Psychology 458/558
Judgment and Decision Making
Prof. Bertram Malle
Fall 1995

Lecture 8: Oct 26
Conditional probabilities and Bayes' theorem

1. Probabilities and frequencies:

You would normally expect a probability of 0.167 for each side. Would you rather bet on odd or even numbers, or would you not care?

2. Conditional probabilities

You have a superstitious friend who claims, for example, that the weather always turns bad after she makes plans. To test her belief, you consider the following two events:

G: She schedules a garden party for Saturday
R: It will rain on Saturday

To test her superstition we have to find the conditional probability that it will rain on Saturdays given that she schedules a party for Saturday, p(rain|Sat) and see whether it is greater than the probability that it will rain Saturday, p(rain).

The event we are considering is quite specific: She schedules the party for Saturday and it will rain that day, p(Sat &rain). By itself, this probability does not tell us much. Conditionalizing on the class of events in which she does schedule the party on Saturday allows us to see in how many cases of scheduling a party on Saturday it actually rains. We can then compare this proportion to the proportion of rainy days overall and see whether it rains more often if she she schedules the party than if she doesn't.

That is, we take p(rain & Sat) and divide it by p(Sat) to get the conditional probability p(rain|Sat). If there is no relation between her party scheduling and rain, then the following must hold: At any probability for her scheduling the party on Saturday and any probability that it will rain Saturday, the conditional probability p(rain|Sat) should be exactly the same as p(rain). Why? Remember chi-square tests. Under the condition of independence between rain and her party scheduling, p(rain & Sat) is the product of p(rain) and p(Sat). This product is If we now divide this product p(rain & Sat) by p(Sat) to get p(rain|Sat), we get p(rain) again. In numbers, this is even more obvious. Below you find an example in which there is no correlation between scheduling and rain, so the chi-square is 0, and the conditional probability, p(rain|Sat) must be identical to the base rate, p(rain). Test: .27/.90 = .30, which is the base rate.

The general relation between conditional probability and the joint probability of the two events is as follows:

Plous points out that people have great difficulites distinguishing between the two conditional probabilities ("conditionals"): p(A|B) and p(B|A). In the equation above you see that the two have one element in common: p(A & B). But they differ in the other element: p(B) for p(A|B) and p(A) for p(B|A). The element in which they differ is the base rate of the event on which you conditionalize. The base rates are therefore crucial for distinguishing between the two conditionalss.

Through simple transformation of the formula in the top part of the previous slide, you can get the ratio in the bottom part of the slide that tells you when the two conditionals must be different--namely, when the two base rates are different.

3. Judgments of diagnosticity

These more technical explanations are important for understanding the subtleties of diagnostic judgments, as discussed next.

In medical diagnosis, personality, psychopathology, and technological risk prevention, people rely on signs (symptoms, indicators) to make inferences about an underlying state, such as a disease, a trait, or a technical failure. We often use tests to diagnose or predict the probability that such an underlying state exists. How should we do that, following the rules of probability and statistics?

Let's look at a first example from the domain of personality.

Given that Harold talks to strangers, is he an extravert?

Assumptions:
p(talks) = .85; that is, 85% of all people talk to strangers
p(ext) = .60; that is, 60% of all people are extraverts
p(talks & ext) = .60; that is, 60% of all people are extraverts who talk to strangers

We can now calculate two conditionals. One is p(talks|ext), which is p(talks & ext)/p(ext), and that is .60/.60 = 1. So if you know that someone is extraverted, you know that he or she talks to stragners. But what about the reverse? You know someone who talks to strangers, what's the probability that he or she is an extravert?

This is the inverse conditional to the one before, and it is far from 1.0:
p(ext | talks) = p(ext & talks)/p(talks) = .60/.85 = .71 If you don't know whether someon talks to strangers or not, your diagnostic judgement is .60; if you know the person talks to strangers, your diagnostic judgment rises to .71. But never do you reach 1.0. Why? Because some of the people who talk to strangers are not extraverts.

In medical diagnosis, we find very similar problems with conditionals. We often know the reliability of a test to indicate a "positive" result if the illness exists. Validity studies would point, for example, to a hit rate of 90%. This is the probability that the test shows positive given that the person has the illness, p(t|i). But all tests sometimes indicate a positive result even if the person doesn't have the illness. This is called the test's false alarm rate, and it might be, say, 11%. It is the probability that the test shows positive given that the person does not have the illness, p(t|~i).

Now let's ask the reverse question: If one of your patients shows a positive test result, what is the probability that the patient has the illness? The answer to this question depends crucially on the prevalence of the illness. So let's look at two different illnesses, one with a prevalence of 0.10, one with 0.70.

Note first that the test's hit rate is 9/10 = 90%, and the test's false alarm rate is 10/90 = 11%. Now, given that the patient had a positive test result, what is the probability that he has the illness? The answer is p(i|t) = p(i & t)/p(t) = 9/19 = 0.53. Why so low? Because the illness itself is rare, so the test result must improve our initially low confidence of 0.10. Or think about it another way. The test rarely has a false alarm relatively speaking: only 11% of all healthy people show a positive result. But because most people are healthy, you will still sample a lot of people who show a positive result even though they are healthy. The large number of false alarms "contaminates" your sample of people with positive test results.

Now let's look at a common illness.

We have a test with the same hit rate as before, namely, 63/70 = 90%, and a similar false alarm rate, namely, 3/30 = 10%. Now, given that the patient had a positive test result, what is the probability that he has this common illness? In this case, given that the patient showed a positive test result (she is one out of the 66), the probability that she has the illness is 63/66 = 0.95. This time the estimate is very high because the prevalence of the illness is already high, namely 0.70. Or put another way: The same false alarm as before now samples from a much smaller pool of healthy people, so the absolute number of false alarms is low, thus not contaminating the group of people with positive test results.

Implications

Applications

Bayes theorem: A more formal approach

If we know a test's reliability (or hit rate), i.e., p(t|e), how can we infer the inverse probability, namely, the test's diagnosticity, i.e., p(e|t)? Bayes' theorem, about 200 years old, allows us to do this algebraically if we know the base rate of the event in question, p(i), and the hit rate and false alarm rate of the test, p(t|i) and p(t|~i), respectively. We can derive the theorem ourselves through simple transformations of conditional and joint probabilities: