Lecture 38 21 127 Concepts of Math 12 07 2012 Administrivia Final Exam Prep Questions are posted Extra Credit projects Due on Sunday at noon Exam 3 regrades you have until Sunday afternoon to talk to me I will have office hours starting on Sunday plus a review session I ll send out details about that later today Seriously check your grades on CMG to make sure they re right I want to take care of these things now Probability The Lazy Professor Problem In this problem we are actually counting derangements which are permutations with no fixed points There are n Concepts students who just took an exam Brendan doesn t feel like grading so he passes out the exam randomly so that every student gets one exam that they will grade What is the probability that no one receives their own exam back so that everyone is grading a peer s exam Let U be the set of all ways to redistribute the exams Suppose the students are numbered from 1 to n For every k from 1 to n let Sk be the set of ways to redistribute the exams so that Person k definitely receives his her own exam back Note This definition of Sk does not put any conditions on whether or not anybody else gets their own exam back It just says that Person k definitely does Notice now that we want to find U n Sk k 1 because from all of the ways to redistribute the exams we want to remove all of the cases where at least one person gets their own exam back Notice that U n because we are just permuting n distinct objects To find S1 we can just assign Person 1 their own exam and then permute the remaining n 1 amongst the others Thus S1 n 1 This argument also shows that S1 S2 S3 Sn n 1 Notice that there are n of these sets with the same size To find S1 S2 we can just assign Person 1 their own exam and the same with Person 2 Then we can permute the remaining n 2 exams amongst the other people Thus S1 S2 n 2 Again this argument also applies to find the size of the intersection of any two sets Thus S1 S2 S1 S3 Si Sj n 2 Notice that there are n2 of these sets To find S1 S2 S3 we can just assign Persons 1 and 2 and 3 their own exams and then permute the remaining n 3 exams amongst the other people Generalizing this argument to any 3 chosen people tells us that S1 S2 S3 Si Sj Sk n 3 Notice that there are n3 of these sets with the same size In general the number of sets in an intersection tells us its size not which sets we are intersecting We are now ready to apply a general Inclusion Exclusion Principle Following the same idea we used to consider the 1 cases where we removed just 2 or 3 or 4 sets we just want to remove the sets individually then add back their duplicate intersections then remove their triplicate intersections and so on until we have added subtracted the intersection of all of them n U Sk U S1 S2 Sn S1 S2 Sn 1 Sn k 1 S1 S2 S3 Sn 2 Sn 1 Sn S1 S2 S3 S4 1 n S1 S2 Sn Conveniently we know that all of the terms like Si have the same size so we can replace them with their size times the number of them Likewise all of the terms like Si Sj have the same size so we can replace them with their size times the number of them This applies to all of the groups of terms so we have U n k 1 n n n n n 1 n 2 n 3 1 n 0 1 2 3 n n X k n 1 n k k Sk n k 0 Notice that the first term also fits this pattern because U n Sk k 1 n X n 0 n 1 n k k k k 0 n X 1 k 1 k k 0 n 1 We can simplify further n n k k n k n X 1 k k k 0 Since we can pull the n factor out of the sum this now makes finding the probability of this event really Sn easy Let X be the event that no one gets their own exam back so X U k 1 Sk Then U Pr X n n X Sk k 1 U 1 k k 0 n k n n X 1 k k 0 k There we have it Here s an interesting fact lim n n X 1 k k k 0 1 e If you know any relevant calculus this series is just the Taylor Polynomial approximation for the function ex evaluted at x 1 Monty Hall This is a classic probability problem It is explained in full detail in Chapter 1 of the textbook You should read that section In particular it shows us why we need to be very careful with probability problems our intuitions can lead us astray 2 An Infinite Example For every n N consider the sets Fn 12 which are the sets of flip sequences of length n for a fair coin Now consider the union of all of these sets which is the set of all possible flip sequences of finite length F Fn 21 n N Notice that F is countably infinite We proved this before it is the set of all finite binary strings It is a countable union of finite sets We claim that one can define a probability function on F However notice that it cannot be uniform If we want every outcome to have the same probability call it q then we require the sum of q with itself infinitely many times to be 1 This is not possible Note There are several ways to do this Some of them are trivial For instance just make the event H have probability 1 and the rest 0 Or do something similar where Pr H Pr T 21 and the rest are 0 And so on However there are non trivial ways to do this The one I had in mind is this Consider the sets Fn 21 by themselves For a fixed n that space already has a given probability function where every outcome has probability 21n If we keep this value for every outcome then this is too big To fix it we just want to weight each such space so that the sum of all of its outcomes is smaller and in such a way that the sum of all the spaces is 1 My favorite countable sequence that sums to 1 is n X 1 1 2 n 1 so just use that That is consider each outcome in Fn 21 and give it probability 1 4n which is 1 2n divided by 2n Retrospectively we can describe this probability in …