|


Poker, the Lottery
Date: 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. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/