According to the report, recent research has identified and quantified some risk factors for melanoma (risk factors for

*malignant*melanoma might have been more interesting, but let that pass). Well, I suppose that is all very worthy, especially in Australia. But one finding, at least as reported by Yahoo!, just does not stack up:

The research found that 40 per cent of Australians are three times more likely than average to develop melanoma.

Run that past me again....

*Forty*percent of Australians are

*three times*more likely than

*average*to develop melanoma? Which average is that? One can only assume that the average is over the whole Australian population (no other candidate population was mentioned).

Glossing over what is meant by “likely” for the moment, one can assume at least that this “likelihood” is a numerical quantity (capable of being multiplied by three) and in all probability a probability. So let us suppose that the average probability of getting melanoma is

*p*. So what we are saying here is that there is a section of the population, some 40 percent, whose average probability of getting melanoma is 3

*p*. Fine, so let’s call the average probability of a person’s getting melanoma from the remaining 60 percent

*q*. We know that the average probability (of getting melanoma) for the whole population is

*p*, and also that it is the weighted sum of the average probabilities of the two sub-populations. Putting these two facts together into one equation, we get

orp= .4*3p+ .6*q

i.e.p= 1.2p+ .6*q

q= -p/3.

Now most of us assume that probabilities are non-negative real numbers between 0 and 1 (inclusive). We also assume that the probability of a randomly selected person getting melanoma is non-zero. (If it were zero, the whole research project would have been a waste of time and money, wouldn’t it?) However, the above lower-high-school-level calculation shows that, if the average person’s probability of getting melanoma is positive, the average probability for those not in the 40-percent high-risk group is negative!

## Something malignant to hide?

So perhaps Yahoo! or the dermatologists involved could explain to us the significance of a negative probability. One hears about them in some esoteric discussions of quantum mechanics, but these are beyond the understanding or interest of most people. However, one’s probability of getting melanoma (which is, after all, a form of cancer) could be a matter of life and death. And this probability is something that it would pay most of us, at least if we live in Australia (or in the tropics), to try to reduce. So to be given the meaningless so-called information that, unless one is in a particular group, the probability is negative, is less than no use at all. Should I try to make my probability even more negative?

There is of course the possibility that the “average” referred to in the quote above referred to some other population, such as the world population, or the US population, or the population of New York (State or City?), given that the study is said to be “based on research by New York University Medical Centre”, but that hint is a full three paragraphs back, and it doesn’t tell us much anyway. “Average” figures are meaningless (or worse) without some specification of the quantity being averaged, the type of average and the population over which the average is taken, just as percentage figures are equally unhelpful unless it is clear what the basis of the percentage is.

In general, inadequate or inaccurate reporting of any news, but especially scientific news, is worse than no reporting at all. In this case, the reported information is obviously bullshit, but there is no way to work out from the report what, if anything, the study actually found on the subject. Perhaps Yahoo! could find someone who didn’t flunk junior-high-school mathematics to do their scientific reporting. And no, I don’t want the job.

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