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Dice Games: Probability and Pascal by Lauren McCluskey

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## Dice Games: Probability and Pascal by Lauren McCluskey

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**This power point was made with help from:**• Bayesian Learning Application to Text Classification Example: spam filteringby Marius Bulacu & prof. dr. Lambert Schomaker • Mathematicians bywww.2july-maths.co.uk/powerpoint/mathematicians.ppt • Basic Models of Probability by Ron S. Kenett, Weizmann Institute of ScienceProbability • 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**bywww.2july-maths.co.uk/powerpoint/mathematicians.ppt BlaisePascal 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**bywww.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.**Basic Models of Probability**by Ron S. Kenett, Weizmann Institute of Science The Paradox of the Chevalier de Mere - 1 Game A Success = at least one “1”**Basic Models of Probability**by Ron S. Kenett, Weizmann Institute of Science The Paradox of the Chevalier de Mere - 2 Game B Success = at least one “1,1”**Basic Models of Probability**by Ron S. Kenett, Weizmann Institute of Science The Paradox of the Chevalier de Mere - 3 Game B Game A P (Success) = P(at least one “1”) P (Success) = P(at least one “1,1”) 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 The calculations of Pascal and Fermat Game A Game B P (Failure) = P(no “1”) P (Failure) = P(no “1,1”) P (Success) = .518 P (Success) = .491 What went wrong before?**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…**1/36*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.**• 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**• Take turns rolling 2 dice. • Keep rolling: Add it up. • 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”**from: Introduction to Information Theory by Larry Yaeger**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**Basic Models of Probability**by Ron S. Kenett, Weizmann Institute of Science P(“1” with 2 dice) =? To add or to multiply ?**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 blackdie? 6/36 or 1/6 What 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?