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