinds, one for each pair. We have 13 2       choices there. Then, we have 11 1       kinds left to choose from for the last card. So, the number of kind choices for this case is 13 11 2 1     ×         . As for the suit choices, we have
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## This note was uploaded on 09/07/2010 for the course FAS stat 230 taught by Professor Unknown during the Fall '08 term at American University of Beirut.

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Poker Hands_solutions - STAT 230 Introduction to...

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