# Bayesian reasoners shouldn’t believe logical arguments

Advance warning: If you’re familiar with Bayesian reasoning it is unlikely this post contains anything new unless I’m making novel mistakes, which is totally possible.

Let me present you with a hypothetical, abstracted, argument:

Me: C
You: Not C!
Me: B?
You: *shrug*
Me: A?
You: Yes
Me: A implies B?
You: Yes?
Me: B implies C?
You: … Yes
Me: Therefore C?
You: C. :-(

Does this seem like a likely scenario to you? We have had a disagreement. I have presented a logical argument from shared premises for my side of the disagreement. You have accepted that argument and changed your position.

Yeah, it sounds pretty implausible to me too. A more likely response from you at the end is:

You: Not C!

I will of course find this highly irrational and be irritated by your response.

…unless you’re a Bayesian reasoner, in which case you are behaving entirely correctly, and I’ll give you a free pass.

Wait, what?

Suppose you have propositions $$A$$ and $$B$$, which you believe with probabilities $$a$$ and $$b$$ respectively. You currently believe these two to be independent, so $$P(A \wedge B) = ab$$

Now, suppose I come along and convince you that $$A \implies B$$ is true (I’ll call this proposition $$I$$). What is your new probability for k$$B$$?

Well, by Bayes rule, $$P(B|I) = \frac{P(B \wedge I)}{P(I)} = P(B) \frac{P(I|B)}{P(I)}$$

$$I = A \implies B = \neg\left( A \wedge \neg B\right)$$. So $$P(I) = 1 – a(1 – b)$$.

$$P(I|B) = 1$$ because everything implies a true proposition. Therefore $$P(B|I) = \frac{b}{(1 – a(1 – b))}$$.

This is a slightly gross formula. Note however it does have the obviously desirable property that your believe in B goes up, or at least stays the same. Lets quickly check it with some numbers.

$$a$$ $$b$$ $$P(B | I)$$
0.100 0.100 0.110
0.100 0.500 0.526
0.100 0.900 0.909
0.500 0.100 0.182
0.500 0.500 0.667
0.500 0.900 0.947
0.900 0.100 0.526
0.900 0.500 0.909
0.900 0.900 0.989

These look pretty plausible. Our beliefs do not seem to change to an unrealistic degree, but we have provided significant evidence in favour of $$B$$.

But as a good Bayesian reasoner, you shouldn’t assign probabilities 0 or 1 to things. Certainty is poisonous to good probability updates. So when I came along and convinced you that $$A \implies B$$, you really shouldn’t have believed me completely. Instead you should have assigned some probability $$r$$ to it. So what happens now?

Well we know what the probability of $$B$$ given $$I$$ is, but what is the probability given $$\neg I$$? Well $$\neg I = \neg (A \implies B) = A \wedge \neg B$$, so $$P(B|\neg I) = 0$$. The implication can only be false if $$B$$ is (because everything implies a true statement).

This means that your posterior probability for $$B$$ should be $$r P(B|I)$$. So $$r$$ is essentially a factor slowing your update process.

Note that because my posterior belief in B is $$b \frac{r}{P(I)}$$, as long as my claim that $$A \implies B$$ is at least as convincing as my prior belief in it, my argument will increase your belief in it.

Now. Lets suppose that you are in fact entirely convinced before hand that $$A$$ and that $$\neg B$$, and my argument entirely convinces you that $$A \implies B$$.

Of course, we don’t believe in certainty. Things you are entirely convinced of may prove to be false. Suppose now that in the past you have noticed that when you’re entirely convinced of something, you’re right with about probability $$p$$. Lets be over-optimistic and say that $$p$$ is somewhere in the 0.9 range.

What should your posterior probability for $$B$$ now be? We have $$b = 1 – p$$ and $$a = r = p$$. Then your posterior probability for $$B$$ is $$r P(B | I) = p \frac{1 – p}{(1 – p(1 – (1 – p)))} = p \frac{1 – p}{1 – p^2} = \frac{p}{p+1} = 1 – \frac{1}{p+1}$$.

You know what the interesting thing about this is? The interesting thing is that it’s always less than half. A perfectly convincing argument that a thing I completely believe in implies a thing I completely disbelieve in should never do more than create a state of complete uncertainty in your mind.

It turns out that reasonable degrees of certainty get pretty close to that too. If you’re right about things you’re certain about with probability 0.9 then your posterior probability for $$B$$ should be 0.47. If you’re only right with probability 0.7 then it should be $$0.41$$. Of course, if you’re only right about that often then $$0.41$$ isn’t all that far from your threshold for certainty in the negative result.

In conclusion: If you believe A and not B, and I convince you that A implies B, you should not now go away and believe B. Instead you should be confused, with a bias towards still assuming not B, until you’ve resolved this.

Now, lets go one step further to our original example. We are instead arguing about $$C$$, and my argument proceeds via an intermediary $$B$$. Your prior is that $$A$$, $$B$$ and $$C$$ are all independent. You are certain that $$A$$, certain that $$\neg C$$ and have no opinion on $$B$$ (i.e. you believe it with probability $$\frac{1}{2}$$.

I now provide you with a p-convincing argument that $$A \implies B$$. What is your posterior probability for $$B$$?

Well, plugging it into our previous we get $$b’ = p \frac{b}{1 – p(1 – b)} = \frac{p}{2 – p}$$. Again, checking against some numbers, if $$p = 0.9$$ then $$b’ \approx 0.82$$, which seems reasonable.

Suppose now that I provide you p-convincing evidence that $$B \implies C$$. What’s your posterior for $$C$$?

Well, again with the previous formula only replacing $$a$$ with $$b’$$ and $$b$$ with $$c$$ we have

\begin{align*} c’ &= \frac{p c}{1 – b'(1 – c)} \\ &= \frac{p(1-p)}{1 – \frac{p^2}{2 – p}} \\ &= \frac{p(1-p)(2 – p)}{2 – p – p^2}\\ \end{align*}

this isn’t a nice formula, but we can plug numbers in. Suppose your certainties are 0.9. Then your posterior is $$c’ \approx 0.34$$. You’re no longer certain that $$C$$ is false, but you’re still pretty convinced despite the fact that I’ve just presented you with an apparently water-tight argument to the contrary. This result is pretty robust with respect too your degree of certainty, too. As $$p \to 1$$, this seems to tend to $$\frac{1}{3}$$, and for $$p = \frac{1}{2}$$ (i.e. you’re wrong half the time when you’re certain!) we get $$c = 0.3$$.

In conclusion: An apparently water tight logical argument that goes from a single premise you believe in to a premise you disbelieve in via something you have no opinion on should not substantially update your beliefs, even if it casts some doubt on them.

Of course, if you’re a Bayesian reasoner, this post is an argument that starts from a premise you believe in, goes via something you have no opinion on, and concludes something you likely don’t believe in. Therefore it shouldn’t change your beliefs very much.

This entry was posted in Decision Theory, Numbers are hard on by .

## One thought on “Bayesian reasoners shouldn’t believe logical arguments”

1. Veky

First, your c’ is in fact p(2-p)/(2+p) (except when p=1), which is a bit nicer. And yes, it does converge to 1/3 as p goes to 1. However, the real WTF is that it is not its maximal value for 0<=p<=1. c'(p) is not monotonic, it attains maximum at 2sqrt2-2=~0.83, and its value is 6-4sqrt2=~0.34. Interpretation of _that_ I leave as an excercise to you. :-P