Essentially, this result can be viewed as showing how the probability of an effect given a cause, P(A|Bi), relates to the probability of a cause given an effect, P(Bi|A).
USING STATISTICS 1. COMMON SENSE AND STATISTICS A F M Smith Department of Mathematics Imperial College of Science, Technology and Medicine, London
1.1 Introduction
In the course of a television programme in 1991 the Prime Minister of the day, a Mr John Major, said: 'I have never been over impressed by academic qualifications. I know a lot of people with an armful of qualifications and (if a professor is looking in I hope he will forgive me) they are wholly useless - no common sense at all. One must combine intelligence with common sense; common sense is the most important of all'.
Well, that's all right then! No need to study the elements of Probability Theory in order to guide one's uncertainties and help one to take rational decisions. Just cast a common-sensical eye over the data and all will be clear.
Or will it? We shall present two examples - familiar to "useless" academics, perhaps less so to Prime Ministers - which demonstrate rather conclusively that, at least where probabilities are concerned, common-sense can be positively dangerous.
But first we need to recall two fundamental results from elementary probability theory.
1.2 Two Key Results
1.2.1 The Theorem of Total Probability
Suppose that A is an event of interest and B1, B2 ...,Bk are background events which partition A, in the sense that
A = B1 È B2 È ...È Bk and Bi Ç Bj = f, for all i¹j.
Then P(A) = P(A|B1)P(B1) + ...+ P(A|Bk)P(Bk),
where P(A|Bi) denotes the conditional probability of A given Bi.
If C is some further conditioning event, assumed fixed throughout, the result can be extended to give
P(A|C) = P(A|CÇ B1)P(B1|C) +...+ P(A|CÇ Bk)P(Bk|C).
The essential usefulness of these results is in being able to approach the evaluation of P(A) or P(A|C) 'indirectly' by thinking things through conditional on the background events.
1.2.2 Bayes' Theorem
Given an event A and a partition B1, ..., Bk of A, it can be shown, for i = 1, ..., k, that
Essentially, this result can be viewed as showing how the
probability of an effect given a cause, P(A|Bi),
relates to the probability of a cause given an effect, P(Bi|A).
1.3. Examples
1.3.1 Resolving a paradox
The following example provides an instructive illustration of the way in which the formalism of conditional probabilities, as encapsulated in the Theorem of Total Probability, provides a coherent resolution of an otherwise seemingly paradoxical situation.
Suppose that the results of a clinical trial
involving 800 sick patients are as shown in Table 1.3.1, where T,
Tc denote, respectively, that patients did or did not
receive a certain treatment, and R, Rc denote,
respectively, that the patients did or did not recover.
Table 1.3.1 Trial results for all patients
| R | Rc | Total | Recovery Rate | |
| T Tc |
200 160 |
200 240 |
400 400 |
50% 40% |
From a common-sense perspective, it seems clear that the treatment is beneficial, and that, were one to base probability judgements on these reported figures, it would seem reasonable to specify
P(R|T) = 0.5, P(R|Tc ) = 0.4.
where recovery and the receipt of treatment by
individuals are now represented, in an obvious notation, as
events. Suppose now, however, that one became aware of the trial
outcomes for male and female patients separately, and that these
have the summary forms described in Tables 1.3.2 and 1.3.3.
Table 1.3.2 Trial results for male patients
| R | Rc | Total | Recovery Rate | |
| T Tc |
180 70 |
120 30 |
300 100 |
60% 70% |
Table 1.3.3 Trial results for female patients
| R | Rc | Total | Recovery Rate | |
| T Tc |
20 90 |
80 210 |
100 300 |
20% 30% |
From a common-sense perspective, the results surely seem paradoxical. Tables 1.3.2 and 1.3.3 tell us that the treatment is neither beneficial for males nor for females; but Table 1.3.1 tells us that overall it is beneficial! How are we to come to a coherent view in the light of this apparently conflicting evidence?
The seeming paradox is easily resolved by an appeal to the formal logic of probability, in the form of the Theorem of Total Probability. With M, Mc denoting, respectively, the events that a patient is either male or female, were one to base probability judgements on the figures reported in Tables 1.3.2 and 1.3.3, it would seem reasonable to specify
P(R|MÇT) = 0.6, P(R|MÇTc ) = 0.7,
P(R|Mc Ç T) = 0.2, P(R|Mc Ç Tc) = 0.3
To see that these judgements do indeed cohere with those based on Table 1.3.1, we can simply check that
P(R|T) = P(R|MÇT)P(M|T) + P(R|M cÇ T)P(M c|T)
P(R|Tc) = P(R|MÇTc)P(M|Tc) + P(R|M cÇ Tc )P(M c|Tc ),
where
P(M|T) = 0.75, P(M|Tc) = 0.25,
P(Mc|T) = 0.25, P(Mc|Tc) = 0.75
The probability formalism reveals that the seeming paradox has arisen from the confounding of sex with treatment as a consequence of the unbalanced trial design. Is this really part of most people's 'common sense' tool-kit?
1.3.2 Resolving conditional confusion
Everyone is familiar with the fact that doctors perform diagnostic tests in order to determine the presence or absence of a suspected disease.
Some people are also familiar with the fact that the accuracy of such a test can be characterized by its so-called sensitivity and specificity. If the presence/absence of the disease are denoted by D, Dc, respectively, and positive/negative test outcomes are denoted by T, Tc, respectively (where positive indicates presence of the disease), test sensitivity is the true positive rate, P(T|D), and test specificity is the true negative rate, P(Tc|Dc ).
Suppose that you undergo a test with very high sensitivity and specificity - P(T|D) = 0.99 and P(Tc|Dc) = 0.99, say - and the outcome is T, indicating the presence of the disease.
Experience suggests that most people's "common-sense" reaction is now the following. "A test that hardly ever gets it wrong has said that I've got the disease, so it's overwhelmingly likely that I have!" Alas, completely wrong! To know what to believe about the chance of having the disease, given the positive test, we need to evaluate the conditional probability P(D|T), not P(T|D)!
And to do this coherently we need to use the probability formalism provided by Bayes' Theorem:
Note that P(T|Dc) = 1 - P(Tc|Dc) = 0.01. However, an as yet undefined quantity, P(D), has entered the picture. What is P(D)? It is the prevalence rate of the disease in the population (appropriately defined). For illustration, let us suppose that P(D) = 10-5, a one-in-ten-thousand prevalence rate.
What does this imply for P(D|T)? Substituting into Bayes' Theorem, we have
In other words in complete contradiction of the "common-sense"
answer, it is somewhat unlikely (a chance of about one-in-a-hundred)
that the disease is present!
It is true that a positive test can substantially increase one's evaluation of the chances of having the disease from that before the test was taken (recall P(D) = .0001). However, the prevalence rate of the disease is clearly an important factor in the evaluation and most people's "common-sense" fails to recognise this.
1.4 Conclusions
Beware of thinking of these examples as somehow contrived or artificial. The dangers exemplified by the example in section 1.3.1 are always present when some form of response or success rate is being related to "explanatory" factors. How can you be sure you have not left out a vital piece of conditioning? The fallacy revealed in the example in section 1.3.2 - that of confusing the direction of conditioning - underlies a number of disquieting social phenomena, from child abuse diagnosis to the "sus" law.