## Peirce’s 1870 “Logic Of Relatives” • Comment 11.21

One more example and one more general observation and then we’ll be all caught up with our homework on Peirce’s “number of” function.

#### NOF 4.4

So if men are just as apt to be black as things in general, $[\mathrm{m,}][\mathrm{b}] ~=~ [\mathrm{m,}\mathrm{b}],$

where the difference between $[\mathrm{m}]$ and $[\mathrm{m,}]$ must not be overlooked.

(Peirce, CP 3.76)

The protasis, “men are just as apt to be black as things in general”, is elliptic in structure, and presents us with a potential ambiguity. If we had no further clue to its meaning, it might be read as either of the following:

1. Men are just as apt to be black as things in general are apt to be black.
2. Men are just as apt to be black as men are apt to be things in general.

The second interpretation, if grammatical, is pointless to state, since it equates a proper contingency with an absolute certainty. So I think it is safe to assume the following paraphrase of what Peirce intends:

• Men are just as likely to be black as things in general are likely to be black.

Stated in terms of the conditional probability: $\mathrm{P}(\mathrm{b}|\mathrm{m}) ~=~ \mathrm{P}(\mathrm{b}).$

From the definition of conditional probability: $\mathrm{P}(\mathrm{b}|\mathrm{m}) ~=~ \displaystyle{\mathrm{P}(\mathrm{b}\mathrm{m}) \over \mathrm{P}(\mathrm{m})}.$

Equivalently: $\mathrm{P}(\mathrm{b}\mathrm{m}) ~=~ \mathrm{P}(\mathrm{b}|\mathrm{m})\mathrm{P}(\mathrm{m}).$

Taking everything together, we obtain the following result: $\mathrm{P}(\mathrm{b}\mathrm{m}) ~=~ \mathrm{P}(\mathrm{b}|\mathrm{m})\mathrm{P}(\mathrm{m}) ~=~ \mathrm{P}(\mathrm{b})\mathrm{P}(\mathrm{m}).$

This, of course, is the definition of independent events, as applied to the event of being Black and the event of being a Man. It seems to be the most likely guess that this is the meaning of Peirce’s statement about frequencies: $[\mathrm{m,}\mathrm{b}] ~=~ [\mathrm{m,}][\mathrm{b}].$

The terms of this equation can be normalized to produce the corresponding statement about probabilities: $\mathrm{P}(\mathrm{m}\mathrm{b}) ~=~ \mathrm{P}(\mathrm{m})\mathrm{P}(\mathrm{b}).$

Let’s see if this checks out.

Let $N$ be the number of things in general. In terms of Peirce’s “number of” function, then, we have the equation $[\mathbf{1}] = N.$ On the assumption that $\mathrm{m}$ and $\mathrm{b}$ are associated with independent events, we obtain the following sequence of equations: $\begin{array}{lll} [\mathrm{m,} \mathrm{b}] & = & \mathrm{P}(\mathrm{m}\mathrm{b}) N \\[6pt] & = & \mathrm{P}(\mathrm{m}) \mathrm{P}(\mathrm{b}) N \\[6pt] & = & \mathrm{P}(\mathrm{m}) [\mathrm{b}] \\[6pt] & = & [\mathrm{m,}] [\mathrm{b}]. \end{array}$

As a result, we have to interpret $[\mathrm{m,}]$ = “the average number of men per things in general” as $\mathrm{P}(\mathrm{m})$ = “the probability of a thing in general being a man”. This seems to make sense.

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