Why so negative? Evidence aggregation and armchair philosophy

Abstract

This paper aims to clarify a debate on philosophical method, and to give a probabilistic argument vindicating armchair philosophy under a wide range of plausible assumptions. The use of intuitions by so-called armchair philosophers has been criticized on empirical grounds. The debate between armchair philosophers and their empirical critics would benefit from greater clarity and precision in our understanding of what it takes for intuition-based approaches to philosophy to make sense. This paper discusses a set of rigorous, probability-based tools for determining what we can and cannot learn from intuitions in various conditions. These tools can tell us whether beliefs can be justified by armchair practices, and what empirical findings would have to show to undermine the use of intuitions in philosophy. Using these tools, the paper shows that armchair philosophy makes sense in a broad range of situations, and that it is quite plausible that we are in those situations at the moment.

Appendices

Appendix A: Aggregating independent intuitions

In this section I show that, if intuitions are independent, the absolute sizes of the groups intuiting some proposition and its negation are irrelevant to the epistemic probability one should assign to the proposition. What matters instead is the difference in size between these two groups.

If we have some proposition $P$, and some evidence $E$ which consists of intuitions for and against $P$, we can calculate the credence we should assign to $P$ via Bayes’ theorem:

$$\begin{aligned} \Pr \left( {P|E} \right) =\frac{\Pr \left( P \right) \Pr (E|P)}{\Pr \left( P \right) \Pr \left( {E|P} \right) +\Pr \left( {\sim }P \right) \Pr (E|{\sim }P)} \end{aligned}$$

(1)

Let’s define some useful variables:

$f$ :

is the number of people who intuit that $P$; by stipulation, $P$ will always be the proposition found intuitive by the majority.

$a$ :

is the number of people who find ${\sim }P$ intuitive. I will assume no one is agnostic.

$r$ :

is the on-average reliability of the relevant body of intuitions.

To calculate $\Pr (P{\vert }E)$ via Bayes’ theorem, we need to calculate $\Pr (E{\vert }P)$ and $\Pr (E{\vert } {\sim }P)$. $\Pr (E{\vert }P)$ is the probability that $f$ people find $P$ intuitive and $a$ people find it counter-intuitive, given that $P$ is true. Given that $P$ is true, $f$ people can intuit that $P$ only if they get things right, and $a$ can intuit that ${\sim }P$ only if they get things wrong. For each intuitor, the probability of getting things right is $r$ and the probability of getting things wrong is $1-r$. Because we are assuming independence, the probability that $f$ people get things right and $a$ get things wrong is $r^{f}(1-r)^{a}$, times the number of combinations of people that can result in this split of intuitions. Note that this last factor will appear in $\hbox {Pr}(E{\vert } {\sim }P)$, and so can be cancelled out of the equation.

To find $\Pr (E{\vert } {\sim }P)$, we instead multiply $(1-r)^{f}$ times $r^{a}$, again times the number of possible combinations (which cancels out).

Putting this all together, we get:

$$\begin{aligned} \Pr \left( {P|E} \right) =\frac{\Pr \left( P \right) r^{f}(1-r)^{a}}{\Pr \left( P \right) r^{f}(1-r)^{a}+\Pr \left( {\sim }P \right) r^{a}(1-r)^{f}} \end{aligned}$$

(2)

We can factor $r^{a}$ and ($1-r)^{a}$ out of both the numerator and denominator (since we are assuming $f > a)$, and get the following:

$$\begin{aligned} \Pr \left( {P |E} \right) =\frac{\Pr \left( P \right) r^{f-a}}{\Pr \left( P \right) r^{f-a}+\Pr \left( {\sim }P \right) (1-r)^{f-a}} \end{aligned}$$

(3)

Here we see that only the difference between the number intuiting that $P$ and that $\sim P$, but not the size of either group, matters.

Appendix B: Implementing the narrowing down approach

This section will discuss how to implement the narrowing down approach, assuming that there is no possible causal dependence between intuitions.

For any proposition $P$ and intuitions $I_{1}\ldots I_{n}$, which are intuitions that $P$ or intuitions that ${\sim }P$, by Bayes’ theorem:

$$\begin{aligned} \Pr \left( {P|I_1 \ldots I_n } \right) =\frac{\Pr \left( P \right) \Pr \left( {I_1 \ldots I_n |P} \right) }{\Pr \left( P \right) \Pr \left( {I_1 \ldots I_n |P} \right) +\Pr \left( {\sim }P \right) \Pr (I_1 \ldots I_n |{\sim }P)} \end{aligned}$$

(4)

$\Pr (P)$ and $\Pr ({\sim }P$) are the prior probabilities of $P$ and ${\sim }P$, and I’ll assume that these are known (and rational). So what needs to be calculated are $\Pr (I_{1}{\ldots }I_{n}{\vert }P)$ and $\Pr (I_{1}{\ldots }I_{n}{\vert } {\sim }P)$.

Let’s start with $Pr(I_{1}\ldots I_{n}{\vert } P)$. Since $I_{1} {\ldots }I_{n}$ is a conjunction of intuitions, $\Pr (I_{1}{\ldots }I_{n}{\vert } P)$ is the product of the probability of each intuition in the set given $P$; because these intuitions are not independent, the probability of each intuition in the set must be calculated given all intuitions before it (by the multiplication rule). Formally,

$$ \begin{aligned} \Pr \left( {I_1 \ldots I_n |P} \right) =\mathop \prod \limits _{i=1}^n \Pr (I_i |P \& \left[ {I_1 \ldots I_{i-1} } \right] ) \end{aligned}$$

(5)

$I_{1}{\ldots }I _{i-1}$ is in brackets because, when $i=1$, no term should be substituted in that place.

How do we calculate $\Pr (I_{i}{\vert }P \& I_{1} {\ldots }I_{i-1})$ for each intuition in the set? We don’t know precisely what reliability value each intuition in the group has. However, we do know all the possible reliability values each could have. If the possible reliability values are $r_{1} {\ldots } r_{m}$, the probability of any given intuition occurring, given $P$, is the probability of it occurring and $r_{1}$ obtaining, or it occurring and $r_{2}$ obtaining, or it occurring and $r_{3}$ obtaining, and so forth. Formally,

$$ \begin{aligned} \Pr \left( {I_i |P \& \left[ {I_1 \ldots I_{i-1} } \right] } \right) \approx \mathop \sum \limits _{j=1}^{j=m} \Pr (r_j |P \& \left[ {I_1 \ldots I_{i-1} } \right] )\Pr (I_i |P \& r_j \& \left[ {I_1 \ldots I_{i-1} } \right] )\nonumber \\ \end{aligned}$$

(6)

It should be noted that reliability can take any value from 0 to 1, so there are uncountable possible reliability values and they cannot actually be listed as a series $r_{1} {\ldots } r_{m}$. To be really precise, Eq. 6 should be an integral, rather than a summation. However, since in practice we want a numeric value for $\Pr (I_{i}{\vert }P \& I_{1}{\ldots }I_{i-1})$, we must use some method for approximating the value of this integral. Treating $r$ as a discrete variable and summing the results for each possible value (as in Eq. 6) is one such method. To generate the values discussed in the main text, I treated $r$ as a discrete variable that could take one of 10,000 values between 0 and 1 (this required a computer to do the math for me). This should give us a very close approximation of the value of $\Pr (I_{i}{\vert }P \& I_{1}{\ldots }I_{i-1})$.

We now have two terms to calculate: $\Pr (I_{i}{\vert }P \& r_{j} \& I_{1}{\ldots }I_{i-1})$ and $\Pr (r_{j}{\vert }P \& I_{1}{\ldots }I_{i-1})$. We can simplify $\Pr (I_{i}{\vert }P \& r_{j} \& I_{1}{\ldots }I_{i-1})$. We know the content of the intuition $I_{i}$, and are given a reliability value and the truth of the relevant $P$. With these givens, $I_{1}{\ldots }I_{i-1}$ make no difference to the probability of $I_{i}$. So

$$ \begin{aligned} \Pr \left( {I_i |P \& r_j \& \left[ {I_1 \ldots I_{i-1} } \right] } \right) =\Pr \left( {I_i{\vert } P \& r_j } \right) \end{aligned}$$

(7)

We’ll discuss how to calculate this at the end of the section (following Eq. 19).

How do we calculate $\Pr (r_{j}{\vert }P \& I_{1}{\ldots }I_{i-1})$? By Bayes’ theorem:

$$ \begin{aligned} \mathrm{Pr}(r_j |P \& I_1 \ldots I_{i-1} )=\frac{\mathrm{Pr}(r_j )\mathrm{Pr}(P \& I_1 \ldots I_{i-1} |r_j )}{\mathrm{Pr}(P \& I_1 \ldots I_{i-1} )} \end{aligned}$$

(8)

$\Pr (r_{j})$ is the prior probability of $r_{j}$, and we’ll assume it is known and rational. (In my calculations in the main text, I assumed that the prior probabilities of reliabilities are normally distributed, and used a programming library by L’Ecuyer (2012) to calculate these priors) By the multiplication rule:

$$ \begin{aligned} \mathrm{Pr}(P \& I_1 \ldots I_{i-1} |r_j )=\Pr (P|r_j )\Pr (I_1 \ldots I_{i-1} |P \& r_j ) \end{aligned}$$

(9)

Being given $r_{j}$ doesn’t by itself affect $\Pr (P)$, so $\Pr (P{\vert }r_{j}) = \Pr (P)$. Substituting this into Eq. 9, and Eq. 9 into Eq. 8, we get:

$$ \begin{aligned} \mathrm{Pr}(r_j |P \& I_1 \ldots I_{i-1} )=\frac{\mathrm{Pr}(r_j )\Pr (P)\Pr (I_1 \ldots I_{i-1} |P \& r_j )}{\mathrm{Pr}(P \& I_1 \ldots I_{i-1} )} \end{aligned}$$

(10)

For a moment, I want to focus on the denominator here. By the multiplication rule:

$$ \begin{aligned} \mathrm{Pr}\left( {P \& I_1 \ldots I_{i-1} } \right) =\Pr \left( P \right) \Pr (I_1 \ldots I_{i-1} |P) \end{aligned}$$

(11)

We can substitute this into Eq. 10, and cancel the $\Pr (P)$ in the numerator and denominator, getting:

$$ \begin{aligned} \mathrm{Pr}(r_j |P \& I_1 \ldots I_{i-1} )=\frac{\mathrm{Pr}(r_j )\Pr (I_1 \ldots I_{i-1} |P \& r_j )}{\Pr (I_1 \ldots I_{i-1} |P)} \end{aligned}$$

(12)

Some of these terms will cancel out with other terms from elsewhere in our equation. So, rather than simplifying or expanding Eq. 12 further, let’s see how what we’ve done fits into what we started out trying to do.

Remember, everything we have done so far has been to calculate $\Pr (I_{1}{\ldots }I_{n}{\vert }P)$, which is a term in the numerator of Bayes’ theorem from Eq. 4. As we saw in Eq. 5, $\Pr (I_{1}{\ldots }I_{n}{\vert } P)$ is equal to the product of the probability of each intuition in $I_{1}{\ldots }I_{n}$, given $P$ (since these intuitions are dependent on each other, each must be calculated given all the prior intuitions). If we have $n$ intuitions, we can also think of think of this as the product of the probabilities of intuitions 1 through $n$-1, times the probability of intuition $n$:

$$ \begin{aligned} \Pr \left( {I_1 \ldots I_n |P} \right) =\Pr \left( {I_1 \ldots I_{n-1} |P} \right) \Pr (I_n |P \& I_1 \ldots I_{n-1} ) \end{aligned}$$

(13)

This will allow us to cancel some terms, as we’ll see in a moment. For now, let’s simplify Eq. 13. Equation 6 shows us how to calculate the probability of a single intuition, and we can substitute it for $\Pr (I_{n}{\vert }P \& I_{1}{\ldots }I_{n-1})$; Eq. 7 shows us how to simplify one of the terms in Eq. 6, giving us:

$$ \begin{aligned} \Pr \left( {I_1 \ldots I_n |P} \right) \approx \Pr \left( {I_1 \ldots I_{n-1} |P} \right) \mathop \sum \limits _{j=1}^{j=m} \left[ {\Pr (r_j |P \& I_1 \ldots I_{n-1} )\Pr (I_n |P \& r_j )} \right] \nonumber \\ \end{aligned}$$

(14)

Equation 12 tells us how to expand $\Pr (r_{j}{\vert }P \& I_{1}{\ldots }I_{n-1})$; substituting that into Eq. 14, we get:

$$ \begin{aligned}&\Pr \left( {I_1 \ldots I_n |P} \right) \approx \nonumber \\&\qquad \ \Pr \left( {I_1 \ldots I_{n-1} |P} \right) \mathop \sum \limits _{j=1}^{j=m} \left[ {\frac{\mathrm{Pr}(r_j )\Pr (I_1 \ldots I_{n-1} |P \& r_j )\Pr (I_n |P \& r_j )}{\mathrm{Pr}(I_1 \ldots I_{n-1} |P)}} \right] \quad \end{aligned}$$

(15)

Let’s focus on the summation in Eq. 15. It sums the value of the given fraction for all possible $r$ values. However, since the probability term in the denominator does not take an $r$ value as given, it does not change as the $r$ values considered vary. So every value in this summation has $\Pr (I_{1}{\ldots }I_{n-1}{\vert }P)$ as a common denominator. Thus, rather than considering this as the summation of values of a fraction, we can consider it as a fraction with a summation in its numerator:

$$ \begin{aligned}&\mathop \sum \limits _{j=1}^{j=m} \frac{\Pr (r_j )\Pr (I_1 \ldots I_{n-1} |P \& r_j )\Pr (I_n |P \& r_j )}{\Pr (I_1 \ldots I_{n-1} |P)}\nonumber \\&\quad =\frac{\mathop \sum \nolimits _{j=1}^{j=m} \left[ {\Pr \left( {r_j } \right) \Pr \left( {I_1 \ldots I_{n-1} |P \& r_j } \right) \Pr (I_n |P \& r_j )} \right] }{\Pr (I_1 \ldots I_{n-1} |P)} \end{aligned}$$

(16)

Substituting into Eq. 15, we get:

$$ \begin{aligned} \Pr \left( {I_1 \ldots I_n |P} \right) \approx \Pr \left( {I_1 \ldots I_{n-1} |P} \right) \frac{\mathop \sum \nolimits _{j=1}^{j=m} \left[ {\Pr \left( {r_j } \right) \Pr \left( {I_1 \ldots I_{n-1} |P \& r_j } \right) \Pr (I_n |P \& r_j )} \right] }{\Pr (I_1 \ldots I_{n-1} |P)}\nonumber \\ \end{aligned}$$

(17)

The denominator of the fraction cancels with the previous term in the equation, leaving us with:

$$ \begin{aligned} \Pr \left( {I_1 \ldots I_n |P} \right) \approx \mathop \sum \limits _{j=1}^{j=m} \left[ {\Pr (r_j )\Pr (I_1 \ldots I_{n-1} |P \& r_j )\Pr (I_n |P \& r_j )} \right] \end{aligned}$$

(18)

This can be slightly simplified: the probability of $I_{1}{\ldots }I_{n-1 }$ (given $P$ and some $r$) times the probability of $I_{n}$ (given $P$ and the same $r$) is just the probability of $I_{1}{\ldots }I_{n}$ (given $P$ and that $r$):

$$ \begin{aligned} \Pr \left( {I_{1} \ldots I_{n} |P} \right) \approx \mathop \sum \limits _{{j = 1}}^{{j = m}} \left[ {\Pr (r_{j} ) {\text {Pr}}(I_{1} \ldots I_{n} |P \& r_{j} )} \right] \end{aligned}$$

(19)

To calculate this, we need to know how to calculate $\Pr (I_{1}...I_{n})$ given the truth of $P$ and some reliability value. This is now relatively easy, because we can now act as if the intuitions are independent of each other. When implementing the narrowing down approach by itself, we are assuming that there is no causal dependence between intuitions. The only dependence relation we are concerned with is informational dependence—we learn something about how reliable intuitions are by observing what pattern of intuitions we find on this topic. If we are given a reliability value, however, then this sort of dependence is no longer a factor. Thus, given $P$ and some reliability value $r_{j}$, we can calculate $\Pr (I_{1}{\ldots }I_n)$ by simply multiplying the probability of each intuition in the set together. The probability of each intuition depends on its content. Given $P$, intuitions that $P$ are correct, and so have a probability of occurring equal to the given reliability value $r_{j}$. Intuitions that ${\sim }P$ have a probability of $(1-r_{j})$.

So, if there are $f$ intuitions for $P$ and $a$ intuitions that ${\sim }P$:²²

$$ \begin{aligned} \Pr \left( {I_1 \ldots I_n |P \& r_j } \right) =r_j^f (1-r_j )^{a} \end{aligned}$$

(20)

Substituting into Eq. 19:

$$\begin{aligned} \Pr \left( {I_1 \ldots I_n |P} \right) \approx \mathop \sum \limits _{j=1}^{j=m} \left[ {\Pr (r_j )r_j^f (1-r_j )^{a}} \right] \end{aligned}$$

(21)

If we look back at Eq. 4, we see that we now only have to calculate the denominator of Bayes’ theorem. The term we need to calculate is $\Pr (I_{1}{\ldots }I_{n}{\vert } {\sim }P$). Almost everything I have just said applies to calculating this term; the only difference is how we calculate the probability of each intuition given some $r$ value and ${\sim }P$. For all the intuitions that $P$, the probability of them occurring given some $r$value and ${\sim }P$ is $1-r$; for intuitions that ${\sim }P$, their probability given some $r$ and ${\sim }P$ is $r$. So

$$\begin{aligned} \Pr \left( {I_1 \ldots I_n |{\sim }P} \right) \approx \mathop \sum \limits _{j=1}^{j=m} \left[ {\Pr (r_j )(1-r_j )^{f}r_j^a } \right] \end{aligned}$$

(22)

Plugging all of this in to Eq. 4, we get:

$$\begin{aligned}&\Pr \left( {P|I_1 \ldots I_n } \right) \approx \nonumber \\&\quad \frac{\Pr \left( P \right) \mathop \sum \nolimits _{j=1}^{j=m} \left[ {\Pr (r_j )r_j^f (1-r_j )^{a}} \right] }{\Pr \left( P \right) \mathop \sum \nolimits _{j=1}^{j=m} \left[ {\Pr (r_j )r_j^f (1-r_j )^{a}} \right] +\Pr \left( {\sim }P \right) \mathop \sum \nolimits _{j=1}^{j=m} \left[ {\Pr (r_j )(1-r_j )^{f}r_j^a } \right] }\quad \quad \end{aligned}$$

(23)

Appendix C: Causal dependence

In this section, we’ll discuss how to deal with possible causal dependence, assuming we have a given reliability value for our set of intuitions.

The calculations for the symmetric and asymmetric dependence approaches are derived almost exactly as above. Instead of narrowing down a reliability value, we consider a range of degrees of dependence; we’ll use $d$ to stand for these. This doesn’t make a big difference to the derivations: for most of the discussion in “Appendix B”, the actual meaning of the $r$ term—that it represents a reliability value—didn’t make a difference to the algebra being done. We can simply swap $d$ for $r$ for most of the derivation without issue, as long as we then add a reliability value as a given.

Where things diverge from the discussion in “Appendix B” substantially is in the final steps in the derivation: Eqs. 20 through 23. This is where we finally calculate the probability of an intuition, or of a series of intuitions. Modifying Eq. 19 so that it is relevant to causal dependence, we get:

$$ \begin{aligned} \Pr \left( {I_1 \ldots I_n |P \& r} \right) \approx \mathop \sum \limits _{j=1}^{j=m} \left[ {\Pr (d_j )\Pr (I_1 \ldots I_n |P \& r \& d_j )} \right] \end{aligned}$$

(24)

Note that $r$ is a given on the left side of the equation. However, $r$ need not be a given in $\Pr (d)$, as $\Pr (d{\vert }r) = Pr(d)$.

The calculation of $\Pr (I_{1}{\ldots }I_{n}{\vert }P \& r \& d)$ varies depending on what notion of dependence we use. Consider first symmetric dependence. For intuitions that $P$, the probability that they obtain is equal to the probability that: one intuits correctly and is not influenced by previous intuitions or one has been influenced by previous intuitions that $P$ but not by previous intuitions that ${\sim }P$. The probability of not being influenced by $n$ intuitions is ($1-d)^{n}$, and the probability of being influenced by at least one out of $n$ intuitions is ($1-(1-d)^{n})$. Thus, the probability of intuiting that $P$, given $r$, $P$, $d$, and previous intuitions, is

$$ \begin{aligned}&\Pr \left( {intuiting\;that\;P|P \& d \& r \& previous\;intuitions} \right) \nonumber \\&\quad =r(1-d)^{all\;prior\;intuitions}\nonumber \\&\quad +(1-(1-d)^{prior\;intuitions\;that\;P}) (1-d)^{prior\;intuitions\;that\;\sim P} \end{aligned}$$

(25)

The probability of intuiting that ${\sim }P$, given $r$, $P$, $d$, and previous intuitions is:

$$ \begin{aligned}&\Pr \left( {intuiting\;that\;{\sim }P|P \& d \& r \& previous\;intuitions} \right) \nonumber \\&\quad =(1-r)\left( {1-d} \right) ^{all\;prior\;intuitions}\nonumber \\&\quad +(\left( {1-d} \right) ^{prior\;intuitions\;that\;P})(1-\left( {1-d} \right) ^{prior\;intuitions\;that\;\sim P}) \end{aligned}$$

(26)

The order we consider intuitions in doesn’t matter, so to simplify our calculations, we can consider all intuitions that $P$ first, and then all intuitions that ${\sim }P$. For intuitions that $P$, this allows us to ignore the possibility of influence by intuitions that ${\sim }P$.

To calculate $\Pr (I_{1}{\ldots }I_{n}{\vert }P \& r \& d_{j})$ let $f$ be the number of intuitions that $P$ and $a$ the number that ${\sim }P$:

$$ \begin{aligned}&\Pr \left( {I_1 \ldots I_n |P \& r \& d_j } \right) \nonumber \\&\quad =\mathop \prod \limits _{i=1}^{i=f} \left[ {r\left( {1-d_j } \right) ^{i-1}+\left( {1-\left( {1-d} \right) ^{i-1}} \right) } \right] \nonumber \\&\quad \mathop \prod \limits _{i=1}^{i=a} \left[ (1-r)\left( {1-d_j } \right) ^{f+i-1}+(1-d)^{f}\left( {1-\left( {1-d} \right) ^{i-1}} \right) \right] \end{aligned}$$

(27)

This can be substituted into Eq. 24.

The calculation for $\Pr (I_{1}{\ldots }I_{n}{\vert } {\sim }P \& r)$ can be derived from what I’ve just said, with minor changes.

For asymmetric dependence, one needs to make changes to Eqs. 25 and 26. On this approach, an intuition can only be influenced by the majority view. The probability that an intuition was or was not influenced is simply based on the degree of asymmetric dependence $d$ and is independent of the size of the majority. Given $P$, for intuitions that agree with the majority, their probability given some degree of asymmetric dependence is the chance that they were correct and were not influenced, or that they were influenced by the majority. For intuitions that disagree with the majority (given $P)$, their probability is the probability that they were incorrect and were not influenced.