Poker, the LotteryDate: 11/30/98 at 12:19:40 From: mike Subject: Probability (1) What is the probability that a 5-card poker hand contains cards of 5 different values? (2) 6 integers are selected at random from the set (1,2,...,40). What is the probability that none of the integers 1,...,6 will be selected? (3) To play the Pennsylvania superlottery, a player selects 7 of the first 80 positive integers. What is the probability that the person will win the grand prize by picking 7 integers that are among the 11 integers selected randomly by the Pennsylvania lottery commission? (4) 6 integers are chosen at random from the set of the first 40 positive integers. What is the probability that your 6 choices match exactly 5 of these integers? (5) Which is more likely: rolling a total of exactly 8 using two fair dice or rolling a total of exactly 8 when three fair dice are rolled? Date: 12/01/98 at 10:11:26 From: Doctor Anthony Subject: Re: Probability >(1) What is the probability that a 5-card poker hand will contain >cards of 5 different values? [5 different face values, not in sequence, not all cards in the same suit] There are 13C5 x 4^5 ways with all cards of different face value. There are 4C1 x 13C5 ways where cards are all of the same suit, and 10 x 4 ways where cards are in sequence. So Prob(5 different face values, not in sequence, not in same suit) 13C5 x 4^5 - 4C1 x 13C5 - 4 x 10 = --------------------------------- = 0.50509 52C5 >(2) 6 integers are selected at random from the set (1,2,...,40). >What is the probability that none of the integers 1,...,6 will be selected? You must select 6 from 34 numbers. This can be done in C(34,6) ways C(34,6) Required probability = -------- = 0.35038 C(40,6) >(3) To play the Pennsylvania superlottery, a player selects 7 of the >first 80 positive integers. What is the probability that the >person wins the grand prize by picking 7 integers that are among >the 11 integers selected randomly by the Pennsylvania lottery >commission? C(11,7) Probability = --------- = 3.149 x 10^(-11) C(80,11) >(4) 6 integers are chosen at random from the set of the first 40 >positive integers. What is the probability that your 6 choices match >exactly 5 of these integers? C(6,5) x C(34,1) Required probability = ------------------ = 5.3147 x 10^(-5) C(40,6) >(5) Which is more likely : rolling a total of exactly 8 using two >fair dice or rolling a total of exactly 8 when three fair dice are >rolled? With two dice, possibilities are 6+2 Each row has probability 1/36 5+3 4+4 Total prob. = 5/36 3+5 2+6 With three dice 1+1+6 (3 of these) Each row has probability 1/216 1+2+5 (6 of these) 1+3+4 (6 of these) Total prob. = 21/216 2+2+4 (3 of these) 2+3+3 (3 of these) ------------------ Total = 21 possibilities Comparing the two 5/36 = 0.13888 21/216 = 0.09722 So the two dice have the greater probability. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/