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    Probability-poker hands

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    If it is assumed that all (52 5) poker hands are equally likely, what is the probability of being dealt
    (a) one pair (a,a,b,c,d) ,
    (b) two pairs (a,a,b,b,c)
    (c) three of a kind (a,a,a,b,c).

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    Solution Preview

    See the attached file.

    There are 52 cards in the pack, 4 suits each consisting of ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, 2
    Thus there are 4 aces, 4 kings, 4 queens and so on

    Thus for the purpose of poker there are 13 different cards (ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, 2) and 4 of each of them

    (a) one pair (a,a,b,c,d)

    there are 13 C 1 ways of choosing a card for a pair= 13
    Since there are 4 cards of a kind there are 4 C2 ways of selecting a pair= 6
    Thus the total no of ways of selecting a pair= 78 =13*6

    The remaining cards b c and d are of different cards
    After the pair is selected there are 48 cards left for choosing b( since all 4 a are excluded)
    After b is ...

    Solution Summary

    The solution calculates probability of being dealt one pair, two pairs, three of a kind in a game of poker.