Mth 095 - Coe College



Thursday Forum: Random Lives

Probability (or Chance)

The likelihood of an event occurring.

Probability (or Chance) is expressed as a number ranging from 0 to 1

It may also be expressed as a percentage.

Chance of an event occurring = number of ways the event can occur

number of ways any event can

Chance is the same thing as the long run average

(what we expect to see, on average, if we do something over & over again)

Tossing a Coin



The problem with calculating chances in the real world

Some chances are easy to find

Example: The chance of rolling die & getting a 6

= 1/6

Some chances are hard to find

Example: Suppose in a game of Blackjack (which started with a new deck of cards) you have the following situation. What is the chance the player wins the hands if she hits once?

Can be found but involves a lot of math

If the dealer’s unknown card is 6 or higher then the player has about an 11% chance of winning

Some chances are impossible to find (at this point in time)

Example: What is the chance that it will

rain tomorrow?

If the chance is impossible to find then we find or best approximation for the chance.

Coincidence

When we experience coincidence in our lives the question arises:

Do we really understand the chance of such a coincidence happening?

Misperceived Coincidence (informal definition):

An event that happens where we do not understand the chances involved (Although we think we do).

1st type of Misperception

A Particular Event vs. Any Event

One way people misperceive coincidence is if they think of the chance of the coincidence as a particular event rather than any event.

Example #1: The Birthday Problem

If there are 23 people in a room then the chance that at least two have the same birthday is about 51%

(Assuming that no one is related and ignoring leap years)

This seems high to many people.

The most likely reason for this is that they are thinking of the chance that two people have the same particular birthday (e.g. June 5th), rather than the chance that they have any birthday in common.

If you are in a room with 22 other people, then the chance that at least one other person has the same birthday as you is only about 6%

F:\Gavin\stat java jar\Birthday Experiment.htm

THE MATH

(chance of A happening) + (chance of A not happening) = 1

(chance of A happening) = 1- (chance of A not happening)

So for our birthday problem…

Chance that at least 2 people in a group of 23 have the same birthday

= 1 – (chance that none of the 23 people share the same birthday)

= 1 – ( (365/365)*(364/365)*(363/365)*…*(344/365)*(343/365) )

= 1 – (0.4927…)

= 0.50729 or about a 51% chance

If you have N people in a room then the chance that at least 2 share the same birthday is

1 - ( 365PN / 365 N )

If you are in a group of 23 people, what is the chance that at least one of these other 22 people has the same birthday as you?

Chance that at least one of these 22 people has the same birthday as you

= 1 – (chance that none of the 22 people share your birthday)

= 1 – ( (364/365)*(364/365)*(364/365)*…*(364/365)*(364/365) )

= 1 – (364/365)22

= 1 – 0.9414…

= 0.0585… or about a 6% chance (5.85…%)

If you have N people in a room then the chance that at least one shares your birthday is

= 1 – (364/365)N

You need about 253 people in the room to have about a 50% chance that at least one other person will share your birthday.

Example #2: Amazing Life Coincidences

Suppose that we assume that the chance of An Amazing Life Coincidence happening is 1 in a million (which is 1/1,000,000 )

Next assume that there are about 100 opportunities a day for an amazing life coincidence to happen to you.

Then the chance that you have at least one amazing life coincidence in a day is close to 0%

= 1 – (999,999/1,000,000)100

= 0.0000999… or about 1 chance in 10,000

The chance that you have at least one amazing life coincidence in a year is about 3.6% (about 1 chance in 28)

= 1 – (999,999/1,000,000)100*365

= 1 - 0.9641…

= 0.0358… or about a 3.6% chance

If you have 20 friends…

The chance that at least one amazing life coincidence happens to you or one of your 20 friends in a year is about 53.5%

= 1 – (999,999/1,000,000)100*365*21

= 1 - 0.4646…

= 0.5353… or about a 53.5% chance

Note: This is the same chance as you having at least one amazing life coincidence this year or in the following in the 20 years.

If each of your 20 friends have 20 friends, the chance that you, one of your friends, or one of their friends has an amazing life coincidence in a year is virtually 100% (99.999956..%)

= 1 – (999,999/1,000,000)100*365*401

= 1 - 0.00000044

= 0. 99999956 or about a 100% chance

2nd type of Misperception

Retrospective vs. Prospective

Often when something happens to us in our life we ask ourselves ‘What was the chance of that happening?’

Usually when we try to calculate the chance we do it prospectively.

Really we should do it retrospectively.

Example #1: World Lottery



[pic]

prospectively

What is the chance that Yu Tang (Shanghai, China) wins?

approximately 1 chance in 6,500,000,000 (1/6,500,000,000)

close to a 0% chance

retrospectively

What is the chance that Yu Tang (Shanghai, China) wins?

approximately 1 chance in 1 (100% chance)

Someone had to win, why not Yu Tang

Example #2: Amazing Football’ ‘Predictions’

You get an e-mail message from someone claiming to have great scheme for picking the winners of football games (or this person might claim to have psychic abilities)

They give you who they think is going to win in each of the up coming NFL football games

(14 a week during the season)

After the games are played the person got all 14 correct.

Should you play the $50 fee to get the person’s predictions for the following week?

Chance of getting all 14 correct (just by randomly guessing)

=(1/2)14 = 1/16,384

If the message sender has a list of 1,638,400 e-mail addresses then 100 people will get messages with all 14 correct

(1,400 will get messages with 13 correct)

(9,100 will get messages with 12 correct)

:

:

Example #3: Nostradamus ‘Predictions’

Nostradamus’ predictions are actually postdictions.

No one has used them consistently for predicting the future.

The problem with using them for prediction: They are very general, vague & written in an archaic language.

Example:

Que d'or d'argent fera despendre,

Quand Comte voudra Ville prendre,

Tant de mille & mille soldats,

Tuez, noyez, sans y rien faire,

Dans plus forte mettra pied terre,

Pigmée ayde des Censuarts.



Example #4: Bible Code

Some have suggested that God hid messages in the letters used to write the Bible (that for tell of the future).



Example #5: Repeat Lottery Winner

Friday, November 12, 2004

[pic]Solon man repeats as lottery winner: Veteran wins $100K again

By Deidre Bello

Iowa City Press-Citizen

SOLON — What are the odds of winning $100,000 through the Iowa Lottery?

That’s easy: 1 in 324,632, according to lottery officials.

But what are the odds of doing it twice? There’s no quick answer, but Solon’s K. Morris Richardson is a repeat winner of the six-figure prize.

Richardson, 79, a member of the Solon American Legion Post 460, bought the $100,000 Cash Game ticket Monday at Hy-Vee Foods on North Dodge Street.

Richardson first won $100,000 in 2000 through the same daily game. The Monday win marked the first time he won at the North Dodge Hy-Vee.

$100KCashGame.html

Prospective

Before K. Morris Richardson won on either occasion, what is the chance of predicting correctly the exact two days on which he would win

(Assuming he only buys one ticket each day)

= (1/ 324,632)*(1/324,632) = 1/ 105,385,935,424

about 1 in 100 billion

Before K. Morris Richardson won on either occasion, what is the chance of predicting correctly that he would win at least twice from 2000 through 2004

(Assuming he only buys one ticket each day & plays each time)

= 1- chance he wins 0 or one time from 2000 through 2004

= 1 – { (324,631/324,632)1564 + 1564*(324,631/324,632)1563 *(1/324,632) }

= 1/86499 so about 1 chance in 86,500 of being correct.

But we did not predict that K. Morris Richardson would win twice, we saw that he won twice. Thus we should think of the chance retrospectively.

Retrospective

How do we think about this chance retrospectively?

Since anyone who had won once could be the person to win a 2nd time we have to think of the any winner rather than the particular winner (K. Morris Richardson)

How many & how often the person buys tickets will affect her/his chances of winning.

How many people play & how regularly they play will affect the chance of getting a multiple winner.

Since there are many lottery games this increases the chances of getting multiple winners.

Since we did not predict the multiple winner of a lottery would be K. Morris Richardson playing the Iowa $100,000 cash game from 2000 through 2004,

we need to try to come up with the general group from which this observation came.

If a player of the Iowa $100,000 lottery buys one ticket for each drawing, then like K. Morris Richardson, each has a 1/86499 of winning twice within 5 years.

OR

a 86498/86499 of not winning twice with 5 years.

What happens if we assume that there are about 100 games like the Iowa $100,000 lottery and about 1,000 people play each regularly?

The chance that at least one player wins their lottery twice within a 5 year period is

1-(86498/86499)1,000*100 = 0.685 or about a 68.5%

Instead, if we think each game has about 10,000 regular players, then the chance that at least one wins their lottery twice within a 5 year period is

1-(86498/86499)10,000*100 = 0.99999 or about a 99.999%

3rd type of Misperception

The affect of clustering

How things are clustered (or bunched) can throw off our intuition.

Example #1: Buses

In large cities the arrival times of buses at a bus stop usually are not evenly spread out.

The buses start to bunch up.

Case #1: Bus arrivals are evenly spread out

___________________________________________________

15 mins 15 mins 15 mins

Average time between Bus Arrivals

= ( 15 + 15 + 15 )/3 = 15 mins

If you ‘randomly’ arrive at the bus stop, then how long on average will you have to wait for a bus?

Note: 15/(15+15+15) = 15/45 or 1/3, so…

= (1/3)*(15/2) + (1/3)*(15/2) + (1/3)*(15/2) = 15/2 or 7 ½ mins

If you just missed a bus, then how long on average will you have to wait for a bus?

= (1/3)*(15) + (1/3)*(15) + (1/3)*(15) = 15 mins

Case #2: Bus arrivals are bunched together

___________________________________________________

1 min 1 min 43 mins

Average time between Bus Arrivals

= ( 1 + 1 + 43 )/3 = 15 mins

If you ‘randomly’ arrive at the bus stop, then how long on average will you have to wait for a bus?

= (1/45)*(1/2) + (1/45)*(1/2) + (43/45)*(43/2) = 20.56… mins

If you just missed a bus, then how long on average will you have to wait for a bus?

= (1/3)*(1) + (1/3)*(1) + (1/3)*(43) = 15 mins

Example #2: Who will be your girl friend?

You are seeing two ladies (Alice & Stephanie). You visit both by train. Alice lives north of town & Stephanie lives to the south of town.

The train to Alice’s home The train to Stephanie’s home

Each day you let ‘fate’ decide which one you are going to visit.

You arrive randomly at the Train Station.

Which ever train arrives first, that is the train that you will get on.

After a month you find that you have visited Stephanie 28 times & Alice only twice. Is fate telling you to make Stephanie your girl friend?

Answer NOT REALLY

It turns out that the train to Stephanie’s home leaves on the hour & every 15 mins after that.

The train to Alice’s home leaves one minute after the hour & every 15 mins after that.

noon____12:01______________12:15________12:16____________12:30

1 min 14 min 1 min 14 mins

So…

14/15 chance that you will visit Stephanie

1/15 chance that you will visit Alice

4th type of Misperception

How we think (Psychology)

Some examples…

Creating Causal Theories

Only remembering positive outcomes (Ex. Adoption Fallacy)

Regression Fallacy

(Ex. Sports Illustrated Jinx)

We are not as random as we think (Ex. Mind Test)



5th type of Misperception

Conditional Probability

Sometimes you know certain pieces of information. This information can change the chances of different things happening (or being true)

Example #1: Children

Case #1 (No information about the children)

Suppose you know that a woman has 3 children, what is the chance that two are boys & one is a girl?

(For simplicity, assume an equal chance that any child is a boy or a girl)

ANSWER: 3/8 = 0.375 or a 37.5% chance of 2 boys & 1 girl.

There are 8 possibilities of (1stchild, 2ndchild, 3rdchild), 3 have 2 boys & 1 girl

BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG

Case #2: Suppose that you see one of her children & it is a boy.

Now, what is the chance that two are boys & one is a girl?

ANSWER: 3/7 = 0.429 or a 42.9% chance of 2 boys & 1 girl.

Since you know that one child is a boy then all three can not be girls.

(So throw out GGG)

There are now 7 possibilities, three which have 2 boys & 1 girl

BBB, BBG, BGB, GBB, GGB, GBG, BGG

NOTE: IF you had seen that one of her children was a girl (instead of a boy) you would still get the same answer of 3/7

Case #3: Suppose that you are told that this boy is her oldest child

Now, what is the chance that two are boys & one is a girl?

ANSWER: 2/4 = 0.5 or a 50% chance of 2 boys & 1 girl.

Since you know that the oldest child is a boy then the three children must be B _ _

(So throw out any possibility that starts with a girl)

There are now 4 possibilities, two which have 2 boys & 1 girl

BBB, BBG, BGB, BGG

Example #2:

The Monty Hall Problem

LET'S MAKE A DEAL

(1963-1976)

(80's & 90's revivals)

Host: Monty Hall

There is one contestant.

There are 3 doors, but only one door has GOOD prize behind it (the other two doors have JOKE prizes behind them).

1) The contestant picks a door.

2) Monty will then open up one of the doors that was not selected to show

that there is a JOKE prize behind that door.

3) The contestant then can stay with the door he/she picked or he/she

can switch to the other door (which was not opened).

4) Monty then opens the contestant's chosen door to see if he/she won

the prized

Play the Monty Hall Game

F:\Gavin\stat java jar\Monty Hall Game.htm

Computer Simulation

F:\Gavin\stat java jar\Monty Hall Experiment.htm

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