Fun with Linda

For fun, answer the following in the comment section, without reading what others have left in the comment section:

Linda is 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.

Which is more likely?

1. Linda is a bank teller.
2. Linda is a bank teller and is active in the feminist movement.

1. #1 NE1
May 27, 2008

1. Hooray for PS12!

2. #2 Ian Durham
May 27, 2008

OK, it’s 10 PM and I can barely keep my eyes open, so I may be reading this all wrong. But I’m guessing this is a logic question along the lines of p implies q so which is more likely, r or (r AND q). Unfortunately my brain is toast at the moment, but my instinct tells me r is more likely (i.e. that she’s just a plain old bank teller).

3. #3 Janne
May 27, 2008

Just slightly disingenuous: the way the alternatives are stated, most people will naturally read the first alternative as “1. Linda is a bank teller (but is not active in the feminist movement)”.

Rephrase the alternatives more honestly and see if you’d get the bias to the same degree:

1. Linda is a bank teller and her involvement in the feminist movement is unknown.

2. Linda is a bank teller and is active in the feminist movement.

4. #4 JohnQPublic
May 27, 2008

Well, if it’s a logic question, I’ll fail for sure. But while “issues of discrimination” does not necessarily include the feminist movement I would think it would be highly probable it would include it. That is, if she was concerned about racial discrimination, for example, she would not be also concerned about gender discrimination?? I don’t think so. So, pure logic aside, I go with…B!

5. #5 dreikin
May 27, 2008

Our evidence for option two are that (1) she is female, indicated by the third person feminine pronoun, giving a bias towards issues that affect females (one of her social classes), (2) a philosophy major, indicating she would have likely been exposed to a fair amount of feminist theory, (3) she has a history of concern about discrimination and social justice, two topics which are major topics within feminism, and (4) she is both outspoken and has a history of activism, indicating a predilection towards being active on topics that concern her.

Thus, the most likely option is choice (1), which requires at least four less assumption of cause and effect relations, and which does not preclude choice (2), as choice (2), as worded, is a subset of choice (1).

6. #6 Domenic Denicola
May 28, 2008

I was grinning when I first read this, but I was grinning even more after reading the first comment. I wonder who “NE1″ is…

And, I’m in the middle of a Ma 116 set, so this was _not_ a challenge for me… there’s a completely-automatic mental processing going on at the moment that converts words into logic-symbols. I seriously read that quote as more or less “$\psi(linda)$ holds. Which is more likely? $\phi(linda)$ or $\phi(linda) \wedge \varphi(linda)$?”

7. #7 Dave Bacon
May 28, 2008

I think I took PS 12 from a guy named Darwin. Now why the hell do I remember only his name and the prisoner’s dilemma from that class.

8. #8 Przemek Kaminski
May 28, 2008

I’d pick (2), but then it says she is ‘very bright’. Thus the only good option is (1).

9. #9 Joseph Hewitt
May 28, 2008

Linda is a bank teller. Assuming infinite parallel universes with alternate Lindas, the set of Lindas which are bank tellers is necessarily greater or equal to the set of Lindas which are bank tellers and active in the feminist movement.

Proof by alternate universes… I believe that’s somewhere on the list below Proof by Intimidation.

10. #10 Morgan
May 28, 2008

I’m guessing (1): being deeply concerned with something as a student is no guarantee of active involvement in a movement relating to it when you’re thirty.

11. #11 Morgan
May 28, 2008

…and I was right, but based on faulty reasoning, but the fault was not the fallacy you linked. As Janne points out above, my error was in automatically reading option 1 as implying she was not active in the feminist movement, and my guess based on reasoning that people tend not to be active in movements in general.

12. #12 laserboy
May 28, 2008

1.

Or can we also choose unemployed?

13. #13 lylebot
May 28, 2008

For reasons similar to Morgan, I picked the right answer for the wrong reason. I think they call that a Type III error.

14. #14 speedwell
May 28, 2008

I’m reasonably bright but not formally trained in logic. I am, however, taking a break from writing process documentation for the users of our engineering database. Thus primed, it just seemed trivially obvious to me that it was more likely that the chica in question was one irrelevant thing than that she was two.

Wonder how young kids would do on this, counting the fact that they are not as likely to be terribly familiar with many of the concepts expressed.

15. #15 eric
May 28, 2008

Since 1 is a simple superset of 2, 1 is clearly more probable.

16. #16 Julie Stahlhut
May 28, 2008

Not entirely fair, since I first saw this question over 20 years ago, but:

The first (“Linda is a bank teller.”) The second is the product of two probabilities, so it can’t be larger (and is almost certainly smaller.)

17. #17 Robin Blume-Kohout
May 28, 2008

Er… I hate to say this, Dave, but technically the correct answer is (2).

Reason is, you said “likely” rather than “probable”. “Likelihood” has a technical meaning in probability theory, which is different from “probability”. The likelihood of a theory is P( data | theory ).

So, in this case, if you ask “Which of (1-2) is more probable?”, then you’re implying that (1) and (2) should be interpreted as events. Then, since (2) is a subset of (1), (1) has got to be more probable.

But if you ask “Which of (1-2) is more likely?”, then [technically] you’re implying that (1) and (2) should be interpreted as theories about Linda (which, incidentally, is fairly plausible), and that the statement “Linda is 31 years old…anti-nuclear demonstrations” is data.

In this case, P( data | 2 ) is almost certainly greater than P( data | 1 ) — precisely because (2) is a more specific theory. So Theory #2 is more likely than Theory #1.

You may be thinking, “But doesn’t this pose some problems for maximum-likelihood inference methods?”, and you’d be right. Likelihood-based methods don’t work well for distinguishing between theories with different levels of detail. Bayes factors and Minimum message length methods can deal with this.

18. #18 Dave Bacon
May 28, 2008

Where did I express my opinion?

19. #19 Oldfart
May 28, 2008

What is most likely is that she is married, has two kids, works in the bank and has no time to think about the feminist movement……

20. #20 Robin Blume-Kohout
May 28, 2008

Dave wrote:

Where did I express my opinion?

Assuming that’s in response to “Er… I hate to say this…”, I’ll admit that I did infer that the link to the Wikipedia article (which indicates that [1] is correct) was meant as a tacit endorsement.

I could weasel out of unwarranted assumption, though: The main reason I began with “I hate to say this…” is that it was a horribly pedantic response. Subjects confused by the wording are almost certainly measure-zero!

21. #21 Anders Ehrnberg
May 28, 2008

bank teller

22. #22 Dave Bacon
May 28, 2008

Woot! Wikipedia has been updated to more accurately reflect the problem! Yeah for the readers of this blog

23. #23 Ian Durham
May 28, 2008

Er… I hate to say this, Dave, but technically the correct answer is (2).

Reason is, you said “likely” rather than “probable”. “Likelihood” has a technical meaning in probability theory, which is different from “probability”.

Sniff, sniff… I smell a Bayesian. The aura of Chris Fuchs has wafted into the room…

24. #24 Robin Blume-Kohout
May 28, 2008

Ian: Hey! I resemble that remark! grin

I’ll let Chris and Ruedinger and Carl Caves take care of the religious Bayesianism. I’m just into getting good answers.

Anyway, likelihood is actually the central quantity in frequentist statistics. Bayesians use it too, but it’s more of a supporting actor in that milieu.

25. #25 Ian Durham
May 29, 2008

Robin: ah, so you’re not a Bayesian fundie (fundy?) eh? Not a Caves acolyte? Heh heh.

In all seriousness, I have actually begun trying to teach the difference between Bayesian and frequentist interpretations to students in one of the introductory courses I teach.

26. #26 JM Geremia
May 30, 2008

AND = multiply in a problem like this, and we can probably assume statistical independence, or some independence at the very least. But, since Linda was a philosophy major, maybe I’m underthinking the problem.

27. #27 JM Geremia
May 30, 2008

Okay.. now that I’ve made it through all the posts: I must say, Dave, it’s pretty awesome that your blog is so powerful that even Wikipedia yields to you… not at all shabby.

It’s been quite enjoyable to read through the posts. Arguing about probability theory is possibly my all time favorite activity, second possibly only to skiing while also arguing about probability theory.

Living across the hall from Carl, it can be all too easy to get swept up in Bayesian fundamentalism, which is perhaps not a bad thing. My personal approach (to maintaining my own identity) has been to take a hardline axiomatic stance (in the Kolmogorov sence). Admittedly, this is really the same thing, but it looks different on paper, makes you sound smart to say Radon-Nikodym derivative instead of conditional probability, and has the added benefit of actually being mathematically correct in continuously random processes.

Joking aside, Ian: I support you 100% on your effort to bring this stuff into the classroom. I’ve been trying to do the same thing. Last fall, I asked a loaded question in one of the early quantum lectures: “what is a probability?” After the expected response, I handed the poor person who answered a quarter and requested that she prove to me that the probability of tails is 1/2. That combined with working out the 3-door problem was a blast, and really effective in getting across some basic thinking essential to quantum mechanics IMHO.

28. #28 Robin Blume-Kohout
May 31, 2008

Hmm, I was going to toddle off to other digital fire hydrants, but since JM’s gone and made the discussion interesting again… I’m back.

I agree with JM and Ian (BTW, Ian, my New Mexico heritage is via Los Alamos and Zurek, so while I respect Carl a lot, and enjoy arguing with him, I got interested in probability and statistics independently) that teaching probability theory to physicists is cool. Moreover, it’s good for ‘em.

So, with that in mind, here’s a question. What do you think about teaching quantum mechanics as noncommutative probability theory? In other words, by starting with probability theory and alluding to probabilistic mechanics (e.g., distributions on phase space), and then introducing quantum theory as a generalization of probability.

This is how I think of quantum theory all the time now — and it makes tremendous sense to me. I think it’s how I want to teach it. And I’m curious what y’all think.

29. #29 Ian Durham
May 31, 2008

Robin: wow, now that’s a neat idea! I think, in a certain sense, I have been unconsciously moving in that direction for awhile, particularly since it (probability) has formed the basis of my personal philosophy of physics for a number of years (really it came out of my years of thesis work on Eddington whose philosophy rubbed off on me, I guess). Well, perhaps next year I’ll try something along those lines. If you do it before me, let me know how it goes.

JM: Awesome! Those little demos are really what gets the idea to sink in with the students. In addition to courses for physics majors, I actually teach this stuff in my algebra-based introductory physics course which is mostly pre-med majors. In fact one of the labs for that class, which I insert just before we study quantum and statistical mechanics, is purely on probability (we use dice, though I might switch to little plastic pigs). There also were some interesting pedagogical things I discovered last year that I use sometimes.

30. #30 melior
June 13, 2008

(A and B) cannot be more likely than (A).