Probability & Discrete Probability Distributions

Objectives2

Describe the idea of probability Introduce permutation and combination Describe chance behavior with a probability model Define random variables Describe finite and discrete probability models

Content

Introduction Permutation and Combination Probability Models Random Variables Discrete Probability Distributions

Introduction

Introduction The Renaissance and the Protestant Reformation in

1500-1600 were a time for geographical exploration, and experimentation in art, science, and mathematics, which led to a dramatic acceleration in the growth of trade and commerce.

Before trade occurs, people got rich largely by exploitation or by plundering another's wealth.

Trade is a mutually beneficial process, a transaction in which both parties perceive themselves as wealthier than they were before.

The newly rich were now the smart, the adventuresome, the innovators, mostly businessmen, instead of just the hereditary princes and their minions.

Introduction But trade is a risky business, and requires the

assessment of uncertainty and probability. The growth of trade transformed the principles of

gambling into the creation of wealth. Capitalism requires: 1) bookkeeping; 2) forecasting. The successful business executive is a forecaster first;

purchasing, producing, marketing, pricing, and organizing all follow.

The Allegory of Fortune, by Dosso Dossi (1486-1542)

Life is a lottery for everyone.

Astragalus

The earliest form of gambling was played with dice made from astragalus.

In Greek mythology, three brothers rolled dice for the universe. Zeus dinning the heaven, Poseidon the sea, and Hades, the loser, won the underworld.

A common game involved tossing four astragali. The outcome in which all four astragali came up different is called a Venus throw.

The Venus throw has a probability of about 384 out of 10,000.

Astragalus

The Greeks also employed astragali when making inquiries of their oracles.

But despite the importance of astragali tosses in both gambling and religion, the Greeks made no effort to understand and develop a theory of randomness and probability.

Sandwich

The Earl of Sandwich invented a snack that bears his name so that he could avoid leaving the gambling table to eat.

Randomness

Which Group is Random?

What is RANDOMNESS?

Humans are bad at generating randomness - http://www.youtube.com/watch?v=H2lJLXS3AYM

Can you guess the next number for each one of the following two series of binary digits?

1. 010101010101010101012. 01101100110111100010 What is the probability of obtaining either

series? Random = patternless

What is RANDOMNESS?

Suppose a friend visiting Pluto forgot to take along his trigonometric tables , and that sending him telegrams is very expensive. How would you send him/her the required information?

A much cheaper way to convey the same information would be to transmit instructions for calculating the table values.

Formula to generate trigonometric table valuesc0 = 1s0 = 0cn+1 = cn - (αcn + β sn)sn+1 = sn + (β cn - α sn)

where α = 2 sin2(π/N) and β = sin(2π/N) and sn = sin(2πn/N) and cn for cos(2πn/N)

What is RANDOMNESS?

Suppose my friend is also interested to know all the mark-six drawn results since he left the earth 9+ years before.

Is there a formula for compressing the information into a short message?

The information embodied in a random series of numbers cannot be ``compressed,'' or reduced to a more compact form.

Efficient market hypothesis

In an efficient market, stock prices reflect all relevant information, so tomorrow’s price change will reflect only tomorrow’s news and will be independent of the price changes today.

But news is by definition unpredictable and, thus, resulting price changes must be unpredictable and random.

All subsequent price changes represent random departures from previous prices.

Therefore, it’s impossible to beat the market.

January effect?

Consistent Winners

Risk and Uncertainty

Risk is an event the probability of which can be relatively accurately assessed.

Uncertainty, on the other hand, is risk that is hard to measure.

Prediction, which is from Latin, is associated with fortune-telling, superstition, and interpreting signs so as to gain an advantage. 

The term forecast came from English’s Germanic roots. Making a forecast suggested having prudence, wisdom, and industriousness. 

Risk and Uncertainty

Unlikely things happen - and likely things fail to happen - all the time.

Probability vs. Statistics

Based on assumptions about how the world works, probability quantifies the likelihood of a future event.

Statistics begins with what is observed and draws conclusions about how the world works.

We can view probability as the “inverse” of statistics.

Probability vs. Statistics

Permutation & Combination

Permutation & Combination

"My fruit salad is a combination of apples, grapes and bananas"

"The password is 472". If the order doesn't matter, it is a

Combination. A Permutation is an ordered Combination. http://

www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

Two types of permutations

Repetition is Allowed, e.g. “The password is 333".

When you have n things to choose from ... you have n choices each time! When choosing r of them, the permutations are:n × n × ... (r times) = nr.

Example: in the password above, there are 10 numbers to choose from (0,1,..9) and you choose 3 of them:10 × 10 × 10 (3 times) = 103 = 1,000 permutations

Two types of permutations

No Repetition: for example the first three people in a running race. The same person can't be first and second simultaneously.

How many different ways are there that 3 numbered passwords could be formed out of 0-9 when the same number cannot be used more than once?

The total number of permutations of choosing r of n distinct items without repetition, and order matters is nPr = n!/(n – r)!

Two types of combinations

Combinations without Repetition The Mark Six Lottery is a 6 out of 49 lotto game

which is conducted by HKJC Lotteries Limited, a subsidiary of The Hong Kong Jockey Club.

In each draw, 7 numbers will be drawn out of 49 numbers. The first 6 numbers are Drawn Numbers and the 7th number is the Extra Number.

What is the chance of winning the first prize, that is, you got six drawn numbers right in any order?

Mark Six Lottery

First, assume that the order does matter (i.e. permutations).

6 out of 49 gave us 10,068,347,520 permutations. All we need to do is adjust our permutations to reduce it

by how many ways the objects could be in order (because we aren't interested in the order).

The required number of permutations for winning the first prize = 10,068,347,520/6! = 13,983,816

The chance of winning the first prize = 1/13,983,816

Mark Six Lottery

Southampton University Study

The numbers 7, 17, 23, 32, 38, 42 and 48 were picked by lots of players.

Numbers greater than 31 are chosen less often, probably because players use birthday dates for their numbers.

Picking numbers on the edges and lower part of the ticket will usually increase the size of any prizes that are won.

There is also evidence that players prefer not to choose consecutive numbers, for example 31 with 32.

Cryptography

Cryptography itself can be divided into two branches: transposition and substitution.

In transposition, the letters of the message are simply rearranged, effectively generating an anagram.

Example: Plaintext: alanCiphertext: LAAN

How many possible ciphertext are there in the above example? What if you are asked to decipher the following? How many distinct

arrangements are there? How long does it take to decipher the message if all the people in the world worked day and night, and if one person could check one arrangement per second?

sdw1it2dwiwap3ml0ap3laa7oz9o9onc4nndkddererss

Two types of combinations

Combinations without Repetition Formula: n is the number of things to choose from, and you choose r of them (No repetition, order doesn't matter)

Two types of Combinations

Combinations with Repetition Let us say there are five flavors of ice cream: banana,

chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?

Let's use letters for the flavors: {b, c, l, s, v}. Example selections would be:

1. {c, c, c} (3 scoops of chocolate)

2. {b, l, v} (one each of banana, lemon and vanilla)

3. {b, v, v} (one of banana, two of vanilla) There are n=5 things to choose from, and you choose

r=3 of them. Order does not matter, and you can repeat!

Two types of combinations

Combinations with Repetition For {c, c, c} (3 scoops of chocolate), think about the ice

cream being in boxes, you could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and you will have 3 scoops of chocolate!

You could write this down as O O O (arrow means move, circle means scoop).

{b, l, v} = O O O {b, v, v} = O O O The problem reduces to a simpler problem: "how many

different ways can you arrange the arrows and circles?“

Two types of combinations

In general, the problem is: there are r + (n-1) positions, how many ways are there to choose r of them to have circles?

Combinations with Repetition Formula: n is the number of things to choose from, and you choose r of them (repetition allowed, order doesn't matter)

Counting Rules

Complete suit, 25th November 2011

Statistical Impossibility Odds that flipping a fair coin 10,000 times would

land on “Heads" 10,000 times = 2 x 103010 to one against.

Not impossible in the strictest sense, but effectively impossible.

James Clerk Maxwell’s gas container:

Odds that all gas molecules move back to one side = 1 (150 billion trillion zeroes) to one against

Probability Models

Three Interpretations of Probability The frequency interpretation: The probability of

an event is the proportion of times that events of the same kind have occurred in the past.

Personal or subjective evaluations, Bayesian approach. 

Axiomatic approach: The probability of any outcome is the proportion of times the outcome would occur in a very long series of repetitions.

> source(“largeNo.R”)

Law of Large Number

Outcome of an individual trial is unknown.

Probability deals with the aggregate, the long run.

>source(“sim.R”) >source(“sim1.R”)

Axiomatic Interpretation of Probability Girolamo Cardano

defines probability as a number obtained by dividing the number of favorable cases by the number of possible cases.

44

Axiomatic Approach

Descriptions of chance behavior contain two parts: a list of possible outcomes and a probability for each outcome.

The sample space S of a chance process is the set of all possible outcomes. E.g. tossing a coin twice, S={H,H},{H,T},{T,H},{T,T}

An event is an outcome or a set of outcomes of a random phenomenon. That is, an event is a subset of the sample space. E.g. No heads: A={T,T}

Probability of an event = No. of favorable outcomes/Total number of outcomes. E.g. P(A) = ¼.

Axiomatic Approach

Union U (or) E.g. A=heads on first, B=heads on second,A U B= {H,T},{H,H},{T,H} , P(A U B) = 3/4

Intersection ∩ (and): E.g. A= heads on first, B=heads on second, A ∩ B = {H,H} , P(A ∩ B) = 1/4

Complement of Event A – set of all outcomes not in A. E.g. A={T,T}, Ac={H,H},{H,T},{T,H}

An event A is included in B (A B) if every outcome in A is also in B.

Kolmogorov's Axioms of Probability

BA PBAP then

events)disjoint for rule(addition

events exclusivemutually are B andA If : 3 Axiom

1S : 2 Axiom

0 : 1 Axiom

2 Axiom 1

3 Axiom APAP

AAPSP

:Proof

A of complement theis A where

AP1AP

: 1 Theorem

c

c

c

c

Kolmogorov's 3 Axiom of Probability

Our formal system now contains …

AP1AP : 1 Theorem

BA PBAP then

events, exclusivemutually are B andA If : 3 Axiom

1S : 2 Axiom

0 : 1 Axiom

c

0

2 Axiom 1 - 1

SP1P :Hence

1 Theorem AP1AP :From

A then S, A Set

:Proof

0P : 2 Theorem

c

c

Kolmogorov's 3 Axiom of Probability

Our formal system now contains …

0P : 2 Theorem

AP1AP : 1 Theorem

BA PBAP then

events, exclusivemutually are B andA If : 3 Axiom

1S : 2 Axiom

0 : 1 Axiom

c

3 Axiom BAPBPAP

BAPBAP*BPBAP *AP

3 Axiom *BPBAP *AP

DiagramVenn *BA*APBAP

:Proof

BAPBPAPBAP : 3 Theorem

B

A* B*BA

Kolmogorov's 3 Axiom of Probability

Our formal system now contains …

BAPBPAPBAP : 3 Theorem

0P : 2 Theorem

AP1AP : 1 Theorem

BA PBAP then

events, exclusivemutually are B andA If : 3 Axiom

1S : 2 Axiom

0 : 1 Axiom

c

Our formal system now contains …

....... dcomplicate more and more becomes and

on, andon goesit And

BA PBA PAP : 4 Theorem

BAPBPAPBAP : 3 Theorem

0P : 2 Theorem

AP1AP : 1 Theorem

BA PBAP then

events, exclusivemutually are B andA If : 3 Axiom

1S : 2 Axiom

0 : 1 Axiom

c

c

2002 Washington, D.C., sniper case A random killing spree by a man claimed the

lives of ten innocent people in Washington. The shootings came from a white van or truck.

Eventually, the police arrested a man who owned a white van, a number of rifles, and a manual for snipers.

There are about 4 million people in the Washington area and, assuming there is only one guilty. It is further estimated that ten people (including the guilty one) own all three of the items mentioned above.

Is there enough evidence to suggest that the arrested man is the wanted sniper?

An old 16th century gambling problem

I roll three dice and if the sum of spots is nine you win, if it is ten I win. I offer you even odds, should you play the game?

Another Question

The Hong Kong Hospital Authority reports that around 3% of the Hong Kong population has sleep apnea. They also report that around 10% of the population has restless leg syndrome. Similarly, they report that 58% of adults experience insomnia.

A pharmaceutical firm has discovered a drug that would treat the above sleep disorders. Does this mean that the market size for the drug is 71%?

Still one more question

Tom has been summarily rejected by all top three Hong Kong MSc programmes. Desperate, he sends his transcripts and GMAT to the two least selective schools he can think of, X and Y. Based on the success his friends have had there, he estimates that his probability of being accepted at X is 0.7, and at Y, 0.4. He also suspects there is a 75% chance that at least one of his applications will be rejected. What is the probability that at least one of the schools will accept him?

A probability oddity by Bradley Efron

Imagine four dice, A, B, C, and D: A has 4 on four faces and 0 on two faces; B has 3 on all six faces; C has four faces with 2 and two faces with 6; and D has 5 on three faces and 1 on three faces.

If die A is rolled against die B, die A will win - by showing a higher number - two-thirds of the time.

If die B is rolled against die C, B will win two-thirds of the time.

If die C is rolled against die D, C will win two-thirds of the time.

If die D is rolled against die A, D will win two-thirds of the time.

A beats B beats C beats D beats A, all two-thirds of the time.

Conditional Probability

The conditional probability of B given A has occurred is denoted by P(B/A).

Unless the events A and B are independent, the probability of A, P(A), is different from the probability of A given that B has occurred P(A/B).

The probability that a person chosen at random from Hong Kong is over 250 pounds is quite small. However, if it's known somehow that the person chosen is over six feet four inches tall, then the conditional probability that he or she also weighs more than 250 pounds is considerably higher.

Conditional Probability

The conditional probability of having chosen a king card when it's known that the card is a face is 1/3 . However, the conditional probability that the card is a face card given that it's a king is 1, or 100 percent.

Conditional Probabiity

P(A|B) = P(A∩B)/P(B) P(A∩B) = P(A|B)P(B) E.g. Drawing a card from a deck of 52

cards, P(Heart)= ¼. However, if it is known that the card

is red, then:P(Heart | Red) = P(Heart∩Red)/P(Red) = (13/52)/ (26/52) = ½.

Conditional Probabiity

Independent events: P(A|B)=P(A) If two events are independent then

P(A∩B)=P(A|B)P(B)=P(A)P(B) There are situations in which knowing that

event B occurred gives no information about event A, E.g. knowing that a card is black gives no information about whether it is an ace.

P(Ace | Black) = P(Ace∩Black)/P(Black) = (2/52) / (26/52) = 2/26 = 4/52 = P(Ace).

Independent Events Example

Consider tossing two fair dice. Let E1 denote the event that the sum of the dice is six, E2 denote the event that the first die equals four and E3 denote the event that the sum of the dice is seven.

Is E1 and E2 independent? Is E2 and E3 independent?

Conditional Probability Example

An experiment consists of flipping two fair coins, what is the probability that both are heads given that at least one of them is heads?

Sample space S = {(H,H), (H, T ), (T,H), (T, T )}.

Let E2 denote the event that both coins are heads, and E1 denote the event that at least one of them is heads, then P(E2|E1) = P(E2 ∩ E1)/P(E1) = P({(H,H)})/P({(H,H), (H, T ), (T,H)}) = (1/4)/(3/4) = 1/3

HIV Test

Assume that there is a test for HIV which is 95 percent accurate; i.e., if someone has HIV, the test will be positive 95 percent of the time, and if one doesn't have it, the test will be negative 95 percent of the time.

Assume further that 1 percent of the population have HIV.

Now imagine that you've taken the test and that your doctor informs you that you've tested positive. How depressed should you be?

HIV Test

Imagine that 10,000 tests for HIV are administered.

On the average, 100 of these 10,000 people (1 percent of 10,000) will have HIV, and since 95 percent of them will test positive, we will have 95 positive tests.

Of the 9,900 people without HIV, 5 percent of them will test positive, for a total of 495 positive tests (.05 x 9,900 = 495).

Thus, of the total 590 positive tests (495 + 95 = 590), most (495) are false positives.

So the conditional probability of having HIV given that one tests positive is only 95/590, or about 16 percent.

HIV Testing Example

Two Envelopes Paradox

One envelope contains twice as much money as the other.

I'll let you select one envelope, which you can have after the game is over. But as soon as you select one, I offer you the option to switch envelopes. Should you switch?

Mutliplication Rule

Total probability theorem

Bayesian analysis

P(Ai ∩ B)/P(B)

Posterior probability

Piorr probability Likelihood

Bayesian analysis

Revise your probability estimation based on new evidence.

Posterior α Likelihood×Prior Bayesian statistics posits a prior on the

parameter of interest. The likelihood is the factor by which our

prior beliefs are updated. The posterior is the revised probability.

Bayesian Theorem

Bayesian Theorem

Bayesian Theorem

Bayesian Theorem

 The September 11 terrorist attacks

PRIOR PROBABILITY

Initial estimate of how likely it is that terrorists would crash planes into Manhattan skyscrapers

x 0.005%

A NEW EVENT OCCURS ： FIRST PLANE HITS WORLD TRADE CENTER (LIKELIHOODS)

Probability of first plane hits WTC given that terrorists are attacking Manhattan skyscrapers

y 90%

Probability of first plane hits WTC given that terrorists are not attacking Manhattan skyscrapers (i.e. mere accident)

z 0.008%

POSTERIOR PROBABILITY

Revised estimate of probability of terrorists would crash planes into Manhattan skyscrapers

36%

 The September 11 terrorist attacks

PRIOR PROBABILITY

Revised estimate of probability of terrorists would crash planes into Manhattan skyscrapers

x 36%

A NEW EVENT OCCURS ： SECOND PLANE HITS WORLD TRADE CENTER (LIKELIHOODS)

Probability of second plane hits WTC given that terrorists are attacking Manhattan skyscrapers

y 90%

Probability of second plane hits WTC given that terrorists are not attacking Manhattan skyscrapers (i.e. mere accident)

z 0.008%

POSTERIOR PROBABILITY

Revised estimate of probability of terrorists would crash planes into Manhattan skyscrapers

98%

 The September 11 terrorist attacks

First plane hits WTC

hypothesis prior likelihood prior x likelihood

posterior

terrorist attack 0.005 0.9 0.0045 0.361156

accident 0.995 0.008 0.00796 0.6388441 0.01246 1

Second plane hits WTC

hypothesis prior likelihood prior x likelihood

posterior

terrorist attack 0.361156 0.9 0.32504 0.98452

accident 0.638844 0.008 0.005111 0.01548

1 0.330151 1

Bayesian Theorem Example

Consider two boxes: the first containing two white and seven black balls, and the second containing five white and six black balls. Now flip a fair coin and draw a ball from the first or second box depending on whether you get heads or tails. What is the conditional probability that the outcome of the toss was heads given that a white ball was selected?

Bayesian Theorem Example

A laboratory blood test is 95% effective in determining a certain disease when it is present. However, the test also yields a false positive result for 1% of the healthy persons tested. If 0.5% of the population actually has the disease, what is the probability a person has the disease given that their test result is positive?

Pregnancy tests

The home pregnancy tests site cites the following: The overall sensitivity home pregnancy tests was 75%. Specificity was in the range 52% to 75%. Assume the lower value for the specificity. Suppose a subject has a positive test and that 30% of women taking pregnancy tests are actually pregnant. What number is closest to the probability of pregnancy given the positive test?

A. 10%B. 20%C. 30%D. 40%

Monty-hall Problem

The Monty Hall problem is based on the American television game show “Let's Make a Deal” and named after the show's original host, Monty Hall.

Monty-hall Problem

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.

You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Monty-hall Problem

Suppose you and I play the game 300 times, each time we always choose door 1, but you stick to the first door while I always switch.

Expected number of times you win = 100 vs. expected number of times I win = 200.

> source(“montyHall.R”)

Monty-hall Problem

Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy.

Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999).

The Snow Fun Ski Resort Q1

The Snow Fun Ski Resort Q2 Suppose Alan knows how much it will snow this

winter. If the snow fall is more than 40 inches, Alan

should operate the resort without the snowmaking equipment, which results in a profit of $120,000.

If the snow fall is between 20-40 inches, the best action is to operate the resort with the snowmaking equipment which results in a profit of $58,000.

If the snow fall is less than 20 inches, the best action is to let the larger hotel operate the resort which results in a profit of $45,000.

Expected profit with perfect information = $120,000*.4 + $58,000*.2 + $45,000*.4 = $77,600

The Snow Fun Ski Resort Q2 Without perfect information, the best action for

Alan is to operate the resort with snowmaking equipment. This results in an expected profit of $58,000.

With perfect information, the expected profit is $77,600.

Thus, the value of perfect information (forecast) is: $77,600 - $58,000 = $19,600

Alan should pay no more than $19,600 for the report.

The Snow Fun Ski Resort Q3

FORECAST

EVENT(SNOWFA

LL)

P(EVENT)

P(FORECAST

/EVENT)

P(FORECAST

& EVENT)

P(EVENT /FORECAS

T)

Above normal

Over 40"20"-40"

Under 20"

0.40.20.4

0.90.60.3

.4 .9 = .36.2 .6 = .12.4 .3 = .12

P(above normal) = .60

.36/.60=.6

.12/.60=.2

.12/.60=.2

           

Below normal

Over 40"20"-40"

Under 20"

0.40.20.4

0.10.40.7

.4 .1= 0.4

.2 .4= 0.8

.4 .7= .28P(below normal) = .40

.04/.40=.1

.08/.40=.2

.28/.40=.7

Random Variables

Random Variables92

A random variable is a variable that can take on different numerical values determined by chance. The probability distribution of a random variable gives its possible values and their probabilities.

Example: Consider tossing a fair coin 3 times.Define X = the number of heads obtained

X = 0: TTTX = 1: HTT THT TTHX = 2: HHT HTH THHX = 3: HHH

Value 0 1 2 3

Probability 1/8 3/8 3/8 1/8

Discrete Random Variable93

• There are two main types of random variables: discrete and continuous.

• If we can find a way to list all possible outcomes for a random variable and assign probabilities to each one, we have a discrete random variable.

Probability mass function pmf

die xp(x)

one dot 11/6

two dots 21/6three dots31/6four dots 41/6five dots 51/6six dots 61/6

Sample Space36 Outcomes

Since the dice are fair, each outcome is equally likely.

Each outcome has probability 1/36.

Example: Give a probability model for the chance process of rolling two fair, six-sided dice―one that’s red and one that’s green.

Probability mass function pmf

In a throw of two dice, the discrete random variable X, sum total of dotsX = 2 3 4 5 6 7 8 9 10 11 12f(x) =(1/36)(2/36)(3/36)(4/36)(5/36)(6/36)(5/36)( 4/36)(3/36)(2/36)(1/36)The pmf can be shown graphically:

0 2 3 4 5 6 7 8 9 10 11 12 X

X, sum of dots in throwing two dice

Probability mass function pmf

Cumulative distribution function

The cumulative distribution function (cdf ) of a random variable X defined by F(k) = P(X < k).

Limit of F(k) is 1.

X = k F(k) = Pr(X ≤ k)

2 1/36 3 3/36 4 6/36 5 10/36 6 15/36 . . . . . . 12 36/36 = 1

Expected Value

Expected value = Probability of Outcome x Value of Outcome

On a fair coin toss, expected value of the game = (+$1)(1/2) + (-$1)(1/2) = 0 Fair Game

A Risky Business Venture

3 Alternative “Projects:” Success depends on economic conditions, which cannot be forecasted perfectly.

Assume the manager is indifferent to variance.

Boom Recession Expected

(Probability) (90%) (10%) Value

Fast food -10,000 +12,000 -7,800

Fine food +20,000 -8,000 +17,200

Time deposit +3,000 +3,000 +3,000

Pascal’s Wager (1669)

First argument using cost-benefit analysis from the perspective of religion.

Expected value of betting on God exists = 1/1,000,000 x ∞ = ∞

Expected value of betting against God exists = 999,999/1,000,000 x lifetime loss of happiness arising from restraining indulgence, lustful enjoyment etc. = finite

A rational decision maker should believe in the existence of God.

What are the problems with Pascal’s argument?

Warranty Contract?

Retailers are aggressively selling extended warranty contracts. For example, 4-year contract on a $3000 flat-panel TV costs about $400.

You pay premium = F If you collect on the policy, the payout

= W Probability they pay you = P Retailer’s expected profit = F - P x W >

0 if F/W > P Profit margins are usually between 50%

and 60%, nearly 18 times the margin on the goods themselves.

Blood Test

The lab director are given 50 samples to test. Is it a good idea to pool the blood samples

together, perform one test? If the pooled sample is negative, job finished; otherwise perform fifty tests.

Given: the probability that any one person is healthy is 99/100.

The probability that all fifty people are healthy is (99/100)50 (assume independent events).

The probability that at least one person suffers from the disease and so the probability of having to perform fifty-one tests is [1 - (99/100)50].

Thus, the expected number of tests necessary is (1 test x (99/100) 50) + (51 tests x [1 - (99/100) 50]) - approximately 21 tests.

Gambling

Chuck-a-luck

You pick a number from 1 to 6 and the operator rolls three dice.

If the number you pick comes up on all three dice, the operator pays you $3.

If it comes up on two of the three dice, he pays you $2.

If it comes up on just one of the three dice, he pays you $1.

Only if the number you picked doesn't come up at all do you pay him anything— just $1.

Should you play the game?

American Roulette

Bet $1 on a number (not 0 or 00)

If it comes up, win $35. If not, lose the $1

E[Win] = (-$1)(37/38) + (+$35)(1/38) = -5.3 cents (house edge = 5.3%)

Stay long enough it will grind you down.

The Gambler’s Odds

Craps

Place your bet. Roll two dice. Total = 7,11 - WIN Total = 2, 3,12 - LOSE Total = 4,5,6,8,9,10 - POINT If your point comes up first: WIN If 7 comes up first: LOSE Suppose your point is 4, what is your

winning probability?

Craps

Craps

Craps

Game EV for $1 bet

Chuck-a-Luck -0.0790

Roulette -0.0530

Craps -0.0140

Gambler’s Ruin Problem

Money: $60 Goal: $100 Strategy: Bet $1, until reach goal or go broke. What is your chance of success? In general:

1. P(Win $1) =p

2. P(Lose $1) = 1 -p

3. start: i dollars

4. goal: n dollars

5. P(Reach goal) = a(i)

Gambler’s Ruin Problem – fair game

For a fair game: p = ½ a(i) = i/n If i = 60, n = 100 a(i) = 60%

Gambler’s Ruin Problem – unfair game

Gambler’s Ruin Problem – unfair game

Game p a(60)

Favorable 0.51 92.6%Fair game 0.5 60%Craps 0.493 28%

0.49 18%0.48 4%

Roulette 0.473 1.4%

Blackjack

Blackjack terms

Bankroll - The initial amount of cash a players has when he sits at the blackjack table.

Blackjack - A hand totaling 21. Blackjack traditionally pays 3:2.

Hard Hand - A hand total without an Ace card, or a total hand above 11 with an Ace.

Bust - When your total exceed 21. Doubling Down - After your two initial cards are

dealt, you can choose to take one last card while doubling your original bet.

Hit - Call for another card to be handed to you by the dealer.

Stand/Stay - When a player doesn’t wish to hit anymore and he decides to remain with his hand.

Blackjack terms

Insurance - A bet you place when the dealer's face up card is an Ace. You place half of your initial bet. If the dealer’s down card is not 10 or a face card, you lose your insurance bet.

Soft Hand - A hand that contains an Ace card. Split Hand - When the two initial cards equal in

value you may split them and play them as two separate hands.

Push – tie. Despite the player has many options, the dealer

has a big advantage that offsets all the advantages enjoyed by the player: Player and dealer both bust, dealer still wins.

However, if you play your cards right, blackjack is the fairest game in the casino.

Blackjack optimal basic strategy

Only very slightly affected by the number of decks of cards used by the casino.

Hard 17 (or higher): Always stand Hard 11 (or lower): Always hit Hard 12-16:

Dealer’s up card Hit until you reach

7, 8, 9, 10, A 17

4, 5, 6 12

2, 3 13

Blackjack optimal basic strategy

Soft hand:

1. Always hit soft 17.

2. Dealer’s up card is 9,10, or A: hit until you get soft 19.

3. Dealer’s up card is 8 or below: hit until you get soft 18.

Double down rules:Your first two cards

Double if dealer has

Total 11 10 or below

Total 10 9 or below

Total 9 4, 5, or 6

A2 thru A7 4, 5, or 6

Blackjack optimal basic strategy

Splitting decisions:

When do you split...

If dealer has...

A, 8 Any card

4, 5, 10 Never!

2, 3, 6, 7 2, 3, 4, 5, or 6

9 2, 3, 4, 5, 6, 8, 9

Blackjack is the fairest game in a casino

For a $1 bet:

Game EV

Blackjack -$0.005

two decks -$0.003

one deck even

casino required to hit on soft 17

drop -$0.002 (-$0.007)

only double down 10, 11

drop -$0.002 (-$0.009)

blackjack pays even drop -$0.023 (-$0.032)

dealer wins ties -$0.093 (-$0.125)

Craps -$0.014`

Roulette -$0.053

Chuck-a-luck -$0.079

Blackjack – infinite deck assumption

A Non-10 card has a 1/13 chance of being dealt next independent of what have been dealt. For a 10-card, the probability is 4/13.

Example:

Blackjack – infinite deck assumption

Chance that dealer busts showing 7 is B(7).

What is B(16)? What is the probability that the next

car delt by the dealer is 6, 7, 8, 9 or 10?

B(16) = 1/13 + 1/13 + 1/13 + 1/13 + 4/13 = 8/13 = 0.615

What is B(15), B(14), …., B(7)?

Blackjack – infinite deck assumption

Blackjack

If you have 17, and the dealer has an up card of 7, should you hit?

Win(17) = B(7) + tie if dealer has a total of 17 = 0.262 + 0.369/2 (assume tie means you win half) = 0.447

If you hit, safe cards are A, 2, 3, and 4 with probability = 4/13 = 0.31 < 0.447. Hence, you should stand.

What if you have a total of 16? If you stand, Win(16) = B(7) = 0.262 If you hit, win probability = 0.38.

Blackjack: hi-lo card counting

Card Counting: 10s and Aces benefit the player.

Low cards (two through six): add 1 to counter. High cards (ten through ace): add -1 to

counter. 7, 8, 9: add 0 to the counter. For the player to get an advantage, counter

value > twice the number of decks remaining. So if the casino starts with 6 decks, and have

dealt about 50 cards, then should bet more if counter value > 10. But note that largest bet should not exceed 1% of your bankroll.

Gambler's fallacy

Oftentimes we try to forecast a particular future event on the basis of our knowledge about the behavior of the class.

A surgeon told a patient that thirty out of every hundred undergoing such an operation die.

Then the patient asked whether this number of deaths is already full.

The doctor replied the past 99 patients he had operated on, 30 died.

Is the patient 100% safe now to undergo the operation?

Gambler's fallacy

The patient has confused case probability with class probability.

Each case is characterized by its unique merits. Case probability is not open to any kind of numerical evaluation.

Knowledge of class probability is useless for determining case probability.

Martingale betting strategy

Every time a player loses a bet, he doubles his bet until he wins.

The strategy is doomed to fail because casinos have “table limits”.

Roulette: Maximum Bet $500 Suppose: -5, -10, -20, -40, -80, -160, -320 Down

$635. Chance of losing: 20/38 = 0.526 Chance of losing 7 in a row = (0.526)7 = 1.1% 100 people using the Martingale strategy:

99 win $5, 1 loses $635Expected profit to casino = -$99 x 5 + $635 = $140

If you are playing a game with a positive expected value, how much should you bet?

Suppose P(win) = 0.51, and bet $1, then EV = 0.51*1 – 0.49*1 = $0.2.

If you have $1000 and bet $1 each time, it takes a long time to win $200.

What if bet all $1000? Or 2 bets at $500 each? Or 4 bets at $250 each?

Kelly Criterion: Bet Your Edge = 51%-49% = 2% $1000, bet $20 As you win: $1500, bet $30 … $2000, bet $40 As you lose: $500, bet $10 If position size is > 2*edge, in the long run,

bankroll < $1.

Joseph Jagger in Monaco, 1873

Born in 1830 in Shelf, England. He hired six clerks to watch the six roulette

wheels at the Beaux-Arts Casino in Monte Carlo and to record every winning number that came up.

The sixth wheel seemed to keep showing the same nine numbers more often than was statistically normal.

He won $300,000 in only four days on this specific wheel !

At night, the casino mixed up the wheels so that Joseph Jagger wouldn't know which one to use!

Joseph Jaggerin Monaco, 1873

But Joseph Jagger realized the tiny scratch on the wheel. He found the wheel, and raised his winnings to $450,000.

Then each night, the casino would take the roulette wheels apart and move the parts around.

Joseph Jagger wasn't able to win again. He took the $320,000 he had left and never even played roulette again!

St. Petersburg paradox

A casino offers a game of chance for a single player in which a fair coin is tossed at each stage.

The pot starts at 1 dollar and is doubled every time a head appears. The first time a tail appears, the game ends and the player wins whatever is in the pot.

Thus the player wins 1 dollar if a tail appears on the first toss, 2 dollars if a head appears on the first toss and a tail on the second, 4 dollars if a head appears on the first two tosses and a tail on the third, 8 dollars if a head appears on the first three tosses and a tail on the fourth, and so on.

What would be a fair price to pay the casino for entering the game?

Continuous Random Variable

• A continuous random variable Y takes on all values in an interval of numbers. The probability distribution of Y is described by a density curve.

• A continuous random variable Y has infinitely many possible values.

• The probability of any event is the area under the density curve and above the values of Y that make up the event.

• All continuous probability models assign probability 0 to every individual outcome. Only intervals of values have positive probability.

Discrete Probability Distributions

Binomial Distribution

The binomial distribution assumptions:

1. The number of observations n is fixed. 2. Each observation is independent. 3. Each observation represents one of two

outcomes ("success" or "failure"). 4. The probability of "success" p is the

same for each outcome. If these conditions are met, then X has a

binomial distribution with parameters n and p, abbreviated B(n, p).

Binomial Distribution

The probability that a random variable X with binomial distribution B(n,p) is equal to the value k, where k = 0, 1,....,n , is given by:

The latter expression is known as the binomial coefficient, stated as "n choose k," or the number of possible ways to choose k "successes" from n observations.

Mean and Variance of the Binomial Distribution:

Binomial Distribution

Suppose that there are trials, each with success probability , and that we wish to determine the probability of observing exactly successes. Some examples of experimental outcomes for which include the following:110000 000011 010010

Because the trials are mutually independent, we have:,,

Binomial Distribution

The probability of each outcome for which is the product of factors of and factors of . Furthermore, the number of such outcome is the number of ways of choosing successes from a total of trials. Thus,

The general formula for the binominal probability is: 

It follows that the general formula for the binominal cumulative distribution function is: 

Example

In 50 trials with success probability 0.5, what is the probability that no more than 10 successes will be observed?

Ans:

> pbinom(10,size=50,prob=.5)[1] 1.193067e-05

Airline overbooking

Airlines have a practice of overbooking. Previous record suggests that P(show) =

0.9 & P(no show) = 0.1. Airplane capacity = 188; Booking = 200. P(show > 188) = 1 -

pbinom(188,size=200,prob=.9) = 0.01678953

Is 0.01678953 too high a probability for the airline?

Poisson Distribution

A discrete distribution often used as a model for the number of random and independent events in a given unit of time and space.

If a discrete random variable X  follows a Poisson distribution with parameter λ > 0, then the probability that k events would occur in a given unit of time is:

λ = average number of events occurring in the given time intervale = base of the natural logarithm (e = 2.71828...)

Poisson Distribution Example

Example: A service hotline can handle at most 2 incoming calls a minute. Past experience suggests that the average incoming calls per hour is 30. What is the probability that the hotline would be overloaded during any given minute?

The mean (expected number of calls) is λ = 30/60 = 0.5 call per minute.

Poisson Distribution Solution

p(X = 0) = 0.50e-0.5/0! = 0.6065 p(X = 1) = 0.51e-0.5/1! = 0.3033 p(X = 2) = 0.52e-0.5/2! = 0.0758 Probability that the hotline would not be

overloaded = p(X = 0) + p(X = 1) + p(X = 2) = 0.9856

Probability that it would be overloaded (> 3 calls)= 1 – probability that it would not be overloaded= 1 – 0.9856 = 0.0144

> ppois(2, lambda=0.5)[1] 0.9856123

> ppois(2, lambda=0.5, lower=FALSE) # upper tail [1] 0.01438768

Coincidence And The Law - People v. Collins, 1968

In 1964 in Los Angeles a blond woman with a ponytail snatched a purse from another woman.

The thief fled on foot but was later spotted entering a yellow car driven by a black man with a beard and a mustache.

Police investigation eventually discovered a blond woman with a ponytail who regularly associated with a bearded and mustachioed black man who owned a yellow car.

There wasn't any hard evidence linking the couple to the crime, or any witnesses able to identify either party.

Coincidence And The Law - People v. Collins, 1968 The prosecutor assigned the following

probabilities: 1. yellow car - 1/10; 2. man with a mustache - 1/4; 3. woman with a ponytail - 1/10; 4. woman with blond hair - 1/3;5. black man with a beard -1/10; 6. interracial couple in a car -1/1,000. The prosecutor further argued that the

characteristics were independent, so that the probability that a randomly selected couple would have all of them would be 1/10 x 1/4 x 1/10 x 1/3 x 1/10 x 1/1,000 = 1/12,000,000, a number so low the couple must be guilty. The jury convicted them.

Should the case be appealed to the Supreme Court?

Coincidence And The Law - People v. Collins, 1968

The Supreme Court of California reversed the conviction, and gave three reasons.

The characteristics are not independent The first reason was insufficient empirical

foundation for the estimate of p = 1/12,000,000.

The second reason: In a city the size of Los Angeles, with around 2,000,000 couples, using poisson to approximate the binomial probability distribution, the probability of having exactly 2 such couples in the population is 0.012, small but certainly allowing for reasonable doubt.

REVENUE MANAGEMENT:

BOOKING LIMITS AND PROTECTION LEVELS

The Hyatt

118 King/Queen rooms.

Hyatt offers a rL= US$159 (low fare) discount

fare targeting leisure travelers.

Regular fare is rH= US$225 (high fare)

targeting business travelers.

Demand for low fare rooms is abundant. Let D be uncertain demand for high fare

rooms. Suppose D has Poisson distribution with mean

27.3 per day. Assume it is not illegal or morally

irresponsible to discriminate the customers. Objective: Maximize expected revenues by

controlling the number of low fare rooms sold.

Yield management decisions

The booking limit is the number of rooms to sell in a low fare class.

The protection level is the number of rooms you reserve for a high fare class.

Let Q be the protection level for the high fare class. Since there are only two fare classes, the booking

limit on the low fare class is

1. 118 – Q:0 118

Q rooms protected forhigh fare passengers

Sell no more than the low fare booking limit, 118 - Q

Optimal protection level

Overage cost: If D < Q we protected too many rooms and earn nothing on Q -

D rooms. We could have sold those empty rooms at the low fare, so Co =

rL.

Underage cost: If D > Q we protected too few rooms. D – Q rooms could have been sold at the high fare but were

sold instead at the low fare, so Cu = rH – rL

Choose Q to balance the overage and underage costs. Optimal high fare protection level:

Optimal low fare booking limit = 118 – Q*

H

LH

uo

u

r

rr

CC

CQF

)( *

152

Hyatt example

Critical ratio:

> qpois(0.2933, lambda=27.3) # inverse Poisson distribution with mean 27.3, gives the smallest integer number of high rate rooms that would satisfy the critical ratio[1] 24

Answer: 24 rooms should be protected for high fare travelers. Similarly, a booking limit of 118-24 = 94 rooms should be applied to low fare reservations.

225 159 660.2933

225 225u h l

o u h

C r r

C C r

Monte Carlo Simulation

References

154

• The Basic Practice of Statistics, 6ed. by Moore, D., W. Notz & M. Fligner, Chapters 10, 12, 13.

END