## Peirce’s 1870 “Logic of Relatives” • Comment 11.22

### Peirce’s 1870 “Logic of Relatives” • Comment 11.22

Let’s look at that last example from a different angle.

#### NOF 4.3

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)

Viewed in various lights the formula $[\mathrm{m,}\mathrm{b}] = [\mathrm{m,}][\mathrm{b}]$ presents itself as an aimed arrow, fair sampling, or stochastic independence condition.

Peirce’s example assumes a universe of things in general encompassing the denotations of the absolute terms $\mathrm{m} = \text{man}$ and $\mathrm{b} = \text{black}.$  That allows us to illustrate the case in relief, by returning to our earlier staging of Othello and examining the premiss that “men are just as apt to be black as things in general” within the frame of that empirical if fictional universe of discourse.

We have the following data.

$\begin{array}{*{15}{l}} \mathrm{b} & = & \mathrm{O} \\[6pt] \mathrm{m} & = & \mathrm{C} & +\!\!, & \mathrm{I} & +\!\!, & \mathrm{J} & +\!\!, & \mathrm{O} \\[6pt] \mathbf{1} & = & \mathrm{B} & +\!\!, & \mathrm{C} & +\!\!, & \mathrm{D} & +\!\!, & \mathrm{E} & +\!\!, & \mathrm{I} & +\!\!, & \mathrm{J} & +\!\!, & \mathrm{O} \\[12pt] \mathrm{b,} & = & \mathrm{O\!:\!O} \\[6pt] \mathrm{m,} & = & \mathrm{C\!:\!C} & +\!\!, & \mathrm{I\!:\!I} & +\!\!, & \mathrm{J\!:\!J} & +\!\!, & \mathrm{O\!:\!O} \\[6pt] \mathbf{1,} & = & \mathrm{B\!:\!B} & +\!\!, & \mathrm{C\!:\!C} & +\!\!, & \mathrm{D\!:\!D} & +\!\!, & \mathrm{E\!:\!E} & +\!\!, & \mathrm{I\!:\!I} & +\!\!, & \mathrm{J\!:\!J} & +\!\!, & \mathrm{O\!:\!O} \end{array}$

The fair sampling condition amounts to saying men are just as likely to be black as things in general are likely to be black.  In other words, men are a fair sample of things in general with respect to the predicate of being black.

On that condition the following equation holds.

$[\mathrm{m,}\mathrm{b}] ~=~ [\mathrm{m,}][\mathrm{b}].$

Assuming $[\mathrm{b}]$ is not zero, the next equation follows.

$[\mathrm{m,}] ~=~ \displaystyle{[\mathrm{m,}\mathrm{b}] \over [\mathrm{b}]}.$

As before, it is convenient to represent the absolute term $\mathrm{b} = \text{black}$ by means of the corresponding idempotent term $\mathrm{b,} = \text{black that is}\,\underline{~~~~}.$

Let is next consider the bigraph for the following relational product.

$\mathrm{m,}\mathrm{b} ~=~ \text{man that is black}.$

We may represent that in the following equivalent form.

$\mathrm{m,}\mathrm{b,} ~=~ \text{man that is black that is}\,\underline{~~~~}.$

$\text{Figure 53. Bigraph Product}~ M,B,$

The facts of the matter in the Othello case are such that the following formula holds.

$\mathrm{m,}\mathrm{b} ~=~ \mathrm{b}.$

And that in turn is equivalent to each of the following statements.

$\begin{matrix} m \land b = b \\[6pt] \mathrm{b} \implies \mathrm{m} \\[6pt] \mathrm{b} ~-\!\!\!< \mathrm{m} \end{matrix}$

Those last implications puncture any notion of statistical independence for $\mathrm{b}$ and $\mathrm{m}$ in the universe of discourse at hand but it will repay us to explore the details of the case a little further.  Putting all the general formulas and particular facts together, we arrive at the following summation of the situation in the Othello case.

If the fair sampling condition were true, it would have the following consequence.

$\displaystyle [\mathrm{m,}] ~=~ {[\mathrm{m,}\mathrm{b}] \over [\mathrm{b}]} ~=~ {[\mathrm{b}] \over [\mathrm{b}]} ~=~ \mathfrak{1}.$

On the contrary, we have the following fact.

$\displaystyle [\mathrm{m,}] ~=~ {[\mathrm{m,}\mathbf{1}] \over [\mathbf{1}]} ~=~ {[\mathrm{m}] \over [\mathbf{1}]} ~=~ {4 \over 7}.$

In sum, it is not the case in the Othello example that “men are just as apt to be black as things in general”.

Expressed in terms of probabilities:

$\mathrm{P}(\mathrm{m}) = \displaystyle{4 \over 7}$   and   $\mathrm{P}(\mathrm{b}) = \displaystyle{1 \over 7}.$

If these were independent terms, we would have:

$\mathrm{P}(\mathrm{m}\mathrm{b}) = \displaystyle{4 \over 49}.$

In point of fact, however, we have:

$\mathrm{P}(\mathrm{m}\mathrm{b}) = \mathrm{P}(\mathrm{b}) = \displaystyle{1 \over 7}.$

Another way to see it is to observe that:

$\mathrm{P}(\mathrm{b}|\mathrm{m}) = \displaystyle{1 \over 4}$   while   $\mathrm{P}(\mathrm{b}) = \displaystyle{1 \over 7}.$

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