2.62) In a poker game, five cards are dealt at random from anordinary deck of 52

Question

2.62) In a poker game, five cards are dealt at random from anordinary deck of 52 playing cards. Find the probabilities ofgetting a) two pairs ( any two distinct face values occuring exactlytwice) b) four of a kind ( four cards of equal face value ) CAN YOU PLEASE EXPLAIN BREIFLY WITH FOLLOWINGSTEPS 2.62) In a poker game, five cards are dealt at random from anordinary deck of 52 playing cards. Find the probabilities ofgetting a) two pairs ( any two distinct face values occuring exactlytwice) b) four of a kind ( four cards of equal face value ) CAN YOU PLEASE EXPLAIN BREIFLY WITH FOLLOWINGSTEPS

Explanation / Answer

a) (13C2)*(4C2)*(4C2)*(44C1): In order to get a two pair, you must first pick two ranks outof a possible 13. Then out of one of those ranks (which has 4suits), you choose two cards. Out of the other rank, youagain choose 2 out of 4 cards. For the fifth card, you have44 choices left (you started with 52 and subtract 8 because of theranks in the two pairs). (13C2)=13*12/2=88 (4C2)=4*3/2=6 (44C1)=44/1=44 Now we get that: (13C2)*(4C2)*(4C2)*(44C1) = 78*6*6*44 =123552 combinations The total number of combinations possible is (52C5) =52*51*50*49*48/(5*4*3*2*1) = 2,598,960 combinations So the probability is 123,552/2,598,960 = .047539 b) We can get four of a kind bypicking one rank out of 13. Then choosing four cards out ofthat rank. We have 48 cards remaining tochoose from. This is:(13C1)*(4C4)*(48C1) (13C1) = 13 (4C4) = 1 (48C1) = 48 So, (13C1)*(4C4)*(48C1) = 2496combinations The probability is624 / 2,598,960 b) We can get four of a kind bypicking one rank out of 13. Then choosing four cards out ofthat rank. We have 48 cards remaining tochoose from. This is:(13C1)*(4C4)*(48C1) (13C1) = 13 (4C4) = 1 (48C1) = 48 So, (13C1)*(4C4)*(48C1) = 2496combinations The probability is624 / 2,598,960

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