Dice Games: Probability and Pascal

by Lauren McCluskey

This power point was made with help from: •Bayesian Learning Application to Text Classification Example: spam filtering by Marius Bulacu & prof. dr. Lambert Schomaker

•Mathematicians by www.2july-maths.co.uk/powerpoint/mathematicians.ppt

•Basic Models of Probability by Ron S. Kenett, Weizmann Institute of Science Probability

Introduction to Information Theory by Larry Yaeger, Professor of Informatics, Indiana University

•Access to Math, Probability published by www.pearsonlearning.com

•www.2july-maths.co.uk/powerpoint/mathematicians.ppt

• www.mtsu32.mtsu.edu:11208/Chap9Pres.ppt

Founders of Probability Theory

Blaise Pascal

(1623-1662, France)

Pierre Fermat

(1601-1665, France)

They laid the foundations of the probability theory in a correspondence on a dice game.

From: Bayesian Learning Application to Text Classification Example: spam filtering by Marius Bulacu & prof. dr. Lambert Schomaker

Pascal from: Mathematicians by www.2july-maths.co.uk/powerpoint/mathematicians.ppt

Blaise Pascal 1623 - 1662   

Blaise Pascal, according to contemporary observers, suffered migraines in his youth, deplorable health as an adult, and lived much of his brief life of 39 years in pain.

Nevertheless, he managed to make considerable contributions in his fields of interest, mathematics and physics, aided by keen curiosity and penetrating analytical ability.

Pascal from: Mathematicians by www.2july-maths.co.uk/powerpoint/mathematicians.ppt

Probability theory was Pascal's principal and perhaps most enduring contribution to mathematics, the foundations of probability theory established in a long exchange of letters between Pascal and fellow French mathematician Fermat.  

The Paradox of the Chevalier de Mere - 1

Success = at least one “1”

Basic Models of Probability by Ron S. Kenett, Weizmann Institute of Science

Success = at least one “1,1”

The Paradox of the Chevalier de Mere - 2

Basic Models of Probability by Ron S. Kenett, Weizmann Institute of Science

P (Success) = P(at least one “1,1”)

The Paradox of the Chevalier de Mere - 3

P (Success) = P(at least one “1”)

32

614 3

236124

Experience proved otherwise !Experience proved otherwise !Experience proved otherwise !Experience proved otherwise !

Game A was a better game to play

Basic Models of Probability by Ron S. Kenett, Weizmann Institute of Science

The Paradox of the Chevalier de Mere - 4

P (Failure) = P(no “1”)

482.65 4

What went wrong before?What went wrong before?What went wrong before?What went wrong before?

P (Success) = .518P (Success) = .518

P (Failure) = P(no “1,1”)

509.3635 24

P (Success) = .491P (Success) = .491

The calculations of Pascal and Fermat

Basic Models of Probability by Ron S. Kenett, Weizmann Institute of Science

Sample Space for Dice from: Introduction to Information Theory by Larry Yaeger

• Single die has six elementary outcomes:

• Two dice have 36 elementary outcomes:

1/6*1/6*1/6*1/6= (1/6)4 or .518

• While…

1/6 chances

1/36*1/36*1/36…= (1/36)24 or .491

1/36

chances

Apply it!

When to sit and when to stand…?

How many times can we roll one die before we get a “1”?

Try this:

S.K.U.N.K.

1. Stand up.

2. Someone rolls a die.

3. Sit down: Keep your score. OR: Remain standing: Add it up.

But…

Watch Out!

4. If you’re standing on “1”: Score= “0”!

5. New round: Stand up.

6. Repeat 5 times: one round for each letter in the word S.K.U.N.K.

Reflection:

*What is your winning strategy?

*Why will this work?

Remember:

Sample Space for Dice from: Introduction to Information Theory by Larry Yaeger

• Single die has six elementary outcomes:

• Two dice have 36 elementary outcomes:

Apply It!

• When to roll and when to stop…?• How many times can we roll 2 dice before

we roll a “1” or a “1, 1”?

• Try this:

PIG

1. Take turns rolling 2 dice.

2. Keep rolling: Add it up.

3. Stop: Keep your score.

But…

Watch Out!

4. Roll “1”: Lose your turn.

Roll “1, 1”: Lose it ALL! (Back to “0”!)

5. Get a score of 100: You WIN!

Reflection:

• *What is your winning strategy? • *Why will this work?

• Remember:

Sample Space for Dice from: Introduction to Information Theory by Larry Yaeger

• Single die has six elementary outcomes:

• Two dice have 36 elementary outcomes:1/36

chances

Sample Space for Dice from: Introduction to Information Theory by Larry Yaeger

• Single die has six elementary outcomes:

• Two dice have 36 elementary outcomes:11/36

Chances

To

roll 1

“1”

The Addition Rule• Now throw a pair of black & white dice, and ask: What is

the probability of throwing at least one one?– Let event a = the white die will show a one– Let event b = the black die will show a one

from: Introduction to Information Theory by Larry Yaeger

To add or to multiply ?

P(“1” with 2 dice) P(“1” with 2 dice) ==??

Basic Models of Probability by Ron S. Kenett, Weizmann Institute of Science

Independent Events

• Independent events: two events with outcomes that do not depend on each other.” (from: Access to Math, Probability”)

Independent Events: Either /OR

• When two events are independent, AND either one is favorable, you add their probabilities.

Example: What is the probability that I might roll a 1 on the

black die? 6/36 or 1/6What is the probability that I might roll a 1 on the

white die? 6/36 or 1/6 What is the probability that I will roll either 1 black

OR 1 white “1”? 12/36 or 1/3.

Independent Events: Either /OR

*This is true when the die are rolled one at a time, if, however, you roll them together, then 1W and 1B cannot be counted twice. So the probability of rolling a “1” is 11/36 instead of 12/36.

Sample Space for Dice from: Introduction to Information Theory by Larry Yaeger

• Single die has six elementary outcomes:

• Two dice have 36 elementary outcomes:11/36

Chances

To

roll 1

“1”

Independent Events: And Then…

• When two events are independent, BUT you want to have BOTH of them, you multiply their probabilities.

• Example:• What is the probability that I will roll a “1, 1”?

[P(1) = 1/6 * P(1) = 1/6] or [1/6*1/6= 1/36]. • The P(1,1) = 1/36 because there is only ONE

way that I can do this.

1/36*1/36*1/36…= (1/36)24 or .491

1/36

chances

Dependent Events

• “Dependent events: a set of events in which the outcome of the first event affects the outcome of the next event.”

(from: Access to Math, Probability”)

Dependent Events

• To find the probability of dependent events, multiply the probability of the first by the probability of the second (given that the first has occurred).

• Example: You have the letters : M; A; T; and H in an envelope. What is the probability that you will pull a “M” then a “A”?

Dependent Events

• P(M) = ¼ because there are 4 cards and

• P(A after M) = 1/3 because there are NOW only 3 cards left…

so …

¼ * 1/3= 1/12.

Apply It!

• Put the letters: M; A; T; and H in an envelope and pull them out 1 at a time.

• Replace the card then do it again.

(Repeat 20 times.) • Record your results.

*Think about it: what would happen if you hadn’t replaced the cards each time?