18 Multiple Witnessing

One way in which the diminution effect can be countered is by piling up the number of witnesses. Attempts to quantify the effects of multiple witnesses started very early. An especially interesting example is contained in an anonymous essay entitled “A Calculation of the Credibility of Human Testimony” published in the 1699 volume of the Philosophical Transactions of the Royal Society (London).⁷³ Suppose that each of N “concurrent reporters” gives an assurance of a of the arrival of a ship or a (p. 54 ) gift to me of £ 1200. It is asserted that together the witnesses give an “assurance” (probability) of 1 − (1 − a)^N. For (the reasoning goes), the first gives an expectation of a· £ 1200, leaving (1 − a)· £ 1200 unassured. Of what is left unassured, the second witness gives an assurance of a(1 − a)· £ 1200, leaving (1 − a)(1 − a)· £ 1200 unassured, etc. In the end, £ 1200 − (1 − a)^N· £ 1200 remains unassured. Dividing this expectation by £ 1220 gives a probability of 1 − (1 − a)^N. On this analysis, multiple witnessing is very powerful indeed: for no matter how small a is, as long as it is greater than 0, 1 − (1 − a)^N can be made as close to 1 as you like by making N large enough.

Karl Pearson (1978, 467–468) approved of this result, but stated that it can be obtained more simply in the following way. Suppose for simplicity that the witnesses never misperceive but may lie. Then the event in question fails to occur just in case every one of the N witnesses lies. If the witnesses are independent, the probability of such mass cretinism is said to be (1 − a)^N, and thus, by the negation principle, the probability of the event is 1 − (1 − a)^N. This reasoning is seductive but potentially misleading. That each witness gives an assurance of a for the gift G presumably means that Pr(G/t_i(G)&E&K) = a for i = 1, 2, . . . , N, where t_i stands for the testimony of witness i. If the independence of witnesses meant that Pr(¬ G/t₁(G)& . . . &t_N(G)&E&K) = Pr(¬ G/t₁(G)&E&K)x . . . xPr(¬ G/t_N(G)&E&K), then Pearson's result would be secured. But this is an implausible way to express the assumption that the witnesses testify independently of one another. Why not say in the same spirit that the independence means that Pr(G/t₁(G)& . . . &t_N(G)&E&K) = Pr(G/t₁(G)&E&K)x . . . xPr(G/t_N(G)&E&K), reaching the contrary result that the posterior probability of G is equal to a^N? There are some special circumstances under which Pearson's result holds, and Bayesianism reveals what they are. But I will leave it to the reader to reach the revelation by turning the crank on Bayes' theorem, for there is another route to revealing the power of independent witnessing that does not rely on such specialized assumptions.

The effects of multiple testimonies to the same event were given a systematic Bayesian analysis by Charles Babbage in his Ninth Bridgewater Treatise (1838). The claimed upshot of his discussion is this: “[I]f independent witnesses can be found, who speak the truth more frequently than falsehood, it is ALWAYS possible to assign a number of independent witnesses, the improbability of the falsehood of whose concurring testimonies shall be greater than that of the improbability of the miracle itself” (NBT 202: 212). Here Babbage is accepting Hume's Maxim (see section 15) and using it against him. I take the form of Babbage's claim to be this. Suppose that Pr(M/E&K) = ɛ > 0, and suppose that the witnesses are independent and that each one's testimony is more likely to be true than false. Then no matter how small ɛ is (as long as it is positive), there is an N(ɛ) such that Pr(M/t₁(M)&t₂(M)& . . . &t_N(M)&E&K) > 0.5. And, in fact, N(ɛ) can be chosen so that the posterior probability is as close to 1 as is desired. (p. 55 ) This claim is still vague until the suppositions of independence and reliability of the witnesses are given precise probabilistic form.

I will suppose that on the relevant occasion each of the witnesses testifies, either to the occurrence or the nonoccurrence of the event in question so that ¬ t_i(M) is equivalent to t_i(¬ M). I then take the independence of the testimonies to mean that $${\displaystyle \begin{array}{cc}\text{Pr}\hfill & (\pm {\text{t}}_1(\text{M})\&\dots \&\pm {\text{t}}_N(\text{M})/\pm \text{M}\&\text{E}\&\text{K})\hfill \\ \multicolumn{1}{c}{\hspace{1em}}& =\text{Pr}(\pm {\text{t}}_1(\text{M})/\pm \text{M}\&\text{E}\&\text{K})\times \dots \times \text{Pr}(\pm {\text{t}}_N(\text{M})/\pm \text{M}\&\text{E}\&\text{K})\hfill \end{array}}$$(1) where ± Φ stands for the choice of Φ or its negation, the understanding being that the same choices must be made uniformly on the left‐ and right‐hand sides of the equality. (Note that (I) is much weaker than the generally implausible principle that, on the basis of E&K alone, the testimonies are uncorrelated, that is. Pr(± t₁(M)& . . . & ± t_N(M)/E&K) = Pr(± t₁(M)/E&K)x . . . xPr(± t_N(M)/E&K).) For sake of simplicity I also assume that all the witnesses are equally reliable (or unreliable) in that for all i, Pr(t_i(M)/M&E&K) = p and Pr(t_i(M)/¬ M&E&K) = q. Bayes' theorem then gives the posterior probability of the miracle, conditional on the testimony of the cloud of witnesses: $${\displaystyle \text{Pr}({\text{M/t}}_1(\text{M})\&\dots \&{\text{t}}_N(\text{M})\&\text{E}\&\text{K})=\frac{1}{1+\left[\frac{\text{Pr}(\neg \text{M/E}\&\text{K})}{\text{Pr}(\text{M/E}\&\text{K})}\right]{\left(\frac{\text{q}}{\text{p}}\right)}^N}}$$(15) The implications of (15) are best discussed in cases. Case (a). p = q. Then for any value of N, Pr(M/t₁(M)& . . . &t_N(M)&E&K) = Pr(M/E&K). Thus, no matter how large the cloud of witnesses, their collective testimony has no probative value. Case (b). q > p. Then as N → ∞, (q/p)^N → ∞ and Pr(M/t₁(M)& . . . &t_N(M)&E&K) → 0. Piling one unreliable witness on another only serves to reduce the credibility of the event. Case (c). p > q. Then as N → ∞, (q/p)^N → 0 and Pr(M/t₁(M)& . . . &t_N(M)&E&K) → 1. Here the power of independent witnessing comes into its own. What is remarkable about this power in the above set up is that the witnesses do not have to be reliable in any absolute sense; for example, it could be that they are unreliable in the absolute sense that Pr(t_i(M)/¬ M&E&K) > 0.5 for each i. All that is required is that they are minimally reliable in the comparative sense that Pr(t_i(M)/M&E&K) > Pr(t_i(M)/¬ M&E&K).

In the case where the witnesses are not equally reliable, (15) has to be replaced by $${\displaystyle \begin{array}{cc}\text{Pr}\hfill & ({\text{M/t}}_1(\text{M})\&\dots \&{\text{t}}_N(\text{M})\&\text{E}\&\text{K}\hfill \\ \multicolumn{1}{c}{\hspace{1em}}& =\frac{1}{1+\left(\frac{\text{Pr}(\neg \text{M/E}\&\text{K})}{\text{Pr}(\text{M/E}\&\text{K})}\right)\left(\frac{{\text{q}}_1{\text{q}}_2\dots {\text{q}}_N}{{\text{p}}_1{\text{p}}_2\dots {\text{p}}_N}\right)}\hfill \end{array}}$$(16) where p_i = Pr(t_i(M)/M&E&K) and q_i = Pr(t_i(M)/¬ M&E&K). Now in order to assure that the posterior probability goes to 1 as N → ∞ it is not (p. 56 ) sufficient to assume that each witness is minimally reliable in the comparative sense that p_i > q_i. It is also necessary that the ratio q_i/p_i does not approach 1 too rapidly as N increases.

In the Théorie Analytique des Probabilités (1812, 463) Laplace derived a formula similar to (16). However, he seems to have assumed that q_i = 1 − p_i. Specializing back to the case where the p_i are all equal to p, we see that under Laplace's analysis the power of multiple independent witnesses does not materialize unless the witnesses are reliable in the absolute sense that p > 0.5. If one is not careful, it is easy to fall in with Laplace's assumption, which may help to explain why there has been a general lack of recognition of the power of independent multiple witnessing.

Hume made a nod to the power of independent witnessing. In the case of the hypothetical miracle of eight days of total darkness, he writes:

[S]uppose, all authors, in all languages, agree, that, from the first of January 1600, there was total darkness over the whole earth for eight days: Suppose that the tradition of this extraordinary event is still strong and lively among the people: that all travellers, who return from foreign countries, bring us accounts of the same tradition without the least variation or contradiction: It is evident, that our present philosophers, instead of doubting the fact, ought to receive it as certain. (E 127–128; 151)

But in the hypothetical case of the death and subsequent resurrection of Queen Elizabeth, he proclaimed that he would be unmoved by the supposition that her physicians, the whole court, and all of parliament proclaim the events: “I must confess that I should be surprised at the occurrence of so many odd circumstances, but I should not have the least inclination to believe so miraculous an event” (E 128: 151).

Assuming that Hume could be made to accept the form of Babbage's result derived above, would he be able to maintain his intransigence against believing in a resurrection? I will return to this matter in section 20. But first there is more to be said about multiple witnessing.