Recitation 18 Problem 3. There is a deck of cards on the table. Either John or Mary shuffled it and we have no reason to believe in one case more than the other. Now John is a well-known cheater with well-known preferences: he always steals the ace of diamonds while shuf fling. Mary, on the other hand, is a very honest girl: a deck suffled by her is always a full card deck do any calculations: Who is more likely to have shuffled the deck ExPlain (a)You pick the topmost card on the deck and you see a queen of hearts. Before you Solution. A shuffling by John strictly increases the fraction of non-aces in the deck Hence, between the two worlds (1) the world where John has shuffled the deck and (2) the world where Mary has shuffled the deck it is the first world rather than the second one that favors the event of the topmost card being a non-ace. Since we know this event is a fact and the two worlds are otherwise equally likely, we should bet we live in world (1) (b)Now calculate What is the probability that John has shuffled the deck? What is the probability that it has been Mary? Solution. Let J be the event that John shuffles the deck and A that the topmost card is an ace. We want the probabilities Pr(J|A) and Pr(M| A) J and therefore Pr(M A)= Pr(J| A)=1-Pr(J| A), so that we need only calculate the probability about John. By the definition of conditional probability(first equation) and then the product rule(on the enumerator)and the law of total probability(on the denominator), we know Pr(J∩A) Pr(J)- Pr(AlJ) Pr(④)Pr(J)·Pr(A|J+Pr(M)Pr(A|M) and everything in this last fraction is known (J|A)=2=15 52+51103 which is(slightly, but) strictly greater than 2, as expected Like John, Peter is also a well-known cheater: when he shuffles the deck, he also steals a card from it; but (unlike John) he steals a random card. That is, every card is equally likely to be stolen when Peter is shuffling� � � � � � � � � � � � � � � � Recitation 18 4 Problem 3. There is a deck of cards on the table. Either John or Mary shuffled it and we have no reason to believe in one case more than the other. Now, John is a well­known cheater with well­known preferences: he always steals the ace of diamonds while shuf­ fling. Mary, on the other hand, is a very honest girl: a deck suffled by her is always a full 52­card deck. (a) You pick the topmost card on the deck and you see a queen of hearts. Before you do any calculations: Who is more likely to have shuffled the deck? Explain. Solution. A shuffling by John strictly increases the fraction of non­aces in the deck. Hence, between the two worlds: (1) the world where John has shuffled the deck and (2) the world where Mary has shuffled the deck it is the first world rather than the second one that favors the event of the topmost card being a non­ace. Since we know this event is a fact and the two worlds are otherwise equally likely, we should bet we live in world (1). (b) Now calculate. What is the probability that John has shuffled the deck? What is the probability that it has been Mary? Solution. Let J be the event that John shuffles the deck and A that the topmost card is an ace. We want the probabilities Pr J | A and Pr M | A Clearly, M = J and therefore Pr M | A = Pr J | A = 1 − Pr J A | , so that we need only calculate the probability about John. By the definition of conditional probability (first equation) and then the product rule (on the enumerator) and the law of total probability (on the denominator), we know � � � Pr (J) � · Pr A | J Pr J | A = Pr J � ∩ � A = Pr (J) Pr A | J + Pr (M) Pr � A M � Pr A · · | and everything in this last fraction is known: 1 48 1 51 1 51 + 1 52 52 52 = = 52 + 51 103 51 · 2 Pr J | A = 1 48 1 48 = · + · 2 51 2 52 1 which 2 is (slightly, but) strictly greater than , as expected. Like John, Peter is also a well­known cheater: when he shuffles the deck, he also steals a card from it; but (unlike John) he steals a random card. That is, every card is equally likely to be stolen when Peter is shuffling