Here is a photo of a picture that hangs on the wall of a friend’s home. Karl Popper dons sunglasses. Hanging from his mouth is a broken rose of socialism, the color of congealed blood.
Logik der Forschung is German for logic of research.
Four cards are shown, each card showing a character on the side shown, namely A, B, 2, and 3, respectively.
Consider the following proposition:
Proposition P: If one side of a card has a vowel, then on the other it has an even number.
To establish that none of these four cards falsifies Proposition P, which cards must one turn over?
When my friend put the question to his besotted dinner group, we puzzled over it merrily. Feel free to answer in the comment field. This exercise in falsificationism might help us discover the value and complications of falsification as scientific practice. If reader response seems to justify it, I will follow up.
READER COMMENTS
Joel Pollen
Aug 3 2021 at 2:07pm
I think all but the “2.” If “A” has a 1 on the other side, it’s false. If “B” or “3” have a vowel on the reverse, it’s false. For “2,” if it has a vowel on the other side it’s true, and if it has a non-vowel on the other side then it isn’t relevant to the proposition, since it doesn’t fulfill the first part of the conditional.
Stéphane Couvreur
Aug 3 2021 at 2:44pm
P implies that A has an even number and 3 a consonant on the other side.
Turning both A and 3 is the only way to try to falsify P, by checking whether P’s implications are verified or not.
Turning B or 4 does not give us any information about the validity of P, whatever the result.
Capt. J Parker
Aug 3 2021 at 3:30pm
I humbly submit that Joel Pollen’s answer is better.
If P were true then
The “A” card would have an even number on the other side.
The “3” card would have anything except a vowel on the other side.
The “B” card would have anything except a vowel on the other side.
The “2” card could have anything at all including a vowel on the other side.
The title of the blog post is Presuppositions. Making and assumption that the “B” card has a number of the other side instead of a letter would be a presupposition. Your presupposition might be correct but, there doesn’t seem to be anything in the problem statement that guarantees that it is correct.
Stéphane Couvreur
Aug 3 2021 at 5:12pm
Yup, without even realizing it, I made the implicit assumption that all cards had a letter on one side and a digit on the other side. In fact, the way the problem is posed, this assumption is unsupported.
Joel is right.
Capt. J Parker
Aug 3 2021 at 3:01pm
I don’t thing you need to turn any of them over. You simply need to move them around a little bit to see whats on the reverse side using the reflection in Popper’s sunglasses. Alternatively, just ask Popper what he sees (if you trust him.)
Steve Fritzinger
Aug 3 2021 at 3:56pm
A and 3.
P says nothing about what’s on the other side of a card with a consonant on the visible side. P could be true even if card B has an even number on the other side.
Similarly, P says nothing about what has to be on the other side of a card with an even number visible. P could be true even if card 2 has a consonant on the other side.
If card A has an odd number on the other side, P is false. If card 3 has a vowel on the other side, P is false.
robc
Aug 3 2021 at 4:30pm
You made the same mistake as Stephane. B could have a vowel on the other side. Joel and the Capt are correct.
Steve Fritzinger
Aug 3 2021 at 9:18pm
Oh, I get it now. Had to read that three times. That’s tricky.
KevinDC
Aug 3 2021 at 5:06pm
I deliberately avoided reading any of the other comments, to avoid them influencing my own thought process – if it turns out I’m just rehashing something that’s already been said, then excuse my repetitiveness!
The proposition states that vowels on one side means even numbers on the other side. So if we turn over card A, we will either find an even number, or we won’t. So this card has the potential to falsify the proposition. So we can turn this one.
We can skip card B. The proposition we are testing doesn’t say that only vowels have even numbers on the opposite sides – we have no information about consonants. So nothing we find on the opposite side of this card will have any bearing.
Card 2: This, too, we can skip. On the other side, we will find either a consonant or a vowel. If we find a consonant – well, that doesn’t help us at all. If we find a vowel, then we’d have a even number/vowel pairing like the proposition suggests – but the question here is about what cards could falsify the proposition. Finding a vowel on the other side wouldn’t do anything to falsify the proposition, so we can leave this card unturned.
Card 3: Turn it. If we find a vowel on the other side of this odd numbered card, then we have falsified the proposition that vowels are paired with even numbers. We could also find a consonant and fail to falsify the proposition, but we’d have to turn the card to find out.
KevinDC
Aug 3 2021 at 5:08pm
Aaaaand having read the comments, I see that Joel gave a better answer. I unintentionally smuggled in a false assumption that each card was a letter on one side and a number on the other, but that doesn’t have to be the case.
Francisco
Aug 3 2021 at 5:35pm
You need to turn “A”, “B” and “3.” P is false if:
“A” does not have an even number on the other side.
“B” has a vowel on the other side.
“3” has a vowel on the other side.
Whatever “2” has on the other side it will not contradict P. If it has a vowel then P is true, and if it doesn’t then the premise of P fails and P is not falsified.
Capt. J Parker
Aug 3 2021 at 10:11pm
I find Joel’s logic airtight if the problem statement said we are dealing with four separate physical cards. But, nothing in the problem statement said we should imagine the 4 cards in the painting should be treated as 4 separate physical cards that is a presupposition. Therefore, I now believe the answer is: To ascertain if P is false one must determine what is on the other side of the painting. To do that we must turn over all four cards.
Peter Gerdes
Aug 3 2021 at 10:37pm
This is a familiar example of a common fallacy used frequently in introductory logic classes.
However, it doesn’t really tell us much about falsification. The big issue there is simply the fact that no hypo is ever really falsified on it’s own but only in conjunction with a bunch of background assumptions/bridge laws. This example hides that complication because the background assumptions are super obvious things like: what’s on the card doesn’t change when you flip it over and that you aren’t hallucinating etc…
Now, TBF to Popper he seems to have been well aware of this concern (tho others took his views and ignored this wrinkle) and I think a better interpratation (ok perhaps improvement) of his view would be something more like: we should identify scientific claims with their falsifiable consequences (even if those consequences of T are of the form: T conjoined with background B is falsified if X isnt observed ).
I mean that seems like a better approach than trying to hang everything on a binary falsifiable or not distinction.
Monte
Aug 4 2021 at 11:12am
A is the only card of consequence. What is on the other side of the remaining 3 cards is irrelevant with respect to the proposition.
Capt. J Parker
Aug 4 2021 at 11:24am
What if there were a vowel on the other side of the “3” card?
Monte
Aug 4 2021 at 7:16pm
Hello Capt.
In the strictest sense, the proposition stipulates only that a vowel card must have an odd number on the flip side to be falsifiable. There is no condition stipulating that an even-numbered card must have a vowel on the flip side.
Monte
Aug 4 2021 at 7:17pm
…or that an odd-numbered card can’t have a vowel on the flip side.
Capt. J Parker
Aug 4 2021 at 9:29pm
Monte,
I humbly submit that:
is falsified if one finds a card that has a “3” on one side and a vowel on the other and that it does not matter if you find such a card with the “3” initially face up or with the vowel initially face up.
I submit further that Proposition P is falsified if one finds a card with a “B” on one side and a vowel on the other. Again, this is true even if you find such a card with the “B” face-up at first.
Your solution would make sense if Proposition P said something more like:
But, in the strictest sense, Proposition P does not say that.
Monte
Aug 5 2021 at 2:16am
Damn it! I was trying pull off a 3-card Monte!
Max More
Aug 4 2021 at 1:35pm
My question is: How can I get that picture? Not “get” as in “understand the problem” but get as in acquire? I love it!
Thomas Leahey
Aug 4 2021 at 2:03pm
Hi all,
Congrats.
You’ve just discovered the most important experiment in the psychology of thinking, Peter Wason’s 4-card selection problem. Originally done in 1964 in approximately the form you presented (w/o the Popper embellishment) it’s been repeated with variation more than any other reasoning task. I did my master’s thesis at University of Illinois at Urbana in 1972 on trying to improve reasoning in this task. It’s been performed with subjects given MRI scans while doing it.
I have recently summarized the literature on it for a new book I’m writing. I can send it along to whoever asks. The most striking finding is that this version (the abstract) version can be translated into linguistic versions, such as this: You are a bar bouncer and must enforce a rule: If a person is drinking alcohol, they must be 21 or older. The four cards are: bar patrons drinking gin; drinking soda; 25 years old; 16 years old. You must turn over cards to find out their age or what they are drinking in order to see who to bounce. The findings are that the kind of problem and who it’s assigned to make big differences in performance. E.g., in this case U. of Florida participants did well, U. of Oxford, not. Logic background doesn’t help.
Have fun. This experiment defeats almost everyone and is always interesting to teach.
Thomas Leahey
Aug 4 2021 at 2:11pm
Hi again- I forgot to mention the great historical importance of this experiment. It gave rise to the 2-systems view of the mind made famous by Kahnemann’s Thinking fast, thinking slow, and the whole heuristics and biases literature. Confirmation bias made its appearance in Wason’s student J. St. B. T. Evans’ idea of “matching bias” to explain peoples wrong choice, and he then began to develop the 2-system view.
Capt. J Parker
Aug 4 2021 at 4:40pm
Thanks Thomas! I’m shocked that I don’t seem to remember encountering this problem before.
When I googled Watson’s Selection Problem I found mostly formulations were it was given in the problem statement that there was a letter on one side of each card and a number on the other side. In the formulation given in here by Dan Klein, there is no specification that if there is a letter on one side there must be a number on the other. This latter formulation seems like it would be much more difficult to solve since it introduces the possibility of mistakenly concluding the “B”card is not relevant because it has no vowel. Is it indeed the case the the formulation given by Dan Klein is more difficult?
Capt. J Parker
Aug 4 2021 at 4:45pm
And it would seem that selecting Wason in stead of Watson is also a difficult task. (for me anyway)
Thomas Leahey
Aug 4 2021 at 5:43pm
You are right on both counts. The original formulation was, e.g., “If there is a 7 on the front of the card, then there is an M on the back.” “Matching bias” was to pick the 7 and M cards. Klein/Popper is harder. Interestingly I used the task in class for 30 years, and over time the most common pattern of response has drifted away from the matching one, which used to be dominant, but no new pattern has emerged, including to the correct 7 and X choices. I can’t figure out why.
In the area of logic, I suspect Watson is primed by link to Holmes.
Tom
Daniel Klein
Aug 5 2021 at 10:21am
Thanks to all here for the nice contributions to the discussion. I have written a follow-up post, which will appear soon–finished it when the comment clicker here was at about 12. At the dinner gathering at my friend’s home, I initially answered “A” and “3.” Again, thanks to all for the engagement.
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