Slide 1Slide 2Slide 3Slide 4Slide 5Several years ago Berkeley faced a law suit …Berkeley did a survey of its departments to find out which ones were at faultEvery department was more likely to admit a female than a maleHow can this be ?AnswerSlide 11Try to get a white ballSlide 13Slide 14Slide 15Simpson’s ParadoxDepartment of Transportation requires that each month all airlines report their “on-time record”Different airlines serve different airports with different frequencySlide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Slide 5415-251Great Theoretical Ideas in Computer ScienceforSomeThis Course…Probability RefresherWhat’s a Random Variable?A Random Variable is a real-valued function on a sample space SE[X+Y] = E[X] + E[Y]Probability RefresherWhat does this mean: E[X | A]?E[ X ] = E[ X | A ] Pr[ A ] + E[ X | A ] Pr[ A ]Pr[ A ] = Pr[ A | B ] Pr[ B ] + Pr[ A | B ] Pr[ B ]Is this true:Yes!Similarly:Air Marshal ProblemEvery passenger has an assigned seatThere are n-1 passengers and n seatsBefore the passengers board, an air marshal sits on a random seatWhen a passenger enters the plane, if their assigned seat is taken, they pick a seat at randomWhat is the probability that the last passenger to enter the plane sits in their assigned seat?Several years ago Berkeley faced a law suit … 1. % of male applicants admitted to graduate school was 10%2. % of female applicants admitted to graduate school was 5%Grounds for discrimination? Grounds for discrimination? SUITSUITBerkeley did a survey of its departments to find out which ones were at faultEvery department was more likely to admit a female than a male#of females accepted to department X#of males accepted #of males accepted to department Xto department X#of female #of female applicants to applicants to department Xdepartment X#of male applicants #of male applicants to department Xto department X>>How can this be ?AnswerWomen tended to apply to departments that admit a smaller percentage of their applicantsWomenWomenMenMenDeptDeptAppliedAppliedAcceptedAcceptedAppliedAppliedAcceptedAcceptedAA9999441100BB111199991010totaltotal100100551001001010A single summary statistic (such as an average, or a median) may not summarize the data well !Try to get a white ballChoose one box and pick a random ball from it.Max the chance of getting a white ball…5/11 > 3/7BetterTry to get a white ballBetter6/9 > 9/14 BetterTry to get a white ballBetterBetterTry to get a white ballBetterBetterBetter11/20 < 12/21 !!!Simpson’s ParadoxArises all the time…Be careful when you interpret numbersDepartment of Transportation requires that each month all airlines report their “on-time record”# of on-time flights landing at nation’s 30 busiest airports# of total flights into those airports# of total flights into those airportshttp://www.bts.gov/programs/oai/Different airlines serve different airports with different frequencyAn airline sending most of its planes into fair weather airports will crush an airline flying mostly into foggy airportsIt can even happen that an airline has a better record at each airport, but gets a worse overall rating by this method.Alaska Alaska airlinesairlinesAmerica WestAmerica West% on % on timetime# # flightsflights% on % on timetime# # flightsflightsLALA88.988.955955985.685.6811811PhoenixPhoenix94.894.823323392.192.152555255San San DiegoDiego91.791.723223285.585.5448448SFSF83.183.160560571.371.3449449SeattleSeattle85.885.82142146676.776.7262262OVERALLOVERALL86.786.73773775589.189.172257225Alaska Air beats America West at each airportbut America West has a better overall rating!Geometric Random VariableFlip a coin with probability p of headsLet X = number of times the coin has to be flipped until we get a headsE[X] =E[X | flip 1 is H]Pr[flip 1 is H] + E[X | flip 1 is T]Pr[flip 1 is T]= p + E[X+1](1-p)= p + E[X](1-p) + (1-p)E[X] = 1/pCMULandAll CMU SCS students fly off to space and colonize the moonFaced with the problem that there are more men than women, the authorities impose a new rule:When having kids, stop after you have a girlWill the number of new girls be higher than the number of new boys?What if the rule is: stop after having two girls?Coupon Collector’s ProblemThere are n types of coupons in cereal boxesWant to collect them allOn average, how many cereal boxes do I have to buy to get them all?Random WalksLecture 12 (February 19, 2009)How to walk home drunkNo newideasSolve HWproblemEatWaitWorkWork0.30.30.40.990.01probabilityHungryAbstraction of Student LifeAbstraction of Student LifeExample questions: “What is the probability of reaching goal on string Work,Eat,Work?”No newideasSolve HWproblemEatWaitWorkWork0.30.30.40.990.01Hungry-Simpler:Random Walks on GraphsAt any node, go to one of the neighbors of the node with equal probability-Simpler:Random Walks on GraphsAt any node, go to one of the neighbors of the node with equal probability-Simpler:Random Walks on GraphsAt any node, go to one of the neighbors of the node with equal probability-Simpler:Random Walks on GraphsAt any node, go to one of the neighbors of the node with equal probability-Simpler:Random Walks on GraphsAt any node, go to one of the neighbors of the node with equal probability0 nkRandom Walk on a LineYou go into a casino with $k, and at each time step, you bet $1 on a fair gameYou leave when you are broke or have $nQuestion 1: what is your expected amount of money at time t?Let Xt be a R.V. for the amount of $$$ at time t0 nkRandom Walk on a LineYou go into a casino with $k, and at each time step, you bet $1 on a fair gameYou leave when you are broke or have $nXt = k + 1 + 2 + ... + t, (i is RV for change in your money at time i)So, E[Xt] = kE[i] = 00 nkRandom Walk on a LineYou go into a casino with $k, and at each time step, you bet $1 on a fair gameYou leave when you are broke or have $nQuestion 2: what is the probability that you leave with $n?Random Walk on a LineQuestion 2: what is the probability that you leave with $n?E[Xt] = kE[Xt] = E[Xt| Xt = 0] × Pr(Xt = 0) + E[Xt | Xt = n] × Pr(Xt = n) + E[ Xt | neither] × Pr(neither)As t ∞, Pr(neither) 0, also somethingt < nHence Pr(Xt = n) k/n k = n × Pr(Xt = n) + (somethingt) × Pr(neither)0 nkAnother Way To Look At
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