Why Evidentialism Sucks 
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Ironically enough, mathematics and probability theory pose serious problems for evidentialism. For those whose math is a bit rusty, I recommend reading this quick introduction to Bayes’ theorem before moving forward.
Ready? Good. Here’s some other math stuff you’ll need to know here. E is evidence for hypothesis H if and only if the following is true:
Pr(HE) > Pr(H)Where Pr(H) is the prior probability.[1] E is evidence for hypothesis H if it makes the hypothesis more likely than it would have been without E. To oversimplify it a bit, something being evidence for a hypothesis means it makes the hypothesis more likely to be true.
One form of Bayes’ theorem, when comparing hypotheses, goes like this:
For example, one can use hypothesis H and it’s negation ¬H as follows:
Pr(H1E) Pr(H2E) =
Pr(EH1) Pr(EH2) ×
Pr(H1) Pr(H2)
Notice the importance Pr(EH) and Pr(E¬H) have in changing the likelihood ratio of Pr(HE) to Pr(¬HE). As one might suspect from the equation above, for this to get bigger via evidence E:
Pr(HE) Pr(¬HE) =
Pr(EH) Pr(E¬H) ×
Pr(H) Pr(¬H)
Pr(EH) must be greater than Pr(E¬H), and a bit of math shows that for E to be evidence for hypothesis H, this has to be true:
Pr(HE) Pr(¬HE)
Pr(EH) > Pr(E¬H)[2]Similarly, for something to be evidence against a hypothesis, the following has to be true:
Pr(E¬H) > Pr(EH)[3]These math facts are part of a recipe of doom for evidentialism. Let ES be the set of all memories, sense experiences, and intuitions you now have. Let HB be the hypothesis you are a recently created (say, within the past few years) brain in a vat of chemicals hooked up to a supercomputer that gives you all the memories, sense experiences, and intuitions you now have. Thus, Pr(ESHB) = 1. You are rational to believe that hypothesis HB is false, but here’s the problem: you cannot cite any evidence against it. Thanks to mathematics, for some wouldbe piece of evidence E to be evidence against HB, the following has to be true:
Pr(E¬HB) > Pr(EHB)Hypothesis HB however predicts with 100% certainty any memory, sense experience, and intuition you now have, thus Pr(EHB) = 1, and it is impossible for any evidence E to be such that Pr(E¬HB) > Pr(EHB).
This actually teaches us an important lesson: some things are rational to believe without proof or evidence, e.g. the falsity of hypothesis HB. It also teaches us that evidentialism sucks; if evidentialism were true, we could never reject that crazy braininthevat hypothesis.
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[1] The Pr(HE) > Pr(H) inequality is oversimplifying it a bit; I haven’t mentioned that e.g. the background knowledge for Pr(H) doesn’t already have E (though I later say E is evidence if “it makes the hypothesis more likely than it would have been without E”), because if our background knowledge for Pr(H) included E then Pr(HE) = Pr(E) even when E is evidence for H. Hopefully though the general raisingtheprobabilityofthehypothesis idea is apparent.
[2] This holds for when Pr(E) ≠ 0, which is the situation that interests us (presumably, any evidence E we are interested in is the sort that Pr(E) ≠ 0). For math nerds who are interested in the mathematical proof for this, I will first prove a few lemmas.
Lemma (1): (H ∩ E) and (¬H ∩ E) are disjoint (mutually exclusive). We can show that no element in the universe can be a member of both (H ∩ E) and (¬H ∩ E). Let x be an arbitrary element and let’s suppose x is a member of both (H ∩ E) and (¬H ∩ E). With a bit of math logic, we show that there can’t be any x such that x ∈ (H ∩ E) and x ∈ (¬H ∩ E) by assuming there is such an x and deriving an impossibility, like so:
 x ∈ (H ∩ E) and x ∈ (¬H ∩ E)
 (x ∈ H and x ∈ E) and (x ∈ ¬H and x ∈ E), from (1) and definition of ∩
 x ∈ H and x ∈ E and x ∈ ¬H and x ∈ E, from (2)
 x ∈ H and x ∈ ¬H, from (3)
With this in mind, let ξ be the universal set.
E ∩ ξ = ELemma (2): Pr(HE) = 1 − Pr(¬HE) for Pr(E) ≠ 0.
⇔ E ∩ (H ∪ ¬H) = E
⇔ (E ∩ H) ∪ (E ∩ ¬H) = E
Proof: since (E ∩ H) ∪ (E ∩ ¬H) and are mutually exclusive as proven in Lemma (1), by the rules of probability:
Pr(E ∩ H) + Pr(E ∩ ¬H) = Pr(E)This holds for all Pr(E) ≠ 0. Thus, Pr(HE) = 1 − Pr(¬HE) for Pr(E) ≠ 0.
⇔ Pr(E) × Pr(HE) + Pr(E) × Pr(¬HE) = Pr(E)
⇔ Pr(E) × [Pr(HE) + Pr(¬HE)] = Pr(E)
⇔ 1 × [Pr(HE) + Pr(¬HE)] = 1
⇔ Pr(HE) + Pr(¬HE) = 1
⇔ Pr(HE) = 1 − Pr(¬HE)
Lemma (3): If Pr(HE) > Pr(H), then Pr(¬HE) < Pr(¬H)
Proof: assume for sake of argument that Pr(HE) > Pr(H) is true. We can then see how this leads to Pr(¬HE) < Pr(¬H) using Lemma(2), where Lemma (2) implies that Pr(HE) = 1 − Pr(¬HE). If you’ve understood what’s going on thus far, I’ll assume you’re already aware of the wellknown probability fact that Pr(¬A) = 1 − Pr(A) and that Pr(A) = 1 − Pr(¬A).
Pr(HE) > Pr(H)The components of the odds form of Bayes’ theorem are these:
⇔ 1 − Pr(¬HE) > 1 − Pr(¬H)
⇔ −Pr(¬HE) > −Pr(¬H)
⇔ Pr(¬HE) < Pr(¬H)
posterior odds 
likelihood ratio 
prior odds 

 = 
 × 

Pr(HE) > Pr(H) entails Pr(¬HE) < Pr(¬H). If Pr(HE) > Pr(H) and Pr(¬HE) < Pr(¬H), then the posterior odds are greater than the prior odds, and the only way for this to happen is if the likelihood ratio is greater than one, which requires Pr(EH) > Pr(E¬H). Thus, if Pr(HE) > Pr(H) then Pr(EH) > Pr(E¬H) for all E such that Pr(E) ≠ 0.
[3] To prove this, the task is to show that if Pr(HE) < Pr(H), then Pr(E¬H) > Pr(EH) for all E such that Pr(E) ≠ 0.
One proof for this is quite similar to the one above, and so we can borrow from the same lemmas. Note that Lemma (2) from above entails that Pr(HE) = 1 − Pr(¬HE), for Pr(E) ≠ 0. From this we can construct a new lemma.
Lemma (4): Pr(HE) < Pr(H) entails Pr(¬HE) > Pr(¬H) for Pr(E) ≠ 0.
Pr(HE) < Pr(H)The components of the odds form of Bayes’ theorem are these:
⇔ 1 − Pr(¬HE) < 1 − Pr(¬H)
⇔ −Pr(¬HE) < −Pr(¬H)
⇔ Pr(¬HE) > Pr(¬H)
posterior odds 
likelihood ratio 
prior odds 

 = 
 × 

Pr(HE) < Pr(H) entails Pr(¬HE) > Pr(¬H). If Pr(HE) < Pr(H) and Pr(¬HE) > Pr(¬H), then the posterior odds are less than the prior odds, and the only way for this to happen is if the likelihood ratio is less than one, which requires Pr(E¬H) > Pr(EH). Thus, if Pr(HE) < Pr(H) then Pr(E¬H) > Pr(EH) for all E such that Pr(E) ≠ 0.