Conditional Probability A conditional probability Pr(b| A) is called an a posteriori if event B precedes event A in time. Here are some other examples of a posteriori probabilities The probability it was cloudy this morning, given that it rained in the afternoon The probability that I was initially dealt two queens in Texas No Limit Hold'Em poker, given that I eventually got four-of-a-kind Mathematically, a posteriori probabilities are no different from ordinary probabilities; the distinction is only at a higher, philosophical level. Our only reason for drawing attention to them is to say, "Dont let them rattle you. Let's return to the original problem. The probability that the Halting Problem won their first game, given that they won the series is Pr(B l A). We can compute this using the definition of conditional probability and our earlier tree diagram Pr(B|4)=P(B∩ Pr(a) 1/3+1/18 1/3+1/18+1/9 This answer is suspicious! In the preceding section, we showed that Pr(A B)was also 7/9. Could it be true that Pr(a B)=Pr (b a)in general? Some reflection suggests this is unlikely. For example, the probability that I feel uneasy, given that I was abducted b liens, is pretty large But the probability that I was abducted by aliens, given that I feel uneasy, is rather small Let's work out the general conditions under which Pr(a B)= Pr(b a). By the definition of conditional probability, this equation holds if an only if Pr(A∩B)Pr(A∩B) Pr(B Pr This equation, in turn, holds only if the denominators are equal or the numerator is 0 Pr(B)=Pr(a) The former condition holds in the hockey example; the probability that the Halting Prob- lem wins the series(event A)is equal to the probability that it wins the first game(event B). In fact, both probabilities are 1 /2 1 A Coin problem Suppose you have two coins. One coin is fair; that is, comes up heads with probability 1/ 2 and tails with probability 1/2. The other is a trick coin; it has heads on both sides, and so
6 Conditional Probability A conditional probability Pr (B | A) is called an a posteriori if event B precedes event A in time. Here are some other examples of a posteriori probabilities: • The probability it was cloudy this morning, given that it rained in the afternoon. • The probability that I was initially dealt two queens in Texas No Limit Hold ’Em poker, given that I eventually got fourofakind. Mathematically, a posteriori probabilities are no different from ordinary probabilities; the distinction is only at a higher, philosophical level. Our only reason for drawing attention to them is to say, “Don’t let them rattle you.” Let’s return to the original problem. The probability that the Halting Problem won their first game, given that they won the series is Pr (B | A). We can compute this using the definition of conditional probability and our earlier tree diagram: Pr (B | A) = Pr (B ∩ A) Pr (A) 1/3 + 1/18 = 1/3 + 1/18 + 1/9 7 = 9 This answeris suspicious! In the preceding section, we showed that Pr (A | B) was also 7/9. Could it be true that Pr (A | B) = Pr (B A| ) in general? Some reflection suggests this is unlikely. For example, the probability that I feel uneasy, given that I was abducted by aliens, is pretty large. But the probability that I was abducted by aliens, given that I feel uneasy, is rather small. Let’s work out the general conditions under which Pr (A | B) = Pr (B A| ). By the definition of conditional probability, this equation holds if an only if: Pr (A ∩ B) Pr (A ∩ B) = Pr (B) Pr (A) This equation, in turn, holds only if the denominators are equal or the numerator is 0: Pr (B) = Pr (A) or Pr (A ∩ B) = 0 The former condition holds in the hockey example; the probability that the Halting Problem wins the series (event A) is equal to the probability that it wins the first game (event B). In fact, both probabilities are 1/2. 2.1 A Coin Problem Suppose you have two coins. One coin is fair; that is, comes up heads with probability 1/2 and tails with probability 1/2. The other is a trick coin; it has heads on both sides, and so
Conditional Probability always comes up heads. Now you choose a coin at random so that you're equally likely to pick each one. If you flip the coin you select and get heads, then what is the probability that you flipped the fair coin? This is another a posteriori problem since we want the probability of an event(that the fair coin was chosen) given the outcome of a later event(that heads came up). Intuition may fail us, but the standard four-step method works perfectly well Step 1: Find the Sample Space The sample space is worked out in the tree diagram below l/4 1/2 1/2 fair 1/2 unfa 1/2 event a: event b: ever outcome choice of result probability fair coin? heads? coin flip Step 2: Define Events of Interest Let a be the event that the fair coin was chosen. Let b the event that the result of the flip was heads. The outcomes in each event are marked in the figure. We want to compute Pr(AI B), the probability that the fair coin was chosen, given that the result of the flip was head Step 2: Compute Outcome Probabilities First, we assign probabilities to edges in the tree diagram. Each coin is chosen with prob ability 1/2. If we choose the fair coin, then head and tails each come up with probability 1/2. If we choose the trick coin, then heads comes up with probability 1. By the Product Rule, the probability of an outcome is the product of the probabilities on the correspond ing root-to-leaf path. All of these probabilities are shown in the tree diagram
Conditional Probability 7 always comes up heads. Now you choose a coin at random so that you’re equally likely to pick each one. If you flip the coin you select and get heads, then what is the probability that you flipped the fair coin? This is another a posteriori problem since we want the probability of an event (that the fair coin was chosen) given the outcome of a later event (that heads came up). Intuition may fail us, but the standard fourstep method works perfectly well. Step 1: Find the Sample Space The sample space is worked out in the tree diagram below. fair unfair choice of coin result H flip 1/2 1/2 1/2 1/2 H T 1/2 1/4 1/4 event A: outcome probability fair coin? event B: outcome heads? event chose A B? Step 2: Define Events of Interest Let A be the event that the fair coin was chosen. Let B the event that the result of the flip was heads. The outcomes in each event are marked in the figure. We want to compute Pr(A | B), the probability that the fair coin was chosen, given that the result of the flip was heads. Step 2: Compute Outcome Probabilities First, we assign probabilities to edges in the tree diagram. Each coin is chosen with probability 1/2. If we choose the fair coin, then head and tails each come up with probability 1/2. If we choose the trick coin, then heads comes up with probability 1. By the Product Rule, the probability of an outcome is the product of the probabilities on the corresponding roottoleaf path. All of these probabilities are shown in the tree diagram