Question

1. Saruman chooses a number of single dollar bills _ call it Z so that 2 P(Z = n)=Jp(1-2)(n/3)-1 when n is a positive multiple of 3 otherwise, where 0 〈 pく1. (Austin Ellis-Mohr suggested parametrizing this problem in terms of Z, which Austin called "the total money in the system." The book does it differently.) Saruman divides the dollar bills into three equal piles and flips a fair coin to decide in which of two envelopes _ call them X and Y _to put one third of the dollar bills and in which to put the other two thirds. He invites Gollum to open envelope X, and Gollum does SO ravenously (a) What is the probability that Gollum finds 7 dollar bills in envelope X? (b) What is the probability that Gollum finds 6 dollar bills in envelope X? (c) Suppose Gollum can choose either to keep what he finds in X or instead take the contents of Y. What choice maximizes Gollum's expected payoff given that he finds an odd number dollars in X? (d) Show that given Gollum finds an even number of dollars _call it m- in X, it is more probable that X contains more than Y than it is that Y contains more than (e) As in part (c), suppose Gollum can choose either to keep what he finds in X or instead take the contents of Y. What choice maximizes Gollum's expected payoff given that he finds an even number of dollars_call it m-in X? Your answer might depend on m and p.

Answer 1

Given that if Saruman chooses Z number of single bills, then z is a multiple of 3 Therefore Exist/3 is an integer. Next Saruman divides this z number of single dollars in 2 piles. Bill and randomly put then in 2 Envelopes X & y Now one of piles contain z/3 number of single. Dollars bills other one contains 2 z/3 number of single dollars bills. Now w z/3 is an integer so 2 z/3 is an even integer If Gollum finds 7 dollars bills in Envelope x Then that means it contains Z/3 number of dollars bills z/3 = 7 or z = 21 P(Z = 21) = P(1 - p)^2/3 - 1 = P(1 - P)^6 so the probability that Gollum finds 7 dollars bills in envelope x is given by P(1 - P)^6/2 \r\n If Gollum finds 6 dollar bills in Envelope X, then z/3 = 6 on Z = 18 or 2z/3 = 6 or z = 9 so z can