The raven paradox is roughly:

"The statement All ravens are black is logically equivalent to All non-black entities are not ravens. Whenever we observe a non-black non-raven, the probability for the latter statement to be true rises. In turn, the probability for the former statement must also rise. Therefore, we can infer something about the colour of ravens without looking at them."

I want to attack the paradox from a slightly different angle than the standard bayesian resolution (which states that one can in fact infer a tiny amount of information about ravens from looking at non-ravens).

Let us assume, for simplicity and to increase the magnitude of the expected effect, that only frogs (which, unbeknownst to us, are always green) and ravens are interesting to us. Furthermore, we only recognise the colours black and green. So initially, we hold equally probable that an observation will reveal a green frog, a black frog, a green raven, or a black raven ($GF, BF, GR, BR$). So $p(GF) = p(BF) = p(GR) = p(BR) = \frac{1}{4}$.

We now observe a number of green frogs. This updates our probabilities such that $p(GF) > \frac{1}{4}$. Since we did not make any other sightings, by symmetry all the other probabilities must be $\frac{1 - p(GF)}{3}$.

What is now the probability of a raven being black? Before the observations, it was clearly $\frac{1}{2}$. Now it is:

$$p(B|R) = \frac{p(BR)}{p(R)} = \frac{p(BR)}{p(BR) + p(GR)} = \frac{1}{2}$$

Weird. I would have thought we learned something about black ravens. Let's double check. What's the probability of a non-black entity to be a non-raven? In other words, what's the probability of a green animal being a frog?

$$p(F|G) = \frac{p(GF)}{p(G)} = \frac{p(GF)}{p(GF) + p(GR)} > \frac{1}{2}$$

So while the probability of raven-ness implying blackness did not increase, the probability of non-blackness implying non-raven-ness did increase.

How can this be? Is this a known, different paradox? Is my explanation an acceptable resolution to the raven paradox? Or is the probability of a statement being true not so easily related to its conditional probability?


I don't think tracking changes to $\Pr(B|R)$ captures completely how the probability of the raven hypothesis changes.

All ravens are black means for all things, if a thing has the predicate raven (R), then it has the predicate black (B).

So what is $\Pr(\forall x \,\ Rx \rightarrow Bx)$?

$\forall x \,\ Rx \rightarrow Bx $ is false if and only if $\exists x \,\ Rx\land \neg Bx$.

Therefore $\Pr(\forall x \,\ Rx \rightarrow Bx) = 1 - \Pr(\exists x \,\ Rx\land \neg Bx)$. This is the probability whose changes we need to track as we learn about the world.

It does not seem, in general, to be equal to $\Pr(B|R)$ which is what you've looked it. Instead it is $1 - (1- \Pr(B|R))\Pr(R)$. Rewrite the second term as $\left(1 - \Pr(B|R)\right)\frac{\Pr(BR)}{\Pr(B|R)}$. The conditional probability doesn't change, but because we've seen green frogs, the odds of seeing black frogs, green ravens and also black ravens have gone down. So $\Pr(BR)$ becomes smaller.

But then the term becomes smaller... and the probability that all ravens are black becomes greater.

Ceteris paribus, it's good for the contention that all ravens are black if ravens are rare, because there need to be ravens for it to have any chance of being wrong. That's not captured when focusing on $\Pr(B|R)$ alone and it seems to suffice to restore the paradoxical conclusion.

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  • $\begingroup$ That's a convincing explanation. One detail irritates me: why is $\Pr(\exists x \,\ Rx\land \neg Bx) = (1- \Pr(B|R))\Pr(R)$ when $\Pr(R)$ denotes the probability of the next observation being a raven? If we want to know whether e.g. there exist any black ravens, shouldn't we ask $1-(1-P(BR))^n$ for $n$ the total number of ravens? $\endgroup$ – Turion Sep 5 '19 at 17:39
  • $\begingroup$ You're right, it's not. How to write down the probability that all observations will be of a certain kind depends on how observations are sampled and what we take the population to be, but I think this isn't essential because the direction of the change of that probability will be the same as for the next obs $\endgroup$ – CloseToC Sep 5 '19 at 20:29

You haven't really analyzed the setup of the raven paradox; or rather, you've analyzed an extremely constrained variant of it. You say:

We now observe a number of green frogs...

Since we did not make any other sightings...

You started from a uniform prior and observed a universe consisting only of green frogs. Of course your subjective probability of $P(GF)$ is going to tend towards 1, and of course your subjective probability for the other three cases is going to tend towards zero. That's what I'd think to, if literally every thing I'd seen since birth was a green frog. Since you've never observed either a black frog nor a raven of any kind, you never had an opportunity to update your prior about $P(B|R)$ in any way, so of course it sticks stubbornly at $\frac{1}{2}$.

If you do your analysis again and allow for a non-zero number of observations of BF or BR, you'll get an answer closer to the standard Bayesian analysis you mentioned.

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  • $\begingroup$ Ok, let's assume we've seen a thousand black ravens. Now we see one green frog. Can you do a calculation where this new observation changes the probabilities in such a way that $P(B|R)$ increases? $\endgroup$ – Turion Sep 5 '19 at 18:01
  • $\begingroup$ $P(B|R)$ does not change when you observe a frog, black or green. My point was that you shouldn't be surprised that it hasn't changed from its initial prior when you've never observed any data that has any bearing on $P(B|R)$. In the absence of evidence, our priors dominate: it's like being stuck in an echo chamber. However, the posterior probability that all ravens are black does increase a little when you observe a green frog. $\endgroup$ – olooney Sep 6 '19 at 1:35
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    $\begingroup$ Assume there are $N$ objects in the universe (total, not just the ones we've observed.) All ravens are black iff there are zero non-black ravens. Write down an expression for the probability of this. You should see that is a function of $P(R \cap G)$ and $N$. Is it an increasing or decreasing function? When we observe a green frog, the posterior probability of $P(F \cap G)$ goes up a little. But because the 4 probabilities must sum to 1, the posterior probability for the other three cases, including $P(R \cap G)$, goes down. How does that affect the probability that all ravens are black? $\endgroup$ – olooney Sep 6 '19 at 1:39

So while the probability of raven-ness implying blackness did not increase

I started to write up an answer in which I noted that this was imprecise language, corrected it to "the probability of black-ness given raven-ness", but as I continued my answer, I noticed quite a bit of cognitive dissonance, and eventually gave up and deleted my answer. Now, thinking about a bit more, I think that this is not merely a bit of impreciseness to be "corrected" and moved on from, but the core of the issue.

There are several very different probabilities at play here. There's P(raven and black), P(not raven or not black), P(black|raven), and P(raven->black). The most direct interpretation of "the probability of raven-ness implying blackness" is P(raven -> black). Since "raven -> black" is equivalent to "all x are black or not raven", that translates to P($\forall$ x: x is black or not raven), but you seem to be conflating it with $\forall$ x: P(x is black or not raven), which is quite different. You shown that P(black|raven) isn't changing. However, P(not raven or not black) and P(raven->black) are increasing. To have a counterexample to raven->black, we need to find something that is both a raven and green. By finding something that is not a raven, you have removed one opportunity for a counterexample.

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