Introduction to Discrete ProbabilityWhy Probability?TerminologyFinite ProbabilityExampleSlide 6Slide 7Slide 8Dice probabilitySlide 10Combinations of EventsSlide 12Probability of the union of two eventsSlide 14Slide 15Slide 16Slide 17When is gambling worth it?When is lotto worth it?An aside: probability of multiple eventsSlide 21Probability vs. oddsMonty Hall ParadoxWhat’s behind door number three?Slide 25Decision TreeWhat’s behind door number one hundred?BlackjackSlide 29Blackjack tableBlackjack probabilitiesSlide 32Slide 33So always use a single deck, right?Blackjack probabilities: when to holdBuying (blackjack) insuranceSlide 37Blackjack strategyWhy counting cards doesn’t work well…RouletteSlide 41The Roulette tableSlide 43Slide 44Slide 45Quick surveySlide 471Introduction to Discrete ProbabilityRosen, Section 6.1Based on slides byAaron Bloomfield and …2Why Probability?In the real world, we often do not know whether a given proposition is true or false.Probability theory gives us a way to reason about propositions whose truth is uncertain.It is useful in weighting evidence, diagnosing problems, and analyzing situations whose exact details are unknown.3TerminologyExperiment•A repeatable procedure that yields one of a given set of outcomes•Rolling a die, for exampleSample space•The range of outcomes possible•For a die, that would be values 1 to 6Event•One of the sample outcomes that occurred•If you rolled a 4 on the die, the event is the 44Finite ProbabilityThe probability of an event E isp EES( )•Where E is the set of desired events (outcomes)•Where S is the set of all possible events (outcomes)•Note that 0 ≤ |E| ≤ |S|•Thus, the probability will always between 0 and 1•An event that will never happen has probability 0•An event that will always happen has probability 15ExampleSuppose two dice are rolled. The sample space would be 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x x6 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x xp(sum is 11)|S| = 367 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x xp(sum is 11)|S| = 368 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x xp(sum is 11)|S| = 36|E| = 2p EES( ) 23 69Dice probabilityWhat is the probability of getting “snake-eyes” (two 1’s) on two six-sided dice?•Probability of getting a 1 on a 6-sided die is 1/6•Via product rule, probability of getting two 1’s is the probability of getting a 1 AND the probability of getting a second 1•Thus, it’s 1/6 * 1/6 = 1/36What is the probability of getting a 7 by rolling two dice?•There are six combinations that can yield 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)•Thus, |E| = 6, |S| = 36, P(E) = 6/36 = 1/610Suppose a lottery randomly selects 6 numbers from 40. What is the probability that you selected the correct six numbers? Order is not important.478 39212|E| = 1|S| = C(40,6)p(E) =1C(40,6)13 8 3 8 3 8 0, ,11Combinations of EventsLet E be an event in a sample space S. The probability of the event E, the complementary event of E, is given by p(E) = 1 - p(E)12Combinations of EventsLet E1 and E2 be events in the sample space S. Thenp E E p E p E p E E( ) ( ) ( ) ( )1 2 1 2 1 2   13Probability of the union of two eventsSE1E2p(E1 U E2)14ExampleSuppose a red die and a blue die are rolled. The sample space would be 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x x15p(sum is 7 or blue die is 3)|S| = 36 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x x16p(sum is 7 or blue die is 3)|S| = 36 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x x|sum is 7| = 6|blue die is 3| = 6| in intersection | = 1p(sum is 7 or blue die is 3) =6/36 + 6/36 - 1/36 = 11/3617Probability of the union of two eventsIf you choose a number between 1 and 100, what is the probability that it is divisible by 2 or 5 or both?Let n be the number chosen•p(2|n) = 50/100 (all the even numbers)•p(5|n) = 20/100•p(2|n) and p(5|n) = p(10|n) = 10/100•p(2|n) or p(5|n) = p(2|n) + p(5|n) - p(10|n) = 50/100 + 20/100 – 10/100 = 3/518When is gambling worth it?This is a sta tistical analysis, not a moral/ethical discussionWhat if you gamble $1, and have a ½ probability to win $10?•If you play 100 times, you will win (on average) 50 of those times•Each play costs $1, each win yields $10•For $100 spent, you win (on average) $500•Average win is $5 (or $10 * ½) per play for every $1 spentWhat if you gamble $1 and have a 1/100 probability to win $10?•If you play 100 times, you will win (on average) 1 of those times•Each play costs $1, each win yields $10•For $100 spent, you win (on average) $10•Average win is $0.10 (or $10 * 1/100) for every $1 spentOne way to determine if gambling is worth it:•probability of winning * payout ≥ amount spent•Or p(winning) * payout ≥ investment•Of course, this is a statistical measure19When is lotto worth it?Many lotto games you have to choose 6 numbers from 1 to 48•Total possible choices is C(48,6) = 12,271,512•Total possible winning numbers is C(6,6) = 1•Probability of winning is 0.0000000814•Or 1 in 12.3 millionIf you invest $1 per ticket, it is only statistically worth it if the payout is > $12.3 million•As, on the “average” you will only make money that way•Of course, “average” will require trillions of lotto plays…20An aside: probability of multiple eventsAssume you have a 5/6 chance for an event to happen•Rolling a 1-5 on a die, for exampleWhat’s the chance of that event happening twice in a row?Cases:•Event happening neither time: 1/6 * 1/6 = 1/36•Event happening first time: 1/6 * 5/6 = 5/36•Event happening second time: 5/6 * 1/6 = 5/36•Event happening both times: …