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314 Decisions involving groups of individuals<br />

In the absence of these conditions a simple average of individuals’<br />

judgments will probably suffice.<br />

Aggregating probability judgments<br />

There are particular problems involved when probabilities need to<br />

be aggregated, as the following example shows. The managers of a<br />

construction company need to determine the probability that a civil<br />

engineering project will be held up by both geological problems and<br />

delays in the supply of equipment (the problems are considered to<br />

be independent). Two members of the management team are asked to<br />

estimate the probabilities and the results are shown in Table 12.4. It can<br />

be seen that the ‘group’ assessment of the probability that both events<br />

will occur differs, depending on how the averaging was carried out.<br />

If we multiply each manager’s probabilities together and then take an<br />

average we arrive at a probability of 0.24. However, if we first average<br />

the managers’ probabilities for the individual events and then multiply<br />

these averages together we obtain a probability of 0.225.<br />

Because of these types of problem a number of alternative procedures<br />

have been suggested for aggregating probabilities. One approach is<br />

to regard one group member’s probability estimate as information<br />

which may cause another member to revise his or her estimate using<br />

Bayes’ theorem. Some of the methods based on this approach (e.g.<br />

Morris 5 and Bordley 6 ) also require an assessment to be made of each<br />

individual’s expertise and all are substantially more complicated than<br />

simple averaging.<br />

Another approach is to take a weighted average of individual probabilities,<br />

using one of the three methods of weighting which we referred to<br />

earlier. However, again there appears to be little evidence that weighting<br />

Table 12.4 – Averaging probabilities<br />

p(geological<br />

problems)<br />

p(equipment<br />

problems) p(both)<br />

Manager 1’s estimates 0.2 0.6 0.2×0.6 = 0.12<br />

Manager 2’s estimates 0.4 0.9 0.4×0.9 = 0.36<br />

Average = 0.24 But:<br />

Average of the estimates 0.3 0.75 0.3×0.75 = 0.225

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