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

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

#### 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)

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

The example apparently assumes a universe of “things in general”, encompassing among other things the denotations of the absolute terms $\mathrm{m} = \text{man}$ and $\mathrm{b} = \text{black}.$ That suggests to me that we might well illustrate this case in relief, by returning to our earlier staging of Othello and seeing how well that universe of dramatic discourse observes the premiss that “men are just as apt to be black as things in general”.

Here are the relevant 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 is tantamount to this:  “Men are just as apt to be black as things in general are apt to be black”. In other words, men are a fair sample of things in general with respect to the factor of being black.

Should this hold, the consequence would be: $[\mathrm{m,}\mathrm{b}] ~=~ [\mathrm{m,}][\mathrm{b}].$

When $[\mathrm{b}]$ is not zero, we obtain the result: $[\mathrm{m,}] ~=~ \displaystyle{[\mathrm{m,}\mathrm{b}] \over [\mathrm{b}]}.$

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

Consider the bigraph for the composition: $\mathrm{m,}\mathrm{b} ~=~ \text{man that is black}.$

This is represented below in the equivalent form: $\mathrm{m,}\mathrm{b,} ~=~ \text{man that is black that is}\,\underline{~~~~}.$

Thus we observe one of the more factitious facts affecting this very special universe of discourse, namely: $\mathrm{m,}\mathrm{b} ~=~ \mathrm{b}.$

This is equivalent to the implication $\mathrm{b} \Rightarrow \mathrm{m}$ that Peirce would have written in the form $\mathrm{b} ~-\!\!\!<~ \mathrm{m}.$

That is enough to puncture any notion that $\mathrm{b}$ and $\mathrm{m}$ are statistically independent, but let us continue to develop the plot a bit more. 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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