Lecture 10/20: More on Chances
Probability Theory (Chapter Ten)
Continued from last time:
E. Availability Heuristic
Number of 7-letter words ending in -ing vs. number of words ending in _n_. Because we can think of more words ending in -ing than we can (non -ing) words ending in _n_, we think the former will be more numerous than the latter. This is wrong, however, because all -ing words are _n_ words, so there will be at least as many of the latter.
Another example: who has a better batting average, NY Yankees or Boston Redsox? Many will think of the superstars and forget that the whole team contributes to the overall batting average of the team. The less famous players are not “available” to you, in the sense that you cannot think of them off the top of your head.
Rules of Probability
We write the probability of h, for ‘hypothesis’, as Pr(h). The Pr(h) = the number of outcomes favorable to h over the number of total outcomes; favorable/total.
1. Negation: Pr(~h) = 1 – Pr(h). The probability that a hypothesis is false is equal to 1 minus the probability that h is true. If the Pr(h) = .4, then the Pr(~h) = 1 – .4 = .6.
2. Conjunction with Independence: Pr(h1 & h2) = Pr(h1) x Pr(h2). Given two independent events, the probability of both occurring is figured by conjunction with independence. Independence refers to whether the outcome of one event gives you any information about the outcome of the other event. For example, if you draw a card from a normal deck, put it back and shuffle it, then the outcome of the next draw is independent of the first; both outcomes have a probability of 1/13. However, if you draw a card, keep it out, and draw a second card, then the information from the first event tells you something about the outcome from the second event. The probability of drawing two kings, by drawing a card, putting it back and shuffling, and drawing another, is: Pr (h1 & h2) = Pr(h1) x Pr(h2) = 1/13 x 1/13 = 1/169.
2G. Conjunction in General: To extend the rule to cover events that are not independent, we need the idea of Conditional Probability. This is the probability that something will happen, given that some other thing happen, i.e., dependent on something else happening. If we want the probability of h2, given that h1 happened, we write Pr(h2|h1). For example, we may want to know the probability that we draw a king (h2), given that we just drew the king of diamonds (h1). Conditional probability is figured out by considering the outcomes where both h1 and h2 are true, divided by the total h1 outcomes. The rule for Conjunction in General is: Pr(h1 & h2) = Pr(h1) x Pr(h2|h1). The probability that you draw two kings in a row without replacing the first is 4/52 x 3/51 = 1/221. The probability that you draw a king, given that you’ve just drawn a king, is the conditional probability. It is 3/51, because there are 3 favorable outcomes when you’ve already drawn a king, over 51 total outcomes where you’ve already drawn a king. Conjunction with independence is a special case of conjunction in general.
3. Disjunction with Exclusivity: Pr(h1 or h2) = Pr(h1) + Pr(h2). The probability that one of two mutually exclusive events is the sum of the probability of each. The probability you roll a 5 or an 8 (Jumanji reference!) is Pr(roll a 5) + Pr(roll an 8 ) = 4/36 + 5/36 = 9/36 = 1/4. Pretty decent chances of getting out the jungle.
3G. Disjunction in General: Of course, not all either/or statements are exclusive. Many are inclusive, meaning that it is possible for both to occur. Thus, we need a general formula for figuring out disjunctive probabilities. It is Pr(h1 or h2) = Pr(h1) + Pr(h2) – Pr(h1 & h2). Suppose half the class are male, and half female, and that half are over 19, and half are under or equal to 19. If we want to know the chances that someone is either female or over 19, we figure the Pr(h1) = 2/4, plus the Pr(h2)= 2/4, minus the Pr(h1 & h2) = 1/4. 2/4 + 2/4 – 1/4 = 3/4. So, the probability that someone is either female or over 19 is 3/4. Disjunction with exclusivity is a special case of disjunction in general.
4. At Least: The probability that an event will occur at least once in a series of n independent trials, where n is the number of trials, is 1 – Pr(~h)raised to the nth power. What are the chances of tossing heads at least once in 8 independent flips of a fair coin? Restate the question so that rules 1 and 2 can be used. First, what are the chances that we don’t flip at least one heads? That is 1 – Pr(flip at least one heads). This is the same as saying the probability of flipping 8 tails in a row. That’s Pr(tails) x Pr(tails) x Pr(tails) x Pr(tails) x Pr(tails) x Pr(tails) x Pr(tails) x Pr(tails) = Pr(tails)to the 8th power = 1/256. So, 1 – Pr(flip at least one heads) = 1/256. Those are the chances we don’t flip at least one heads. So, Pr(flip at least one heads) = 255/256. Pretty good chances! So, to calculate ‘At Least x’, you start by asking the chances that you DON’T get at least x: this is 1 – Pr(at least x). This is the same as asking the chances that the alternatives to x happen n times in a row, which is just an application of rule 2 or 2G above, depending on whether it is independent or not. Then, remember to reconvert it to the original question by figuring 1 minus whatever you got.