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Suppose that you see Linda go to the bank every single day. Presumably this supports the hypothesis H = Linda is a banker. But this also supports the hypothesis H = Linda is a Banker and Linda is a librarian. By logical consequence, this should increase my credence in the hypothesis H = Linda is a librarian.

Note that by the same logic, this also supports the hypothesis H = Linda is a banker and not a librarian. Thus, this should increase my credence in the hypothesis H = Linda is not a librarian by logical consequence since it is directly implied by the former.

But this is a contradiction. You cannot increase your credence both in a position and the consequent. How does one resolve this?

Presumably, the response would be that seeing Linda go to the bank doesn’t tell you anything about her being a librarian. That would be true but under Bayesian ways of thinking, why not? If we’re focusing on the proposition that Linda is a banker and a librarian, clearly her being a banker makes this more likely that it is true.

One could also respond by saying that her going to a bank doesn’t necessitate that she is a librarian. But neither does her going to a bank every day necessitate that she’s a banker. Perhaps she’s just a customer. (Bayesians don’t attach guaranteed probabilities to a proposition anyways)

This example was brought about by David Deutsch on Sean Carroll’s podcast here and I’m wondering as to what the answers to this are: https://youtu.be/ldgK7EhEnto?si=eypIGUOmcMzxNww-. He uses this example and other reasons to completely dismiss the notion of probabilities attached to hypotheses and proposes the idea of focusing on how explanatorily powerful hypotheses are instead

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  • “Logic” takes on a lot more power as a fully formalized, rule-based system. In a way, that’s why it’s so valuable as compared to intuitive “logical thinking” (in the sense of “appealing to your sense of reasonableness”). The logical deductions you presented aren’t formalized. It isn’t clear what the “rules” are, in this system. Thus, all of the above claims about Linda are subjective and debatable, not “necessary” or “obligatory”, as derivations in a formal system would be. If I see Linda go to the bank every day, I personally may disagree that this “supports the hypothesis that Linda is a ban Commented Apr 8 at 2:55
  • Having listened to the podcast (or at least the relevant part of it), I think I see the problem. Yes the evidence increases credence in P(B,L) and P(B,~L), but it reduces credence in P(~B,L) and P(~B,L). The probability that Linda is a librarian is P(L) = P(B,L) + P(~B,L), i.e. she is a librarian either because she is a librarian/banker or a librarian/non-banker. P(L) doesn't change because the increase in P(B,L) is canceled by an equivalent decrease in P(~B,L). There are four compound events here, and Deutsch is only looking at two of them.
    – user6527
    Commented Apr 10 at 18:19
  • You’ve highlighted how P(L) doesn’t change based on the rules of probability. Deutsch doesn’t disagree with that. He says, as mentioned in our other comment thread, that one must increase P(L) as a matter of logic if one takes it as credence. This is because, again, B&L imply L. Thus an increase in credence of B&L should result in an increase in credence of L. But that is not what happens. By the way, Deutsch actually doesn’t make an error. His original pronouncement is correct and then when trying to correct himself he makes an error which is funny but that’s a different story Commented Apr 10 at 18:24
  • Logic is probability theory where probabilities are limited to 0 and 1. If you have probabilities that are not 0 or 1 somewhere in your analysis, logic doesn't apply. Logic is a subset of probability theory, so it is an error to assume that the rules of logic universally apply to probability. If probabilities are constrained to be 0 and 1 then P(B,L) = 1 does imply that P(L) = 1, but if P(B,L) is between 0 and 1, that is no longer a valid use of logic. At least one Deutsch's errors is only looking at half the problem and forgetting that he is conditioning on B.
    – user6527
    Commented Apr 10 at 18:33
  • Again, logic does apply to probability theory. Bayesianism != probability theory. Bayesianism is a reasoning scheme that identifies credences with probabilities. And as shown, it violates logic. Probability does not. Logic applies everywhere and is inescapable. It is not impossible for you to employ systems that are incoherent. But it doesn’t change their incoherence. For example, I can say “I love my child implies I do not love my child”. I can say that. But I would be incoherent. Commented Apr 10 at 18:40

8 Answers 8

6

Suppose that you see Linda go to the bank every single day. Presumably this supports the hypothesis H = Linda is a banker. But this also supports the hypothesis H = Linda is a Banker and Linda is a librarian. By logical consequence, this should increase my credence in the hypothesis H = Linda is a librarian.

Note that by the same logic, this also supports the hypothesis H = Linda is a banker and not a librarian. Thus, this should increase my credence in the hypothesis H = Linda is not a librarian since it is directly implied by the former.

But this is a contradiction. You cannot increase your credence both in a position and the consequent. How does one resolve this?

Bayesianism is about shifting probabilities. It might help to have a full blown example.

We have a society with 1000 people. 50 of them (5%) are Bankers. 50 of them (5%) are librarians. 1 of them (0.1%) is a librarian and banker (1 of the existing 50 listed before, not a new one).

I have a person behind a closed door. Knowing nothing about this person, you can assume a 5% chance they're a banker, a 5% change they're a librarian, and a 0.1% chance that they're a librarian and a banker.

Now, I tell you this: the person behind the door is a banker. How do those probabilities change?

Your space of possible people shrunk by 950 people - there's 950 people you know this person isn't.

You now know there's a 100% chance that they're a banker - so that probability went up from 5% to 100%.

You know there's a 2% chance they're a librarian - because 1 of the 50 bankers are librarians - so that probability went down from 5% to 2%.

You know there's ALSO a 2% chance they're a librarian and a banker - this probability went up from 0.1% to 2%.

This is how learning they're a banker can be evidence against them being a librarian, but evidence for them being a banker-and-librarian.

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A few preliminary points, for information:

  1. The discussion within the interview that you are referring to starts at timestamp 1:06:55 in the linked video.

  2. In the example, Deutsch refers to Linda being a banker and a feminist, rather then a banker and a librarian.

  3. Deutsch's reference to Hempel relates to a demonstration that Bayesian confirmation is non-transitive. I.e. it is possible to have a situation whereby some evidence E provides confirmatory support for a hypothesis H; and H entails K; but E provides disconfirmatory evidence for K. The simplest case of this is where K is logically equivalent to H ∨ ¬E.

  4. The Popper and Miller paper mentioned by Deutsch is "Why Probabilistic Support is not Inductive", K. Popper and D. W. Miller, Phil. Trans. Royal Society of London. A 321, 569-591 (1987). There have been many responses to this by various defenders of Bayesian confirmation theory.

Deutsch in the interview mentions Hempel's result that Bayesian confirmation is nontransitive. He proceeds to claim that somehow this nontransitivity is illogical. This appears to be confusing two different relations. The logical consequence relation is transitive (in most systems of logic at least). But evidential support is not the same relation as logical consequence and it is not guaranteed to be transitive. There is really nothing illogical about it; they are just different relations.

The fact that Linda is seen frequently in a bank supports the hypothesis that she is a banker. But in the absence of any correlation between being a banker and being a feminist it provides equal support for the hypothesis that she is a banker-and-feminist and the hypothesis that she is a banker-and-not-feminist. So it does not support the hypothesis that she is a feminist. In the interview, Sean Carroll correctly points this out and Deutsch makes a mistake in his reply but offers no response.

There are different ways to represent the degree of probabilistic support of a piece of evidence for a hypothesis. One common one is due to Jack Good, which he calls the weight of evidence (e.g. Bayesian Statistics 2, pp 249-270, 1985). It is defined by:

                P(E|H) 
W(H:E)  =  log --------
                P(E|¬H)

By this formula, the weight of evidence that Linda is a banker given that she is frequently seen in a bank is positive. The weight of evidence that Linda is a feminist banker given that she is a banker is zero (assuming no correlation). The weight of evidence that Linda is a feminist given that she is frequently seen in a bank is zero (in the absence of any other correlation). There is nothing 'illogical' about this combination.

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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Philosophy Meta, or in Philosophy Chat. Comments continuing discussion may be removed.
    – Geoffrey Thomas
    Commented Apr 10 at 11:18
  • +1 FWIW I've added a Bayesian analysis of the problem in my answer below, but mostly just to show what Bayesian analysis actually says
    – user6527
    Commented Apr 10 at 17:10
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There is no contradiction. The probability of two independent things being true is the product of their individual probabilities, p1p2. If you increase the probability of p1, you also increase the probability of the combination of p1 and p2, but that doesn't have any affect on p2 itself. It might help to take a worked example...

Suppose I have two dice, one red and one blue, and I roll them out of your sight and ask you to guess what they show. You might guess that the red dice shows a five and the blue dice a three. The probability of each guess being correct is one in six, and of both being correct one in thirty-six. There are thirty five different ways in which you could be wrong, and each of them has the same probability of one in thirty six. Now suppose I reveal the red dice and it does indeed show a five as you guessed. The probability of both your guesses being right has now increased to one in six. The probability of you being wrong has decreased from thirty five in thirty six to five in six.

Now, the fact that the red dice has indeed shown a five has not just increased the probability of your guess- it has also increased the probability of each of the remaining alternatives to your guess, namely that the red dice shows a five and the blue dice shows a one, or a two, etc. In the second paragraph of your question you seem to be confused by that effect- ie that establishing Linda is a banker increases both the likelihood she is a banker and a librarian and the likelihood that she is a banker and not a librarian. But crucially what has been raised are the probabilities of the combinations- it has no effect whatsoever on the probability of whether Linda is or isn't a librarian.

To generalise, increasing the probability of p1 increases the probability of both p1 and p2 being true. It also increases, to exactly the same degree, the probability of both p1 and not p2 being true. Again, that is because it is the probability of the combination that is being affected, not the relative probabilities of p2 and not p2.

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  • But the person doesn’t know if they are independent. Some bankers can be librarians as well and vice versa. The Bayesian who simply sees her go to a bank doesn’t know if they are independent. Your examples are of knowledge that the person does not possess Commented Apr 9 at 8:31
  • Sorry, you know they are independent in the sense that you don't have to be a banker to be a librarian, and you don't have to be a librarian if you are banker. Commented Apr 9 at 9:19
  • If I’m more confident in A, then it implies that I’m more confident in everything that makes up A. Linda is a woman = Linda has a vagina and Linda has XY Chromosomes.Now, if I’m more confident in Linda being a woman, can I be less confident in her having a vagina? Can I be less confident in her having XY chromosomes? No. There is no case where it makes sense to somehow become more confident that Linda is a woman while simultaneously being less confident that Linda has a vagina or being less confident that Linda has XY chromosomes Commented Apr 9 at 10:10
  • Hi Stella, I'll reply the chain of comments under Bumble's answer. Commented Apr 9 at 12:08
  • Cogito, you've addressed the main issue in the OP's question. She did bold the and in "H = Linda is a Banker and Linda is a librarian". The probability of the compound event went up, but that was due to increase in only one of the component probabilities.
    – Hudjefa
    Commented Apr 10 at 3:53
1

Let B = "Linda is a banker", L = "Linda is a librarian", and E = "Linda entered a bank every day".

Suppose that you see Linda go to the bank every single day. Presumably this supports the hypothesis H = Linda is a banker.

If we assume that, then P(B|E) > P(B)

But this also supports the hypothesis H = Linda is a Banker and Linda is a librarian.

That doesn't follow. P(B ∧ L|E) = P(L|B,E) P(B|E) and P(L|B) P(B) = P(B ∧ L). P(B|E) > P(B), but we can satisfy these equations while keeping P(B ∧ L|E) = P(B ∧ L) if we also let P(L|B,E) decrease (relative to P(L|B)) in proportion to how much P(B|E) increased relative to P(B). (It would follow if L and E were independent.)

A good way to see why this couldn't follow (without additional assumptions) would be to consider the case where L = "Linda is NOT a banker". The structure of the math would be unchanged, and no amount of evidence can increases P(B ∧ ~B).

By logical consequence, this should increase my credence in the hypothesis H = Linda is a librarian.

That also doesn't follow, even if P(L ∧ B | E) > P(L ∧ B).

P(L) = P(L ∧ B) + P(L ∧ ~B) and P(L|E) = P(L ∧ B | E) + P(L ∧ ~B | E).

We can satisfy P(L ∧ B | E) > P(L ∧ B) either by increasing the probability of L or by moving probability mass from P(L ∧ ~B) to P(L ∧ B) (or a combination of the two). More formally, we can infer that (P(L|E) > P(L)) v (P(L ∧ ~B | E) < P(L ∧ ~B)).

The only assumption we made was P(B|E) > P(B), and that isn't sufficient to determine which of those options is correct.

0

Let r be the fraction of the population that are banker-librarians who go to the bank every day, r' the fraction that are banker-librarians who don't go to the bank every day, and a the fraction who go to the bank everyday.

Then prior to your observation that Linda goes to the bank everyday, the probability she is a banker-librarian is r+r'. After your observation, the Bayesian-updated probability is r/a. That could be either an increase or a decrease.

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  • Can you edit your comment first? As it stands, the formatting is atrocious and I don’t think that was intended Commented Apr 8 at 21:07
  • I just read your comment. In no scenario should there be a contradiction. So just plug in values for those such that it results in an increase for the credence in her being a banker and the credence in her being a banker-librarian and the credence in her being a banker-not librarian. This now results in a contradiction in the OP if you follow the steps. Commented Apr 10 at 14:58
  • @Stella if you perform the analysis, there is no contradiction, as demonstrated in my answer.
    – user6527
    Commented Apr 10 at 17:31
  • @Stella : Your mistake is here: "Presumably, the response would be that seeing Linda go to the bank doesn’t tell you anything about her being a librarian." In fact, seeing her go to the bank could either increase or decrease the probability she is a librarian.
    – WillO
    Commented Apr 10 at 21:08
  • @WillO If it could increase or decrease it, then it literally means it doesn’t tell you anything about whether she’s a librarian. The original example was the term feminist. Not a mistake Commented Apr 11 at 1:05
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Suppose that you see Linda go to the bank every single day. Presumably this supports the hypothesis H = Linda is a banker. But this also supports the hypothesis H = Linda is a Banker and Linda is a librarian. By logical consequence, this should increase my credence in the hypothesis H = Linda is a librarian.

Your conclusion is wrong, your credence that Linda is a librarian should be unchanged.

The way this problem is stated, Linda being a Banker, Linda being a Librarian, or Linda being a Feminist are all independent of each other.

If Linda being a banker is independent from Linda being a librarian, then having a 100% probability that Linda is a banker means the probability that Linda is both a Banker and Librarian is now completely dependent on the probability that Linda is a Librarian:

P(Both) = P(Banker)*P(Librarian)

If P(Banker) = 1 then

P(Both) = P(Librarian)

There are no observations that Linda is a Librarian making the probability that Linda is a Librarian undetermined.

Your knowledge of the probability that Linda is both Librarian and Banker is unchanged despite having full knowledge that Linda is a Banker.

0

I think the problem here is that we are only looking at a fragment of the problem and only waving hands towards Bayesian analysis without actually performing one. So lets try actually performing one and see what it actually says.

Let B be the proposition that Linda is a banker, and L be the proposition that Linda is a librarian; furthermore let E (the evidence) be the event that Linda was seen walking into a bank.

Now, if Linda is a banker, then P(E|B) = 1 (bankers can reliably be spotted walking into a bank), but if she isn't then P(E|~B) = 2/10 (arbitrary value capturing the idea that banks have customers as well, who can also be spotted walking into banks, but less frequently). If we don't have any prior information about whether Linda is a banker, we might say P(B) = P(~B) = 1/2, however in practice we know that bankers are no were near as common as that, but it will do for an exploratory exercise. So lets plug that into Bayes' rule:

P(B|E) = P(E|B)P(P)/PE = P(E|B)P(B)/(P(E|B)P(B) + P(E|~B)P(~B)) = (1 x 1/2)/(1 x 1/2 + 2/10 x 1/2) = 5/6

So observing Linda walking into a bank does increase our belief that she is a banker from 1/2 to 5/6.

So what about librarians. In the original problem, we can have librarians that are also bankers, so we have a few more probabilities to set out. If Linda is a banker and a librarian, then P(E|B,L) = 1, if she is a banker but not a librarian, P(E|B,~L) = 1 (note P(E|B,L) = P(E|B,~L) = P(E|B) because B and L are independent). If she is a librarian but not a banker, or not a librarian or a banker, then she may still be a customer, so P(E|~B,L) = P(E|~B,~L) = 2/10 (note P(E|~B,L) = P(E|~B,~L) = P(E|~B) as B and L are independent). Similarly, for the prior, we have P(B,L), P(B,~L), P(~B,L) and P(~B,~L). To keep life simple, lets make them all equiprobable, i.e. 1/4.

Lets run that all through Bayes' rule:

P(B,L|E) = P(E|B,L)P(B,L)/P(E)

Lets work out P(E) first because we don't have LaTeX on this forum.

P(E) = P(E|B,L)P(B,L) + P(E|B,~L)P(B,~L) + P(E|~B,L)P(~B,L) + P(E|~B,~L)P(~B,~L)

P(E) = 1/4*(1 + 1 + 2/10 + 2/10) = 1/4 x 24/10 = 24/40 = 3/5

So

P(B,L|E) = P(E|B,L)P(B,L)/P(E) = (1x1/4)/(3/5) = 1/4 x 5/3 = 5/12

and

P(~B,L|E) = P(E|~B,L)P(~B,L)/P(E) = (2/10 x 1/4)/(3/5) = 1/4 x 1/3 = 1/12

Marginalising

P(L|E) = P(B,L|E) + P(~B,L|E) = 5/12 + 1/12 = 1/2

In other words, observing Linda walking into a bank raises the probability that she is a librarian from the prior probability of P(L) = P(B,L) + P(~B,L) = 1/4 + 1/4 = 1/2, to, err... 1/2! It tells us precisely nothing about whether Linda is a librarian.

1
0

Suppose that you see Linda go to the bank every single day. Presumably this supports the hypothesis H = Linda is a banker. But this also supports the hypothesis H = Linda is a Banker and Linda is a librarian. By logical consequence, this should increase my credence in the hypothesis H = Linda is a librarian.

The highlighted bit is simply incorrect, the problem is that logic applies to propositions that are either true or false, but it can't be applied to "credences" (uncertain beliefs about propositions), which are likely to be real-values. Let c(A) be the credence regarding the truth of proposition A. Note that I have not made it p(A) so that it is not specific to probability, but that would be one possible implementation.

So if we have B is the proposition that Linda is a banker and L is that Linda is a librarian, then we can apply logic to those propositions and say that L AND B implies L. Totally uncontroversial but not very useful. It just says "if we are certain that Linda is a banker and Linda is a librarian then we know that Linda is a librarian". However that does not mean that an increase in c(B,L) implies an increase in C(L).

This is easier to see if we apply it to an example where we can have no non-trivial credence about one of the propositions because it is an independent random event. Let A be the event that a coin A lands heads and B be the event that coin B lands heads and both coins are known to be fair. If we have some evidence that A is true, i.e. c(A) increases, then it is reasonable for c(A AND B) to increase. This is because A being true can be thought of as a "cause" of A AND B being true. However we can't say that the resulting increase in c(A AND B) should cause an increase in c(B) - that would mean that evidence regarding A would change our credence about B. But that is obviously nonsense because B is the outcome of flipping a fair coin. The only justifiable credence would be that c(B) = c(~B), and they can't both increase.

As usual many problem posed here come down to not expressing the question properly. In this case, the question is conflating B and L with c(B) and c(L), the propositions and their credences. As soon as you clearly separate those two things, the error is easily spotted. Logic only applies to booleans, things that are true or false, so we can use it to reason about the truth of the propositions given what we know to be true, but we can't use it to reason about the credences regarding the propositions.

So it all comes down to undisciplined use of language, and using some mathematical notation helps when dealing with mathematical concepts (like logic and credences/probabilities).

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  • Thank you for your detailed answer again. Two confusions. One: I’m not sure credence gets around this since you’re already assuming a measure. Note that it is possible to talk about me being more confident than yesterday without an absolute measure for each. Two: Imagine confidence as a sort of vividness of a picture. Now, it becomes impossible to make that picture vivid without making each constituent more vivid. And yet in your coin example, c(B) doesn’t increase. So it seems to create contradictory attitudes between treating probability as confidence and the logical meaning of statements. Commented Apr 11 at 8:13
  • I do wonder if part of the issue here is that even though credences can perhaps mark uncertainty the statements/propositions themselves can only be correct or incorrect (two values). So there seems to be a fundamental mismatch there. Commented Apr 11 at 8:15
  • @Stella if credance doesn't have a measure, and is just like "vividness of a picture" then it doesn't need to conform to logic. It is perfectly possible for people to hold illogical levels of confidence. For example having confidence about coin B based on coin A is possible - people do that sort of thing all the time - it is called "the gambler's fallacy" and it is not logical.
    – user6527
    Commented Apr 11 at 8:22
  • "So it seems to create contradictory attitudes between treating probability as confidence and the logical meaning of statements" this is ignoring what I wrote. Logic applies to the propositions themselves, it doesn't apply to credances about propositions. It can't because it only applies to things that are either true or false and credences are not just "true" or "false".
    – user6527
    Commented Apr 11 at 8:24
  • "I’m not sure credence gets around this since... " credence was YOUR term, not mine: "By logical consequence, this should increase my credence in the hypothesis H = Linda is a librarian."
    – user6527
    Commented Apr 11 at 8:26

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