Recitation 18 (c)Suppose you know that Mary shuffled the deck and you are about to pick the topmost card. what is the probability that you will see an ace? Solution. If we know Mary shuffled the deck, we know the deck if a full 52-card deck. So, easily, the probability is 5? (d) Suppose you know that Peter shuffled the deck and you are about to pick the topmost card. What is the probability that you will see an ace?(Hint: What is this probability if Peter steals an ace? What if Peter steals a non-ace Solution. Suppose Peter shuffles the deck. Then there are two cases about what card he steals: it's either an ace or a non-ace. let sa be the event that he steals an ace. Since he steals a card at random, we know he steals an ace with probability Pr(Sa)=2 and a non-ace with probability Pr(SA)=52 Now, as before, let A be the event that the topmost card is an ace. If we know what case we are in with respect to the stolen card, it is easy to calculate the probability 小: Pr(AISA=3 and Pr(AISA So, we know the probability in each case and we also know the probability of each case. This rings the bell of the law of total probability Pr(4)=Pr(SA)Pr(4|SA)+P(SA)Pr(A|S4)=是·晶+盏·==是 So the probability that the topmost card is an ace is a2 (e) Anything strange with the answers to parts(c)and( d)? Solution The two answers are identical. In other words whether the deck is miss- ing a card or not does not affect the probability that the topmost card is an ace! How can that be? Here is an explanation. The situation of part(c)is identical to the following what is the probability that the first card is an ace?Ck we pick the top two cards of a well-shuffled 52-card deck; (because the selection of the second card is irrelevant). Similarly, the situation of part(d)is identical to the following we pick the top two cards of a well-shuffled 52-card deck That is the probability that the second card (because the effect of Peter stealing a card at random and shuffling is identical to the effect of us drawing the topmost card). Now the two situations we have just introduced are identical, because the number of pairs where the first card is an ace is equal to the number of pairs where the second card is an ace, for obvious reasons of symmetry� � � � � � Recitation 18 5 (c) Suppose you know that Mary shuffled the deck and you are about to pick the topmost card. What is the probability that you will see an ace? Solution. If we know Mary shuffled the deck, we know the deck if a full 52­card 4 deck. So, easily, the probability is . 52 (d) Suppose you know that Peter shuffled the deck and you are about to pick the topmost card. What is the probability that you will see an ace? (Hint: What is this probability if Peter steals an ace? What if Peter steals a non­ace?) Solution. Suppose Peter shuffles the deck. Then there are two cases about what card he steals: it’s either an ace or a non­ace. Let SA be the event that he steals an ace. Since he steals a card at random, we know he steals an ace with probability 4 48 Pr (SA) = 52 and a non­ace with probability Pr SA = . 52 Now, as before, let A be the event that the topmost card is an ace. If we know what case we are in with respect to the stolen card, it is easy to calculate the probability of A: 3 � � 4 Pr (A | SA) = and Pr A | SA = 51 51 So, we know the probability in each case and we also know the probability of each case. This rings the bell of the law of total probability: 4 3 4 4·(3+48) = 4 Pr (A) = Pr (SA) · Pr (A | SA) + Pr SA Pr A | SA = 52 · 51 + 48 = . 52 · 51 52·51 52 · 4 So the probability that the topmost card is an ace is . 52 (e) Anything strange with the answers to parts (c) and (d)? Solution. The two answers are identical. In other words, whether the deck is miss￾ing a card or not does not affect the probability that the topmost card is an ace! How can that be? Here is an explanation. The situation of part (c) is identical to the following: we pick the top two cards of a well­shuffled 52­card deck; what is the probability that the first card is an ace? (because the selection of the second card is irrelevant). Similarly, the situation of part (d) is identical to the following: we pick the top two cards of a well­shuffled 52­card deck; what is the probability that the second card is an ace? (because the effect of Peter stealing a card at random and shuffling is identical to the effect of us drawing the topmost card). Now the two situations we have just introduced are identical, because the number of pairs where the first card is an ace is equal to the number of pairs where the second card is an ace, for obvious reasons of symmetry