patrick's casino
Post on 06-Jan-2016
39 Views
Preview:
DESCRIPTION
TRANSCRIPT
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Fre
quen
cy
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Fre
quen
cyWhat is the probability of picking an ace?
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Fre
quen
cy
Probability =
0
1
2
3
4
5
Ace 2 3 4 5 6 7 8 9 10 J Q K
Card
Fre
quen
cyWhat is the probability of picking an ace?
4 / 52 = .077 or 7.7 chances in 100
0
1
2
3
4
5
Ace
(.0
77)
2 (.
077)
3 (.
077)
4 (.
077)
5 (.
077)
6 (.
077)
7 (.
077)
8 (.
077)
9 (.
077)
10 (
.077
)
J (.
077)
Q (
.077
)
K (
.077
)
Card
Fre
quen
cyEvery card has the same probability of being picked
0
1
2
3
4
5
Ace
(.0
77)
2 (.
077)
3 (.
077)
4 (.
077)
5 (.
077)
6 (.
077)
7 (.
077)
8 (.
077)
9 (.
077)
10 (
.077
)
J (.
077)
Q (
.077
)
K (
.077
)
Card
Fre
quen
cyWhat is the probability of getting a 10, J, Q, or K?
0
1
2
3
4
5
Ace
(.0
77)
2 (.
077)
3 (.
077)
4 (.
077)
5 (.
077)
6 (.
077)
7 (.
077)
8 (.
077)
9 (.
077)
10 (
.077
)
J (.
077)
Q (
.077
)
K (
.077
)
Card
Fre
quen
cy(.077) + (.077) + (.077) + (.077) = .308
16 / 52 = .308
0
1
2
3
4
5
Ace
(.0
77)
2 (.
077)
3 (.
077)
4 (.
077)
5 (.
077)
6 (.
077)
7 (.
077)
8 (.
077)
9 (.
077)
10 (
.077
)
J (.
077)
Q (
.077
)
K (
.077
)
Card
Fre
quen
cyWhat is the probability of getting a 2 and then after replacing the card getting a 3 ?
0
1
2
3
4
5
Ace
(.0
77)
2 (.
077)
3 (.
077)
4 (.
077)
5 (.
077)
6 (.
077)
7 (.
077)
8 (.
077)
9 (.
077)
10 (
.077
)
J (.
077)
Q (
.077
)
K (
.077
)
Card
Fre
quen
cy(.077) * (.077) = .0059
0
1
2
3
4
5
Ace
(.0
77)
2 (.
077)
3 (.
077)
4 (.
077)
5 (.
077)
6 (.
077)
7 (.
077)
8 (.
077)
9 (.
077)
10 (
.077
)
J (.
077)
Q (
.077
)
K (
.077
)
Card
Fre
quen
cyWhat is the probability that the two cards you draw will be a black jack?
0
1
2
3
4
5
Ace
(.0
77)
2 (.
077)
3 (.
077)
4 (.
077)
5 (.
077)
6 (.
077)
7 (.
077)
8 (.
077)
9 (.
077)
10 (
.077
)
J (.
077)
Q (
.077
)
K (
.077
)
Card
Fre
quen
cy10 Card = (.077) + (.077) + (.077) + (.077) = .308
Ace after one card is removed = 4/51 = .078
(.308)*(.078) = .024
Practice
• What is the probability of rolling a “1” using a six sided dice?
• What is the probability of rolling either a “1” or a “2” with a six sided dice?
• What is the probability of rolling two “1’s” using two six sided dice?
Practice
• What is the probability of rolling a “1” using a six sided dice?1 / 6 = .166
• What is the probability of rolling either a “1” or a “2” with a six sided dice?
• What is the probability of rolling two “1’s” using two six sided dice?
Practice
• What is the probability of rolling a “1” using a six sided dice?1 / 6 = .166
• What is the probability of rolling either a “1” or a “2” with a six sided dice?(.166) + (.166) = .332
• What is the probability of rolling two “1’s” using two six sided dice?
Practice
• What is the probability of rolling a “1” using a six sided dice?1 / 6 = .166
• What is the probability of rolling either a “1” or a “2” with a six sided dice?(.166) + (.166) = .332
• What is the probability of rolling two “1’s” using two six sided dice?(.166)(.166) = .028
Cards
• What is the probability of drawing an ace?
• What is the probability of drawing another ace?
• What is the probability the next four cards you draw will each be an ace?
• What is the probability that an ace will be in the first four cards dealt?
Cards
• What is the probability of drawing an ace?• 4/52 = .0769• What is the probability of drawing another ace?• 4/52 = .0769; 3/51 = .0588; .0769*.0588 = .0045 • What is the probability the next four cards you
draw will each be an ace? • .0769*.0588*.04*.02 = .000003• What is the probability that an ace will be in the
first four cards dealt?• .0769+.078+.08+.082 = .3169
Probability
1.00.00
Event will not occur
Event must occur
Probability
• In this chapter we deal with discreet variables– i.e., a variable that has a limited number of
values
• Previously we discussed the probability of continuous variables (Z –scores)– It does not make sense to seek the probability
of a single score for a continuous variable• Seek the probability of a range of scores
Key Terms
• Independent event– When the occurrence of one event has no
effect on the occurrence of another event• e.g., voting behavior, IQ, etc.
• Mutually exclusive– When the occurrence of one even precludes
the occurrence of another event• e.g., your year in the program, if you are in prosem
Key Terms
• Joint probability– The probability of the co-occurrence of two or
more events• The probability of rolling a one and a six• p (1, 6)• p (Blond, Blue)
Key Terms
• Conditional probabilities– The probability that one event will occur given
that some other vent has occurred• e.g., what is the probability a person will get into a
PhD program given that they attended Villanova– p(Phd l Villa)
• e.g., what is the probability that a person will be a millionaire given that they attended college
– p($$ l college)
Example
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
What is the simple probability that a person will own a video game?
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
What is the simple probability that a person will own a video game? 35 / 100 = .35
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
What is the conditional probability of a person owning a video game given that he or she has children? p (video l child)
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
What is the conditional probability of a person owning a video game given that he or she has children?25 / 55 = .45
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
What is the joint probability that a person will own a video game and have children? p(video, child)
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
What is the joint probability that a person will own a video game and have children? 25 / 100 = .25
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
25 / 100 = .25.35 * .55 = .19
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
The multiplication rule assumes that the two events are independent of each other – it does not work when there is a relationship!
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55
Total 35 65 100
Practice
Republican Democrat Total
Male 52 27 79
Female 18 65 83
Total 70 92 162
p (republican) p(female)p (republican, male) p(female, republican)p (republican l male) p(male l republican)
Republican Democrat Total
Male 52 27 79
Female 18 65 83
Total 70 92 162
p (republican) = 70 / 162 = .43p (republican, male) = 52 / 162 = .32p (republican l male) = 52 / 79 = .66
Republican Democrat Total
Male 52 27 79
Female 18 65 83
Total 70 92 162
p(female) = 83 / 162 = .51p(female, republican) = 18 / 162 = .11p(male l republican) = 52 / 70 = .74
Republican Democrat Total
Male 52 27 79
Female 18 65 83
Total 70 92 162
Foot Race
• Three different people enter a “foot race”
• A, B, C
• How many different combinations are there for these people to finish?
Foot Race
A, B, CA, C, BB, A, CB, C, AC, B, AC, A, B
6 different permutations of these three names taken three at a time
Foot Race
• Six different people enter a “foot race”
• A, B, C, D, E, F
• How many different permutations are there for these people to finish?
Permutation
Ingredients:
N = total number of events
r = number of events selected
NrPrN
N
)!(
!
Permutation
Ingredients:
N = total number of events
r = number of events selected
A, B, C, D, E, F Note: 0! = 1
720)!66(
!6
Foot Race
• Six different people enter a “foot race”
• A, B, C, D, E, F
• How many different permutations are there for these people to finish in the top three?
• A, B, C A, C, D A, D, E B, C, A
Permutation
Ingredients:
N = total number of events
r = number of events selected
NrPrN
N
)!(
!
Permutation
Ingredients:
N = total number of events
r = number of events selected
120)!36(
!6
Foot Race
• Six different people enter a “foot race”
• If a person only needs to finish in the top three to qualify for the next race (i.e., we don’t care about the order) how many different outcomes are there?
Combinations
Ingredients:
N = total number of events
r = number of events selected
NrCrNr
N
)!(!
!
Combinations
Ingredients:
N = total number of events
r = number of events selected
20)!36(!3
!6
Note:
• Use Permutation when ORDER matters
• Use Combination when ORDER does not matter
Practice
• There are three different prizes– 1st $1,00– 2nd $500– 3rd $100
• There are eight finalist in a drawing who are going to be awarded these prizes.
• A person can only win one prize• How many different ways are there to
award these prizes?
Practice
336 ways of awarding the three different prizes
336)!38(
!8
Practice
• There are three prizes (each is worth $200)
• There are eight finalist in a drawing who are going to be awarded these prizes.
• A person can only win one prize
• How many different ways are there to award these prizes?
Combinations
56 different ways to award these prizes
56)!38(!3
!8
Practice
• A shirt comes in four sizes and six colors. One also has the choice of a dragon, alligator, or no emblem on the pocket. How many different kinds of shirts can you order?
Practice
• A shirt comes in four sizes and six colors. One also has the choice of a dragon, alligator, or no emblem on the pocket. How many different kinds of shirts can you order?
• 4*6*3 = 72
• Don’t make it hard on yourself!
Practice
• In a California Governor race there were 135 candidates. The state created ballots that would list candidates in different orders. How many different types of ballots did the state need to create?
Practice
2.6904727073180495e+230
Or
)!135135(
!135
26,904,727,073,180,495,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
Bonus Points
• Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?
• http://www.nytimes.com/2008/04/08/science/08monty.html?_r=1
You pick #1
Door 1 Door 2 Door 3 Results
GAME 1 AUTO GOAT GOATSwitch and you lose.
GAME 2 GOAT AUTO GOATSwitch and you win.
GAME 3 GOAT GOAT AUTOSwitch and you win.
GAME 4 AUTO GOAT GOATStay and you win.
GAME 5 GOAT AUTO GOATStay and you lose.
GAME 6 GOAT GOAT AUTOStay and you lose.
Practice
• The probability of winning “Blingoo” is .30
• What is the probability that you will win 20 of the next 30 games of Blingoo ?
• Note: previous probability methods do not work for this question
Binomial Distribution
• Used with situations in which each of a number of independent trials results in one of two mutually exclusive outcomes
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Game of Chance
• The probability of winning “Blingoo” is .30
• What is the probability that you will win 20 of the next 30 games of Blingoo ?
• Note: previous probability methods do not work for this question
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(30.)!(!
!)( XNX q
XNX
NXp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(70.30.)!(!
!)( XNX
XNX
NXp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)30(70.30.)!30(!
!30)( XX
XXXp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)2030(20 70.30.)!2030(!20
!30)(
Xp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)2030(20 70.30.)!2030(!20
!30)(
Xp
p = .000029
What does this mean?
• p = .000029
• This is the probability that you would win EXACTLY 20 out of 30 games of Blingoo
Game of Chance
• Playing perfect black jack – the probability of winning a hand is .498
• What is the probability that you will win 8 of the next 10 games of blackjack?
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)810(8 502.498.)!810(!8
!10)(
Xp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)810(8 502.498.)!810(!8
!10)(
Xp
p = .0429
Excel
Binomial DistributionWhat is this doing?
Its combining together what you have learned so far!
One way to fit our 8 wins would be (joint probability):W, W, W, W, W, W, W, W, L, L =
(.498)(.498)(.498)(.498)(.498)(.498)(.498)(.498)(.502)(.502)=
(.4988)(.5022)=.00095
pX q(N-X)
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Binomial Distribution
Other ways to fit our question
W, L, L, W, W, W, W, W
L, W, W, W, W, L, W, W
W, W, W, L, W, W, W, L
L, L, W, W, W, W, W, W
W, L, W, L, W, W, W, W
W, W, L, W, W, W, L, W
Binomial Distribution
Other ways to fit our question
W, L, L, W, W, W, W, W = .00095
L, W, W, W, W, L, W, W = .00095
W, W, W, L, W, W, W, L = .00095
L, L, W, W, W, W, W, W = .00095
W, L, W, L, W, W, W, W = .00095
W, W, L, W, W, W, L, W = .00095
Each combination has the same probability – but how many combinations are there?
Combinations
Ingredients:
N = total number of events
r = number of events selected
45 different combinations
NrCrNr
N
)!(!
!
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Binomial Distribution
• Any combination would work• . 00095+ 00095+ 00095+ 00095+ 00095+
00095+ 00095+ 00095+. . . . . . 00095• Or 45 * . 00095 = .04
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Practice
• You bought a ticket for a fire department lottery and your brother has bought two tickets. You just read that 1000 tickets were sold.– a) What is the probability you will win the grand
prize?– b) What is the probability that your brother will
win?– c) What is the probably that you or your bother
will win?
5.2
• A) 1/1000 = .001
• B)2/1000 = .002
• C) .001 + .002 = .003
Practice• Assume the same situation at before
except only a total of 10 tickets were sold and there are two prizes.– a) Given that you didn’t win the first prize, what is the
probability you will win the second prize?
– b) What is the probability that your borther will win the first prize and you will win the second prize?
– c) What is the probability that you will win the first prize and your brother will win the second prize?
– d) What is the probability that the two of you will win the first and second prizes?
5.3
• A) 1/9 = .111
• B) 2/10 * 1/9 = (.20)*(.111) = .022
• C) 1/10 * 2/9 = (.10)*(.22) = .022
• D) .022 + .022 = .044
Practice
• In some homes a mother’s behavior seems to be independent of her baby's, and vice versa. If the mother looks at her child a total of 2 hours each day, and the baby looks at the mother a total of 3 hours each day, and if they really do behave independently, what is the probability that they will look at each other at the same time?
5.8
• 2/24 = .083
• 3/24 = .125
• .083*.125 = .01
Practice
• Abe ice-cream shot has six different flavors of ice cream, and you can order any combination of any number of them (but only one scoop of each flavor). How many different ice-cream cone combinations could they truthfully advertise (note, we don’t care about the order of the scoops and an empty cone doesn’t count).
5.296
)!61(!1
!6
15)!62(!2
!6
20)!63(!3
!6
15)!46(!4
!6
6)!56(!5
!6
1)!66(!6
!6
6 + 15 + 20 +15 + 6 + 1 = 63
Extra Brownie Points!
• Lottery
• To Win:• choose the 5 winnings numbers
– from 1 to 49
• AND• Choose the "Powerball" number
– from 1 to 42
• What is the probability you will win?– Use combinations to answer this question
top related