1) We are creating a new card game with a new deck. Unlike the normal deck that
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Question
1) We are creating a new card game with a new deck. Unlike the normal deck that has 13 ranks (Ace through King) and 4 Suits (hearts, diamonds, spades, and clubs), our deck will be made up of the following. Each card will have: i) One rank from 1 to 11 ii) One of 8 different suits. Hence, there are 88 cards in the deck with 11 ranks for each of the 8 different suits, and none of the cards will be face cards! So, a card rank 11 would just have an 11 on it. Hence, there is no discussion of "royal" anything since there won't be any cards that are "royalty" like King or Queen, and no face cards! The game is played by dealing each player 5 cards from the deck. Our goal is to determine which hands would beat other hands using probability. Obviously the hands that are harder to get (i.e. are more rare) should beat hands that are easier to get. a) How many different ways are there to get any 5 card hand? The number of ways of getting any 5 card hand is DO NOT USE ANY COMMAS b)How many different ways are there to get exactly 1 pair (i.e. 2 cards with the same rank)? The number of ways of getting exactly 1 pair is DO NOT USE ANY COMMAS What is the probability of being dealt exactly 1 pair? Round your answer to 7 decimal placesExplanation / Answer
There are 88 cards in the deck which belong to 8 different suits (Each suit numbered from 1 to 11)
a)
No of ways of getting a 5 card hand(T) = 88C1 * 87C1 *86C1 *85C1 *84C1 = 88*87*86*85*84 = 4701090240
b)
Ways to pick one face and two suits of the paired cars :11C1*8C2
Ways to pick three different faces and a suit for each as: 10C7*(8C1) ^3
Exactly 1 pair in 5 card hand(F) = 11C1*8C2 *10C7*(8C1) ^3 = 11*28*120*8*8*8 = 18923520
Probability of 1 exactly 1 pair (P) = Favourable Outcomes / Total Outcome = 18923520 /4701090240 = 0.00402535 or 0.4%
c)
Exactly 2 pair in 5 card hand(F) = 11C2*8C2*8C2*9C1*8C1 = 55*28*28*9*8 = 3104640
Probability of Exactly 2 pair (P) = Favourable Outcomes / Total Outcome = 3104640/4701090240 = 0.00066 or 0.07%
d)
Exactly 3 of a kind in 5 card hand (F) = 11C1*8C3*10C2*(8C1) ^2= 11*56*120*8*8 = 4730880
Probability of Exactly 3 of a kind in 5 card hand (P) = Favourable Outcomes / Total Outcome = 4730880/4701090240 = 0.001006 or 0.10%
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