This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mathematics Department, UCLA T. Richthammer winter 09, sheet 4 Jan 23, 2009 Homework assignments: Math 170A Probability, Sec. 1 037. An ordinary deck of 52 cards (containing 4 aces) is randomly distributed to 4 players (each getting a hand of 13 cards). We are only interested in the number of aces every player gets. Let A be the event that every player gets one ace. (a) Choose a sequential model consisting of 4 experiments, each describing the number of aces a certain player gets (you dont have to calculate all conditional probabilities). Determine P ( A ) by calculating p 1  1 , p 1  1 , 1 and p 1  1 , 1 , 1 . (b) Choose a probabilistic model describing the position of the aces in the deck of cards immediately before the distribution. Determine P ( A ) using this model. Answer: (a) S = { , . . . , 4 } 4 . For sequential models we have to specify the probabilities p i (proba bility that first player gets i Aces), p j  i (probability that second player gets j Aces, given that first player gets i Aces), and so on. We get p 1 = ( 4 1 )( 48 12 ) / ( 52 13 ) , p 1  1 = ( 3 1 )( 36 12 ) / ( 39 13 ) , p 1  1 , 1 = ( 2 1 )( 24 12 ) / ( 26 13 ) , p 1  1 , 1 , 1 = 1. A = { (1 , 1 , 1 , 1) } , so the probability for every player to get an ace is p 1 p 1  1 p 1  1 , 1 p 1  1 , 1 , 1 = 13 4 / ( 52 4 ) 10 . 5%. (b) Let S be the set of all possible choices of 4 elements of the set { 1 , . . . , 52 } , i.e. S = { C { 1 , . . . , 52 } :  C  = 4 } . It is plausible to assume that every choice of positions is equally probable, so P = equidistribution on S . The event A considered is described by the situa tion that the choice C of positions for the aces includes one position in { 1 , . . . , 13 } (cards for the first player), one in { 14 , . . . , 26 } (cards for the second player), one in { 27 , . . . , 39 } , one in { 40 , . . . , 52 } , i.e. A = { C S :  C { 1 , . . . , 13 } = 1 , . . . ,  C { 40 , . . . , 52 } = 1 } . By (MP) we have  A  = 13 4 , so P ( A ) = 13 4 / ( 52 4 ) 10 . 5%. 038. What is the probability that a fivecard poker hand (formed from a 52 card stack) has: (a) 4 aces (b) 4 of a kind (c) full house (d) no pair (e) just 2 pairs (Full house is 3 of a kind and a pair. (e) is meant to be 2 pairs, but nothing better, i.e. not 4 of a kind or a full house.) Answer: We use the same model as in the lecture.  S  = ( 52 5 ) = 2598960. Let A 1 , . . . , A 5 denote the events of the subproblems. (a) The number of hands with four aces is the number of choices for the fifth card:  A 1  = ( 48 1 ) = 48. The probability is thus 48 /  S  . 002%....
View
Full Document
 Winter '10
 RICHTHAMMER
 Pallavolo Modena, Sisley Volley Treviso, Associazione Sportiva Volley Lube, Piemonte Volley, M. Roma Volley

Click to edit the document details