## A reasoning puzzle

by Greg Mayer

In today’s New York Times, there is an opinion piece by Manil Suri, a mathematician at the University of Maryland, Baltimore County, entitled “Does math make you smarter?”

Four cards are laid in front of you, each of which, it is explained, has a letter on one side and a number on the other. The sides that you see read E, 2, 5 and F. Your task is to turn over only those cards that could decisively prove the truth or falsity of the following rule: “If there is an E on one side, the number on the other side must be a 5.” Which ones do you turn over?

I’ll post tomorrow (Monday) the answer and a discussion, along with the link.

1. Craw
Posted April 15, 2018 at 3:13 pm | Permalink

E,2

2. Mike Anderson
Posted April 15, 2018 at 3:14 pm | Permalink

Cards showing “E” and “2” might falsify the claim.

3. jaxkayaker
Posted April 15, 2018 at 3:15 pm | Permalink

E and 2

4. John
Posted April 15, 2018 at 3:17 pm | Permalink

The 5 and the F.

• abram
Posted April 15, 2018 at 3:24 pm | Permalink

Rebel!

• Don
Posted April 15, 2018 at 3:27 pm | Permalink

I almost started explaining why you are wrong.

• John Taylor
Posted April 17, 2018 at 5:56 am | Permalink

All of them.

5. freiner
Posted April 15, 2018 at 3:18 pm | Permalink

E, 2. There are isomorphic variations of this puzzle that place it in more “real life” settings. Comparing results of how people do in different versions is best left for later discussion.

• freiner
Posted April 15, 2018 at 3:25 pm | Permalink

Okay, I just checked the article in my (real paper) copy of the Times. He mentions a variation that people do much better at.

• Ted Burk
Posted April 15, 2018 at 3:43 pm | Permalink

Cosmides and Tooby use this (Wason selection test) to support the brain modules idea in Evolutionary Psychology. Presented as logic puzzle, most miss it. But if you substitute an age and a drink, and ask which cards need to be turned over to see if someone is drinking under age, most get it right. C & T suggest we don’t have a “pure logic” brain module, but we have a highly developed “cheating detection” module.

• Posted April 16, 2018 at 12:17 pm | Permalink

On the other hand, see _Human Reasoning and Cognitive Science_, for a very interesting counterargument using a non-classical logic.

6. Acolyte of Sagan
Posted April 15, 2018 at 3:20 pm | Permalink

E and 2. It doesn’t say that no other letter can also have 5 on the reverse, so it doesn’t matter what is on the reverse of 5 or F. So, if either E has anything but 5 on the reverse, the statement is false, but it might have a 5 on the reverse, so turning the 2 is essential; if the reverse of 2 is E then the statement is false.
In short, if the reverse of E is 5, and the reverse of 2 is anything other than E, the statement is true.

• Acolyte of Sagan
Posted April 15, 2018 at 3:23 pm | Permalink

Whoops, superfluous ‘either’ in there. (So, if either E has anything but 5 on the reverse, the statement is false)

7. abram
Posted April 15, 2018 at 3:23 pm | Permalink

E & 2

I spent about 15-20 seconds on this and it struck me as so easy there must be some sort of dirty trick or something we are missing. And now I see other commenters arrived at the same answer and we’re probably all wrong and will feel very foolish tomorrow.

• Craw
Posted April 15, 2018 at 3:48 pm | Permalink

You got it right of course. My bet is that those who got it wrong will not feel the slightest bit foolish.

• BJ
Posted April 15, 2018 at 4:51 pm | Permalink

Yeah, I don’t see how one can come to any other conclusion. The last puzzle Jerry posted was much harder. Now I am convinced there’s some trick I’m missing, but I don’t see how that can possibly be.

• Posted April 16, 2018 at 3:47 am | Permalink

Not all! 😦 clearly I am not as logical as you…

8. Don
Posted April 15, 2018 at 3:24 pm | Permalink

I know the answer because I have seen it before, but I did get it right the first time I saw it. On the other hand, I was tipped off that the correct answer was not as completely intuitive as one might think so I thought it about it more closely than I otherwise might have.

9. Daniel
Posted April 15, 2018 at 3:25 pm | Permalink

Turning the E and finding something other than 5 would falsify the claim. Finding a 5 would not prove it though, because in another hand there could exist an E with a number other than 5. Turning the 2 and finding an E would falsify the claim. I don’t see how the rule can be proven.

• Don
Posted April 15, 2018 at 3:36 pm | Permalink

I think it proves it if there are no other cards or other hands, otherwise you are correct.

• Posted April 15, 2018 at 3:44 pm | Permalink

I think the unspoken assumption is that the four cards constitute the universe of cards so turning the E and 2 will suffice.

• Jimbo
Posted April 15, 2018 at 6:29 pm | Permalink

Yes, this. I find many of these brain teaser questions leave out critical information thereby leaving ambiguity and count solutions as incorrect by forcing a solver to make certain assumptions. If only 4 cards exist, that’s important. Variants of the Monty Hall Problem and the Trolley Problem run into these issues.

• infiniteimprobabilit
Posted April 15, 2018 at 7:42 pm | Permalink

Yes, certainly. I assumed only 4 cards, but it’s not explicitly stated.

Also, “only those cards that could decisively prove the truth or falsity” caused me some trouble.

Does it mean proof of either truth or falsity, one or the other, or does it mean proof of either? – what I mean is, the ‘E’ could prove the falsity but not the truth.

Does that mean “any card by itself”? If so, the answer is – no cards.
Or does it mean “cards taken together” – which is E and 2?

cr

• infiniteimprobabilit
Posted April 15, 2018 at 8:36 pm | Permalink

“Does it mean proof of either truth or falsity, one or the other, or does it mean proof of either?”

Ack! Damn confusing words. Try again…

Does it mean ‘proof of one option will do’, or does it mean ‘conclusive proof one way or the other’?

cr

• Posted April 16, 2018 at 1:48 am | Permalink

That’s the point. You can decisively prove the hypothesis false but not decisively prove it true. It’s meant to be a model of how science works where all theories are provisional.

10. ThyroidPlanet
Posted April 15, 2018 at 3:28 pm | Permalink

Marcus DuSautoy put this puzzle in his book Symmetry – good puzzle.

• Posted April 15, 2018 at 4:05 pm | Permalink

It is an old puzzle (circa 1966) from a cognitive psychologist named Wason. Known as the Wason Card Problem, although there would be many logical equivalents and probably some historical antecedent.

11. BobTerrace
Posted April 15, 2018 at 3:31 pm | Permalink

The 2 and the F. If they respectively have a B and a 6 then it shows that the letters have their ordinal number on the reverse.

• abram
Posted April 15, 2018 at 3:44 pm | Permalink

No, it doesn’t.

• freiner
Posted April 15, 2018 at 4:00 pm | Permalink

I just turned over my cards. Sure enough, the 2 had a B and the F had a 6. But the E had a 4 and the 5 had a picture of dachshund.

• Posted April 15, 2018 at 7:12 pm | Permalink

The answer is 42. As in, comment number 42. (I like how you questioned the other rule.)

• Posted April 15, 2018 at 9:44 pm | Permalink

I like a refutation with panache!

• infiniteimprobabilit
Posted April 15, 2018 at 7:44 pm | Permalink

That crossed my mind but, though it’s suggestive, it wouldn’t prove that scenario.

cr

12. Lambert
Posted April 15, 2018 at 3:32 pm | Permalink

The question is, ‘which cards”could” be decisive when turned over. Only the E fits the bill. If it does not have a 5 on the other side then the proposition is proved false. Nothing else works.

• Alan Clark
Posted April 15, 2018 at 4:06 pm | Permalink

What if the 2 has E on the other side? That would also prove it false.

13. Posted April 15, 2018 at 3:48 pm | Permalink

I wrote an piece in three parts about screening for latent disease that addresses the kind of logic one must apply to solve this puzzle except my discussion addresses matters of probability, not certainty as in NY Times puzzle:

https://www.skepticink.com/health/2013/12/21/predictive-value-screening-tests-phone-calls-guys-want-love-part-1/

https://www.skepticink.com/health/2013/12/31/predictive-value-screening-tests-phone-calls-guys-want-love-part-2/

https://www.skepticink.com/health/2014/01/19/predictive-value-screening-tests-phone-calls-guys-want-love-part-3/

The key (I think relatable) analogy to the screening test problem is:

The probability that a guy will call you if he loves you is not the same as the probability that a guy loves you if he calls you.

14. Posted April 15, 2018 at 3:53 pm | Permalink

E

15. Posted April 15, 2018 at 3:55 pm | Permalink

I would love to be able to participate, but learned back in grade school that I have no ability to solve such problems. I am a math dunce. Oh, if only I weren’t a bear of little brain!

16. John Frum
Posted April 15, 2018 at 4:07 pm | Permalink

If there are no identical cards in the set then you don’t have to turn any cards over to falsify the claim because you can already see the E and the 5 cards so neither of them can be E and 5.

17. littleboybrew
Posted April 15, 2018 at 4:07 pm | Permalink

The “F” card is not relevant, as the rule only says an E card must have a 5 on the opposite side. Turning over any of the other three cards would test the rule.

• µ
Posted April 15, 2018 at 5:15 pm | Permalink

Bingo! I agree.

• BJ
Posted April 15, 2018 at 8:40 pm | Permalink

The rule only requires that cards with an E have a 5, but it doesn’t require that cards with a 5 have an E, so the card showing a 5 doesn’t factor into the analysis.

18. Saul Sorrell-Till
Posted April 15, 2018 at 4:08 pm | Permalink

I came to the conclusion E and F, but only after thinking about did I realise I chose F because I thought to myself(quite cheerily without any part of my brain going ‘no Saul, that’s wrong’) that if you turn over the card with F on it and find an E that would falsify the original proposition. At no point did I think to myself that there was anything wrong with the logic of those steps, and in a sense I was right. If you turned over E and found an F on the other side it really would falsify the proposition. I was so pleased with having figured it out I forgot that one of the rules is that you can’t turn over a letter card and find another letter on the back. My brain just thought ‘that’s enough work for today, give me crisps’, and I went to the comments.

19. Posted April 15, 2018 at 4:08 pm | Permalink

There is the Indiana Jones solution. Just turn them all over.

20. Kyle B.
Posted April 15, 2018 at 4:16 pm | Permalink

E and 2.

I find this puzzle is easier to understand if it is not a card example, but a “real world” example.
(I’m sure I read this variation from some psychology experiment)

With “If a person is drinking alcohol, then that person must be over 21”, then I find it incredibly easy to see which people would need to be checked (i.e., what could falsify the rule).

You have 4 people drinking beverages (may or may not be alcoholic unless specified).

Person A is drinking alcohol. (Equivalent to E above)

Person B is 20 years old. (Equivalent to 2 above)

Person C is 40 years old. (Equivalent to 5 above)

Person D is drinking water. (Equivalent to F above)

You must check A (for age) and B (for beverage type) to see if it is true since B cannot drink alcohol. I’m not sure why I find this version much easier to understand than with cards since it is exactly the same argument. But I’m not at all tempted to test person C (5 in the above puzzle) in this scenario.

• nicky
Posted April 15, 2018 at 11:34 pm | Permalink

I think Tooby and Cosmides (?) used both examples to show how good we are at cheat detection.
The card game is a ‘difficult’ problem, the drinking one, albeit the same, ‘easy’. They propose the reason is that in the latter cheating is involved….

• nicky
Posted April 15, 2018 at 11:56 pm | Permalink

That was the whole purpose of their card ‘n beer problem, how our brains evolved certain ways.

• Kyle B.
Posted April 16, 2018 at 11:14 am | Permalink

Thanks. That’s probably where I heard about it, and seems like a good possible explanation for the difference.

• Posted April 16, 2018 at 12:19 pm | Permalink

And there is a competing one. See above.

21. Posted April 15, 2018 at 4:28 pm | Permalink

Unless “ones” (“which ones do you turn over?”) makes the plural obligatory, only the E.

22. freiner
Posted April 15, 2018 at 4:31 pm | Permalink

I used to use this problem (together with the drinking age variation) with math majors who were prospective math teachers. (Admittedly, this gave me a leg up in solving it here.) Apart from the logical reasoning aspect of it — which included using counterexamples to respond to faulty conjectures — it also showed that even though two problems might be the “same” up to isomorphism, this did not mean they were necessarily the same psychologically. Also, they did pretty well with the logic puzzle version, they I never took score.

• freiner
Posted April 15, 2018 at 4:32 pm | Permalink

“though” I never took score

23. Posted April 15, 2018 at 4:31 pm | Permalink

Just E

• Posted April 15, 2018 at 5:01 pm | Permalink

No. If you turn the E and it is a 5 you cannot know if the statement is true without also turning the 2 to make sure it is not an E.

24. Posted April 15, 2018 at 4:40 pm | Permalink

Any letter can appear on the reverse of the 5 without proving or disproving the rule, and any number can appear on the reverse of the F without proving or disproving the rule.

But if there is any number other than 5 on the reverse of E, the rule is disproven, and if the is an “E” on the reverse of the 2, the rule is disproven.

Turning over just E doesn’t rule out the possibility that E also appears on the reverse of 2, so you have to turn them both over.

• Steve Pollard
Posted April 15, 2018 at 4:50 pm | Permalink

My reasoning too.

But I “saw” the answer almost at once. I had to carry out the above chain of reasoning to justify it, which took a bit longer. Interesting.

25. BJ
Posted April 15, 2018 at 4:44 pm | Permalink

Turn over 2 and F first. There could be an E on any of those. If you turn over the E and it has a 5, that proves nothing except that an E sometimes might have a 5. If you turn over the 2 and F and either of them has an E on the other side, that proves the claim false. Turning over the 5 and finding an E proves nothing.

If neither the 2 nor F have an E on the other side, there’s nothing left to be done. Turning over the E and finding a 5 only proves the “sometimes” idea, and turning over the 5 and finding an E proves nothing.

• BJ
Posted April 15, 2018 at 4:44 pm | Permalink

Wait, I didn’t read it properly AT ALL. That was terrible on my part!

• Posted April 15, 2018 at 4:59 pm | Permalink

I think we made the same mistake. I thought these were four cards out of a pack of some arbitrary number – in which case it would actually be impossible to prove there isn’t a card with E on one side and a number other than 5 on the other. But reading it again are there only four cards in all? In that case it’s E and 2.

• BJ
Posted April 15, 2018 at 6:34 pm | Permalink

Nah, I wasn’t paying attention and thought you could have cards with any letter or number on either side. So you could have an F with an E on the back. I only skimmed the problem and failed to understand that it’s always a letter on one side and a number on the other.

Though the problem does need to be explicit that these four cards are the only cards that exist in the world of the problem. The way the problem is written suggest that there might be other cards outside these four (in which case nothing can be proven by any of these four), but that’s clearly not the intention.

• BJ
Posted April 15, 2018 at 8:38 pm | Permalink

” The way the problem is written suggest that there might be other cards outside these four (in which case nothing can be proven by any of these four)…”

If the 2 has an E, it proves the rule false regardless of whether there are other cards outside the four presented.

• BJ
Posted April 15, 2018 at 4:49 pm | Permalink

OK, turn over the E and the 2. If the E has any number but 5 on the back, you’ve disproved the claim. If the 2 has an E on the back, you’ve disproved the claim. The other two cards aren’t relevant.

• infiniteimprobabilit
Posted April 15, 2018 at 7:53 pm | Permalink

Ah, but neither card can ‘prove’ the rule true. And the wording was ‘prove the truth or falsity…’
So does that wording demand proof either way, or does proving one contingency (but not the other) satisfy the requirement?

The devil is in the wording, isn’t it?

(Actually, the two cards, *taken together*, could prove the truth; since the 5 and F have nothing to say on the subject).

cr

• BJ
Posted April 15, 2018 at 8:35 pm | Permalink

No, neither can prove it true (unless the problem is taken to mean that the rule applies only to these four cards), but they can prove it false, if they turn out to demonstrate the properties I suggested. And, since the problem asks which cards you turn over to prove the “truth or falsity” of the rule, well…

• BJ
Posted April 15, 2018 at 8:36 pm | Permalink

And, if the problem is indeed meant to suggest that it applies only to these four cards, then flipping the two and not finding an E, and flipping the E and finding a five will prove it true as well.

• BJ
Posted April 15, 2018 at 8:37 pm | Permalink

But both have to happen to prove it true (again, also assuming that the problem applies only to these four cards).

Like I said in another comment, the problem needs to be written better, but it still works well enough as it is.

• infiniteimprobabilit
Posted April 15, 2018 at 8:44 pm | Permalink

Yes.

It’s all in the way the problem is worded (which is somewhat ambiguous, or at least, doesn’t fully specify the conditions.)

I struggled with that ‘prove truth or falsity’ too – would either one suffice, or was conclusive proof either way required.

cr

26. Betty Meiners
Posted April 15, 2018 at 4:45 pm | Permalink

Turn over the card showing a 2 and the card showing the F.

27. phoffman56
Posted April 15, 2018 at 5:00 pm | Permalink

Didn’t look; I’ll leave it a day before doing so, except for replies to this. Here I’ll just be a fussbudget:

Firstly the question is overly verbose and slightly ambiguous.
It probably means to decide the truth or falsity of the statement:
‘All E-cards are 5-cards’
But it could possibly mean to decide about:
‘There is at least one E-card which is a 5-card’

I suppose the use of the word “rule” makes it likely the first, universally quantified, statement–rather than the other one, existentially quantified.

And I suppose part of the trick is to seduce the reader into thinking it means

‘E-card if and only if 5-card’; whereas it seems clear it’s the “only if”.

Secondly, nothing is said about whether in effect you do all your turns simultaneously; or whether you turn one, see what you get, then decide whether (and which) you do a second turnover, etc. I haven’t thought about it enough to know whether there is any real difference there. I won’t spend the time for fear of embarrassing myself by assuming the wrong interpretation.

As a 16 year-old, I tended to feel not so good at solving this type of combinatorial problem. Only later I realized that I tended to think of too many possibilities for different meanings in those low level math textbooks with stuff about permutations and combinations. And the text writers were quite loose on this: they tried to be ‘nice’ at English rather than being perfectly accurate in their logical expressions.

You got the marks on the math exam partly by remembering exactly what a rather inexact question always means.

People have trouble with probability sometimes because of ambiguity in what they are asked.

It will be interesting if there is some disputation in replies here which is caused by some such ambiguities.

• Posted April 15, 2018 at 6:08 pm | Permalink

“You got the marks on the math exam partly by remembering exactly what a rather inexact question always means.

People have trouble with probability sometimes because of ambiguity in what they are asked.”

I agree, and this is something I always try to remind myself of when teaching. Inevitably though, what is obvious to the author will sometimes be ambiguous to the reader. I always encourage students to either ask for clarification or, if this isn’t practical, make their interpretation of the problem explicit in their solution. Mathematics should be about sound deductive reasoning and critical thinking; not guessing at what someone else means.

• infiniteimprobabilit
Posted April 15, 2018 at 7:57 pm | Permalink

“Only later I realized that I tended to think of too many possibilities for different meanings in those low level math textbooks with stuff about permutations and combinations.”

Yes!!!

It can be quite diabolically difficult to express in words, a number of alternatives, and whether they’re mutually exclusive; or just what constitutes the inverse of a proposition.

cr

• Garnetstar
Posted April 16, 2018 at 12:05 pm | Permalink

I agree with those comments. This again illustrates the inherent ambiguity of texts, and falsifies the claim that answering questions on, say, an IQ test, does not depend on one’s interpretation. Even the text “See Dick run” is subject to interpretation.

I had trouble with the phrase “could decisively prove”. What does the author mean by that?

To turn over the E requires no interpretation of that phrase: that would “decisively prove” whether the given condition is true or not.

It is the 2 that requires interpretation of the question’s text. If 2 has an E on the other side, that “decisiely proves” that the condition is false. But, if the 2 has any other letter on it, that does not prove the statement either true or false.

So, what does the author mean by asking “could decisively prove”? Do they mean “what cards being turned *will* give information about the statement’s truth or falsity? Or does it mean “what cards being turned *have the possibility* of giving us information about the statement’s truth or falsity, but might also not provide any information about the statement’s truth of falsity”?

I, being from probably a similar culture as the author of the question, would (probably) look at the “could” as the deciding word, and take that to mean the second interpretation is what the author meant, as did most of the commenters here. “Could” in English often signifies possibility, not definiteness. Although, it also can signify only possible *action* (turning the card), not possible proof. However, others may not read the “could” as strongly as the “decisively” which, confusingly, signals the opposite of the “possible” meaning of “could”, and leans one to choose the first interpretation I wrote above.

So, getting the “right” answer depends wholly on your interpretation of the author’s text and on your best guess as to what they intended. Thus again demonstrating that all questions rely to some degree on such factors, and not wholly on innate intelligence.

• Garnetstar
Posted April 16, 2018 at 12:28 pm | Permalink

Note too that merely changing the word order would make readers lean towards a different interpretation and a different “right” answer. If the question had been written “could prove decisively”, it would be widely interpreted that only cards which would *definitely* prove or disprove the statement are being asked for. The answer that commenters would lean toward would then be E only.

• phoffman56
Posted April 16, 2018 at 12:41 pm | Permalink

Let me give an ‘as precise as reasonable’ proof of the most likely precise version of the problem:

Prove that:

(1) With 2 turns, one may always determine for certain the truth or falsity of the statement ‘Every E-card is a 5-card’. (Here of course by “card” we mean exactly the 4 cards as described.)
(2) With less than 2 turns, the determination above cannot be made.

Proof of (1):
Turn the E-card and the 2-card (as I say in 29., and as do many others). If and only if both the E-card is a 5-card, and the 2-card is not an E-card, then it should be clear that the statement is true. Just think about each of the four cards separately. Surely no one will expect me to elaborate any more here.

Proof of (2):
Examine what happens in each of the four possibilities of turning a single card:
If you only turn the E-card, and if not 5 on the other side, the statement is false. But if a 5, you do not know whether the 2-card has an E on the back or not, i.e. whether the statement is false or true, respectively. So turning only that E-card will not necessarily decide truth or falsity.
If you only turn the 2-card, and if you get an E, the statement is false. But if not an E, you don’t know whether the E-card has a 5 on the back or not, i.e. whether the statement is true or false, respectively. So turning only that card will not necessarily decide truth or falsity.
If you only turn one of the other 2 cards, either argument in the two paragraphs above shows that turning only that card will not necessarily decide truth or falsity.

If you must, examine what happens when not turning any card, but surely that’s trivial enough to be ignored.

Finally as a joke, I note that 1 and 0 are the only relevant numbers less than 2 here. E.g. turning –17 cards or turning 1 ½ cards is not terribly meaningful!

I do think this is at most a high school level problem, but ONLY IF the darn thing is stated precisely, not the way it is done in NYTimes, as I blathered about above. And apologies for the tediousness of the proof, but a proof has to say what you are thinking. At least Fermat’s last theorem took a bit more!

• phoffman56
Posted April 16, 2018 at 12:45 pm | Permalink

The “–” should have sat in front of the 17, that is, a negative number of turns is presumably no contemplated by the poser (or is it poseur?) I was unlucky.

• phoffman56
Posted April 16, 2018 at 7:00 pm | Permalink

I said: “..not the way it is done in NYTimes,”
Actually, I think I’m unfair here, except for the absolutely infuriating use of the word “rule”. And Matthew just quoted Suri in NYT explicitly.
A rule is an imperative sentence, a command. It is not assertive/declarative. So calling it true or false as he does is just bad grammar/logic.
But of course he means: Decide whether or not the choice of the 4 cards followed the rule.
so I now think it’s not really ambiguous despite what I blathered about earlier. I’d fault the readers who misinterpret it more.

And implicitly it seems to be asserted that the order of turning is immaterial, which is correct. It is also implicitly asserted that there is only one minimal set of turnings which do the trick. Why not include those as part of the problem?

And I’ve always hated the impression math teachers often give that learning math is just learning the rules and how to fokllow them! Okay for 8 year olds and younger, so they become familiar with arithmetical examples. That’s now called mathematics. In my ancient day irt was called arithmetic.

• Garnetstar
Posted April 17, 2018 at 11:35 am | Permalink

This is a whole lot better, and thanks for it! You see the problem, which is that, to be as unambiguous as possible, the poseur (!!LOL!) must add more and more words to explain his or her words! This sometimes leads to verboseness, as you say, or inelegant writing, and sometimes it seems to the poseur that the many words are obscuring the question more!

It’s just an inherent problem with language. But your phrasing is excellent!

28. Posted April 15, 2018 at 5:00 pm | Permalink

Is the table made of glass?

• Wonderer
Posted April 15, 2018 at 5:26 pm | Permalink

It doesn’t matter, because I glued all of the cards to the table with opaque glue. You won’t be able to find out what is on the other side anyway.

29. phoffman56
Posted April 15, 2018 at 5:15 pm | Permalink

On second thought, assuming it’s “all E-cards are 5-cards”, you turn the E and turn the 2 and it makes no difference which order or simultaneous.

But would make a difference if it was whether at least one E-card was a 5-card. But then answer is not unique.

30. Mark Sturtevant
Posted April 15, 2018 at 5:20 pm | Permalink

Just the card showing E, if I take the wording exactly. It must have a 5 for the statement to be true. The card with the 5 showing could have anything and it would not violate the statement, as said.

31. Posted April 15, 2018 at 5:21 pm | Permalink

Reblogged this on The Logical Place.

32. freiner
Posted April 15, 2018 at 5:22 pm | Permalink

The problem in this world isn’t that there is too much negative thinking or too little positive thinking. The sad thing is there’s hardly any contrapositive thinking.
So it’s a quick check to see “If E then 5” but it takes longer to come up with “If not 5 then not E.”

33. Jamescor
Posted April 15, 2018 at 5:43 pm | Permalink

E,2

34. Robert Bate
Posted April 15, 2018 at 5:45 pm | Permalink

Doesn’t seem you could prove or disprove the rule given the set up. The E could have a five but that doesn’t mean another E (there are other Es somewhere I assume)has to. Same logic with the 5 – it could have an E but another 5 could have a Q. Are there some other parameters of the puzzle missing?

• Robert Bate
Posted April 15, 2018 at 5:49 pm | Permalink

Ooops, I fell. Could not would.

• BJ
Posted April 15, 2018 at 6:36 pm | Permalink

That’s what I thought, but some people are saying that these four cards are the only four cards in this card-universe. There are no others but these four.

The problem really needs to be written better.

• BJ
Posted April 16, 2018 at 9:37 am | Permalink

Regardless of whether there are only these four cards or not, the 2 can still disprove the rule. If the 2 has an E, it disproves the rule.

35. Andy Nelson
Posted April 15, 2018 at 5:53 pm | Permalink

Just the E, it doesn’t matter what is on the other side of even the 5, nobody promised that would be E

36. Posted April 15, 2018 at 5:59 pm | Permalink

E and 2.

37. Nikos APOSTOLAKIS
Posted April 15, 2018 at 6:15 pm | Permalink

You need to turn E and 2. The reason for turning E is obvious, the reason for turning 2 requires a bit more thought: if you turn 2 and the other side is (say) A then that means your conjecture is false.

Of course there is no need to turn F, because the conjecture doesn’t say that the number side is 5 if and only if the letter side is E.

I think that this would be easier for mathematicians to understand, not necessarily because we are smarter but because we are trained to carefully distinguish between if and if and only. We are also trained to look for counterexamples to our conjectures.

• Craw
Posted April 15, 2018 at 6:21 pm | Permalink

No, it’s because we’re smarter.

• Posted April 15, 2018 at 7:44 pm | Permalink

if you turn 2 and the other side is (say) A then that means your conjecture is false.

I don’t think so

• BJ
Posted April 15, 2018 at 8:32 pm | Permalink

Yeah, you only turn over the 2 because it will disprove the statement if has an E on the other side.

• Nikos APOSTOLAKIS
Posted April 16, 2018 at 7:30 am | Permalink

Yep. I meant “If you turn 2 and the other side is E, your conjecture is disproved”.

This probably disproves Craw’s assertion.

🙂

38. Jake Sevins
Posted April 15, 2018 at 6:23 pm | Permalink

I first saw this in a Steve Pinker talk a while back and I gave it to my two sons (ages 11 and 13) who both instantly solved it. But I understand that “normal” people (ie, not readers of this blog) find it tricky and challenging.

• Thanny
Posted April 15, 2018 at 7:31 pm | Permalink

Part of that may be the ill fit between such logic puzzles and standard English.

Consider the phrase, “that animal has two legs”. If you’re going to be strictly logical, you will say that a cow has two legs, because it has four legs total, and two legs fits into four legs.

But in English, “that animal has two legs” means “that animal has exactly two legs”.

That’s because there are many phrases which implicitly include clarifying words that are omitted due to being commonly understood. English speakers trend towards frugality in the currency of both words and syllables.

Another example is the answer, “I have ten dollars”, which means two different things depending on what question was asked. If it’s “The cover charge is \$10; do you have it?”, then the answer means, “I have at least ten dollars on my person”. If it’s “How much money do you have on you?”, the answer mans “I have exactly ten dollars on my person”.

So when an adult fluent in the language reads the puzzle, it’s no leap at all to insert some implicit qualifiers, such as making “if” really mean “if and only if”, which mandates checking the card with a “5” on it.

Children are less versed in these linguistic shortcuts, so are more likely to read things literally as written, with no assumptions.

• infiniteimprobabilit
Posted April 15, 2018 at 8:23 pm | Permalink

Yes.

Assumptions – not explicitly stated in the question – were:
There are just 4 cards (not a pack)?
Turn over cards that *might* (not necessarily will) prove…
And if a card can ‘prove false’ but not ‘prove true’ – do you turn it over?
And, do you turn over only cards that *taken individually* prove something, or do you turn over cards that *taken together* prove it?

cr

• phoffman56
Posted April 16, 2018 at 7:14 pm | Permalink

“If you’re going to be strictly logical, you will say that a cow has two legs…” :

I’d say, if you’re going to be logical, you make sure your English, if you use that, can be very directly translated into a symbolic (mathematical is all that means) language.
So you will say,
‘For all cows, its number of legs is at least 2’, or
‘For all cows, its number of legs is at most 2’ , or
‘For all cows, its number of legs is exactly 2’.
All of these are false, of course, recalling that occasionally a cow is born alive but with severe genetic problems.

39. Jim
Posted April 15, 2018 at 6:35 pm | Permalink

E and 2.

40. Karst
Posted April 15, 2018 at 6:48 pm | Permalink

Turn over E and 2.

I have seen other versions of this in several books on reasoning.

41. Posted April 15, 2018 at 6:54 pm | Permalink

E and 2.

42. Posted April 15, 2018 at 7:03 pm | Permalink

“cards are laid in front of you, each of which, it is explained, has a letter on one side and a number on the other.”

Do we get to test this rule too, or do we have to take this one on faith? 😉

43. infiniteimprobabilit
Posted April 15, 2018 at 7:33 pm | Permalink

As formulated, there is no restriction on the number of cards you can turn over.

Also, it is not clear whether the statement relates to just those four cards, or they are taken from a pack. I’ll assume it’s just those four.

It says an E must have a 5, but not that all 5’s must have a E.

Umm, “those cards that could decisively prove the truth or falsity”.

You must turn over the 2 and E – or, none, depending how you interpret the sentence I quoted.

If the E has a 5 the other side, that is compatible with (but does not prove) the rule; any other number disproves.

If the 2 has an E the other side, that would decisively disprove the rule.

So either card could disprove the rule; neither (alone) could prove it. But the two, taken together, could prove it.

(The 5, whether it has an E the other side or not, is neutral – i.e. not incompatible with the rule.
The F is irrelevant whatever is on the other side)

My brain hurts.

cr

44. Dagoberto Flores
Posted April 15, 2018 at 8:20 pm | Permalink

The E card.

If the other side is 5, the claim is true
If the other side is other than 5, is false.

The rest of the cards are irrelevant. “If one side is E the other must be 5” but that doesn’t imply that being 5 on one side, the other must be E.

• Posted April 16, 2018 at 9:18 am | Permalink

What if there is an E on the other side of the 2 card, after finding a 5 on the other side of the E card?

45. Kevin Damron
Posted April 15, 2018 at 8:32 pm | Permalink

Why would you turn over any card except the card marked “E”? If there is a “5” on the back then the statement is true. If not, it is false.

46. eplommer@icloud.com
Posted April 15, 2018 at 8:48 pm | Permalink

All four of them: that’s the only way.

• nicky
Posted April 16, 2018 at 12:03 am | Permalink

check post 20.

47. Scott Little
Posted April 15, 2018 at 8:51 pm | Permalink

E & 5, duh…

48. Loosh64
Posted April 15, 2018 at 8:56 pm | Permalink

Turn over the first two cards. If the “E” has a number other than 5, the rule has been disproven. If the “2” has an “E” on the other side, the rule has been disproven. The other two cards contribute nothing and can be ignored. Note that I am assuming that these four cards are the whole set covered by the rule; if the “E” card reveals a “5” *and* the “2” card reveals a letter that is not “E”, then the rule has been proven for this set of cards.

49. infiniteimprobabilit
Posted April 15, 2018 at 9:08 pm | Permalink

Of course another assumption that could be made, is that each letter or number only occurs once.

In which case you don’t need to turn over any cards, the rule is disproved immediately. 🙂

cr

50. Posted April 15, 2018 at 9:23 pm | Permalink

unsolvable without sampling the entire universe of cards, me thinks. There may always exist an “exception to the rule” unless it is a universal law’ or a finite universe of cards.

• infiniteimprobabilit
Posted April 16, 2018 at 5:34 am | Permalink

There’s an unspoken assumption that these 4 cards are all there is. Otherwise, they could disprove the rule but never prove it.

cr

• Craw
Posted April 16, 2018 at 2:41 pm | Permalink

No. Read it again, more carefully. Prove or disprove. You are only addressing prove.

51. JonLynnHarvey
Posted April 15, 2018 at 9:26 pm | Permalink

Turn over all except the 5.
You must establish that there is NOT an ‘E’ on the non-fives, and that there is a 5 on the other side of the E. A non-E or E on the other side of the 5 is OK, so don’t flip that one.

52. Dave meyers
Posted April 15, 2018 at 9:47 pm | Permalink

E and 5 SAME ANSWER EITHER THE STATEMENT IS TRUE OR FALSE NOT BOTH

53. another fred
Posted April 15, 2018 at 10:08 pm | Permalink

E & 2.

Turning over the F proves nothing as the presence (or not) of a 5 means nothing to the statement, likewise turning over the 5 proves nothing as the statement does not say that 5s can only be obverse to Es.

The 2 must be turned over to prove that an E is not obverse and the E must be turned over to prove that there is a 5 obverse.

• another fred
Posted April 15, 2018 at 10:15 pm | Permalink

Further to the above, if one turns only the 2 and finds other than an E the truth is not proven, likewise, if one turns only the E and finds a 5 there still may be an E obverse to the 2, therefore both must be turned.

Posted April 15, 2018 at 11:04 pm | Permalink

Turn over the E card. If it is not 5 you are finished. The proposition is false. If is is a 5 you are not finished. This is consistent with the proposition being true but does not prove it.

Turn over the 2 card. If it is E you have found a counter example, the proposition is false. If it is not E then it is irrelevant to the truth of the proposition. The proposition is now proved to be true because neither the 5 card or the F card can be counterexamples, they are irrelevant. In other words, you have demonstrated that there is NO CARD with an E and a number other than 5

Bottom line: the statement that every E card has a 5 is NOT the same as the statement that every 5 card has an E.

55. David
Posted April 16, 2018 at 12:21 am | Permalink

Turn over the E

56. Posted April 16, 2018 at 1:35 am | Permalink

The E, as a non 5 is immediate disproof.
The 2, as an E is immediate disproof.
If neither of these negates the assertion, then the assertion is definitively true, irrespective of the other 2 cards. The reason is that the other cards do not speak to the assertion at all. Anything could be on the other side of an F including (but irrelevantly) a 5. Anything could be on the other side of a 5, including an E, but importantly an E is not required by the assertion.

57. Posted April 16, 2018 at 2:57 am | Permalink

This is an old problem – turn over cards E and 2. If the E has anything other than a 5 on the other side then the claim is falsified, and also if the 2 has an E on the other side the claim is falsified.

58. Posted April 16, 2018 at 3:30 am | Permalink

Turn the E and the 2 cards.

The rule says nothing about 5 being paired with another letter so that’s of no interest. The rule says nothing about F either, so that can be ignored too.

• Posted April 16, 2018 at 3:41 am | Permalink

Ah!

59. Posted April 16, 2018 at 3:40 am | Permalink

I would start with 2 – if there is an E on the other side then the rule is wrong. If not, ie any other letter, we turn the 5 over next.

On the other hand, why not just turn the E & the 5?

In a sense there is too little information – are we supposed to think that E is the 5th letter of the alphabet, so 2 would be b & f 6?

I have come across these types of puzzle before… & it really is not my strength, if I have any 😦

60. Mark Ayling
Posted April 16, 2018 at 4:42 am | Permalink

We are testing the proposition E => 5 and by implication also ¬5 => ¬E (“=>” is “implies”, “¬” is “not”). We should not test anything else, certainly not 5 => E, nor ¬E => ¬5.

Firstly, the ‘E’ must be backed by a ‘5’ or else that would disprove the rule, so I would turn over the ‘E’. I would also turn over the ‘2’, because that must NOT be backed by an ‘E’. And that’s all.

61. Posted April 16, 2018 at 5:17 am | Permalink

E and 5 the rest are irrelevant.

62. Valeriu
Posted April 16, 2018 at 6:08 am | Permalink

By turning E and 2 I would prove the truth or falsity of the rule “If there is an E on one side, the number on the other side IS a 5,” assuming these are all the cards in the Universe. I cannot prove the rule “If there is an E on one side, the number on the other side MUST BE a 5.” Necessity cannot be proved by turning the cards.

63. Gil
Posted April 16, 2018 at 7:10 am | Permalink

E and 5. The other two cards cannot prove anything.

64. Mark Sordahl
Posted April 16, 2018 at 7:55 am | Permalink

Turn over the cards marked 2 and f. If they have 5 or e on the other side, then the rule is disproved.

65. J. Quinton
Posted April 16, 2018 at 8:03 am | Permalink

People handle this sort of question better when you personalize the cards.

So instead of “E has to have a 5 on its back” you change it to “You’re a police officer in a bar. People with drinks in their hand have to be 21 and over” then a lot more people get the correct answer.

So let E be “has a drink in their hand” and 5 is “21 and over”.

66. chaorex
Posted April 16, 2018 at 8:10 am | Permalink

First turn over 2. If it is not E, the claim still stands. If it is E, then the claim is disproven.
Second turn over E. If it is 5, the claim still stands, and cannot be disproven, if it is not 5, then the claim is disproven.

• Posted April 16, 2018 at 9:12 am | Permalink

Good strategy, but the claim has no prior standing. We’re using the cards to either prove or falsify the statement, with no presupposition that it is to be considered true and until proven false, or that it is considered false and until proven to be true.

67. SSE
Posted April 16, 2018 at 8:25 am | Permalink

All of them

68. Posted April 16, 2018 at 8:58 am | Permalink

I assume that these are the only four cards of this type, that there is not another 48 cards in a deck of such cards.

The Rule says “If there is an E on one side, the number on the other side must be a 5.” This does not exclude that a 5 might be on a card with another side, just if there is an E, then there is a five, so if the E card is turned over and there is no 5, then the rule is disproved. Turn over the 2, if it is an E, the rule is disproved, similarly, if the F is turned over and an E is found, the Rule is disproved.

If the E is turned over and a 5 is found, or the 5 turned over and an E is found, the rule is not proved, just instances of the rule are found, all of the rest of the challenges must be met.

So, the rule can be proven only by all of the cards meeting the rule and that all of the cards are the only ones to which the rule applies.

69. Posted April 16, 2018 at 9:04 am | Permalink

E2 and F.

• Posted April 16, 2018 at 9:07 am | Permalink

F only if letter/number has to be confirmed

70. Posted April 16, 2018 at 9:05 am | Permalink

There isn’t enough information on the 4 cards to conclusively prove or falsify the rule in all scenarios.

1. Cards: E|? 2|? 5|? and F|?

Flip E|?: If E|5 then go to 2.
else – falsified. Stop.

2. Cards: 5|E 2|? 5|? and F|?

Flip 5|?: If 5|E – proved. Stop.
else go to to 3.

3. Cards: 5|E 2|? W|5 and F|?

Flip 2|?: If 2|E – falsified. Stop.
else – inconclusive. Stop

71. Don
Posted April 16, 2018 at 10:23 am | Permalink

It is interesting how many people bring with them the assumption that the card universe is larger than the 4 cards on the table. Understandable I guess since we are used to decks of cards and who ever saw a card deck with only 4 cards.

I think that in these types of problems and in both science and engineering one should make as few assumptions as possible. One should also try to identify and understand the assumptions one is making. That, of course, is not always easy and probably the cause of many failed experiments and failed engineering projects.

• Posted April 16, 2018 at 10:39 am | Permalink

You have to make an assumption either way. These are the only cards or these are not the only cards.

• Don
Posted April 16, 2018 at 12:46 pm | Permalink

Is it an assumption to just go with the bare facts as stated, “Four cards are laid in front of you,…”? It is obviously an assumption to think the universe of cards is larger. But to just accept the facts as given is not really an assumption as such or so it seems to me. I guess you can say I am assuming that I have been give all the facts, but that is kind of my point that we should assume that as a start. I think assumptions should only be added as necessary until you have a solution or proved that no solution is possible. Now in science and engineering I think you keeping adding assumptions even after you have a solution to see where that takes you but you still start with as few assumptions as possible.

• Leigh
Posted April 16, 2018 at 12:36 pm | Permalink

Wouldn’t the solution be the same even if there were more cards. You would have to turn over all cards with an E to verify that a 5 was on the other side. You would also have to turn over any number card, other than those with a 5, since any of these cards having an E would falsify the rule. Does it matter if there are more than the four cards on the table? I don’t think it does. The solution does not change. What changes as you add more cards are the number of cards turned over, not the reason why you need to turn over a card.

• Posted April 16, 2018 at 12:46 pm | Permalink

Yes, as long as you know that the cards you see are all the cards there are.

72. Bob
Posted April 16, 2018 at 10:44 am | Permalink

Turn over the 2 and the F

73. Posted April 16, 2018 at 12:25 pm | Permalink

The original Wason test was done with actual cards, IIRC, so some of the “universe of cards”, etc. questions are solved easier than trying to render an imaginary situation in words.

As mentioned above, there’s a wonderful book (2011, IIRC) _Human Reasoning and Cognitive Science_ which suggests that the difficulty with the task is that we implicitly reason with a default logic, rather than a classical one. I don’t know whether this is the case or not. It is, however, interesting to contrast that with the Tooby-Cosmides-Pinker line (which the book is definitely aware of). The most important aspect of the book to me is the reminder that many cognitive psychologists work assuming “classical logic or we’re irrational”, for whatever reason, and that’s just a false dichotomy.

It is a shame that my old colleague from CMU, Horacio Arlo-Costa is no longer with us; it is the sort of book I think he would have adored.

74. Dean Dest
Posted April 16, 2018 at 1:03 pm | Permalink

Turn over the E to prove or disprove the existence of a 5 on the other side, and turn over the 2, which might disprove the statement (but cannot prove it). No other card(s) can lead to proof either way.

75. nicky
Posted April 16, 2018 at 2:22 pm | Permalink

It is an excessively simple problem as shown by Kyle B at 20.
As a card problem it is a ‘difficult’ one (as many of the posts above illustrate), as a ‘beer’ problem, a ‘cheating problem, I do not think anybody would have the slightest hesitation in a fraction of as second, the very point Cosmides and Tooby (?) were trying to make.

76. Wayne Robinson
Posted April 16, 2018 at 2:51 pm | Permalink

The ‘E’ and the ‘2.’

If there’s another number other than ‘2’ on the ‘E’ card, or the ‘2’ card has an ‘E’ then it disproves the statement.

The statement says nothing about other lettered cards having a ‘5’ or ‘5’ cards having other letters than ‘E’ so turning over the ‘5’ or the ‘F’ tells nothing.

77. Don
Posted April 16, 2018 at 3:44 pm | Permalink

No, the solution wouldn’t change, but if the 4 cards shown are only a subset of all the cards then you could falsify the rule but you couldn’t prove it. This is because there could be a card somewhere in the larger set that is not on the table that breaks the rule.

• Don
Posted April 16, 2018 at 3:45 pm | Permalink

Whoops, that was supposed to be in response to Leigh under #71.

78. Curtis from Texas
Posted April 16, 2018 at 4:03 pm | Permalink

I’m sure it has already been answered, but the E must be flipped (it must have a 5 on the other side) and the 2 must be flipped (it must not have an E on the other side).

The 5 is irrelevant (something like a tautology) and the F is irrelevant.

I guess the tricky part, if it is tricky, is that the rule is only concerned with cards that have an E on one side; a 5 by itself is not important and could have any letter on the other side. Interestingly, with a math and probability background, the answer was immediately obvious (assuming I have the correct answer).

79. Steve T
Posted April 16, 2018 at 6:46 pm | Permalink

e, 5; the other 2 might or might not be relevant

80. Rupinder Sayal
Posted April 17, 2018 at 2:24 am | Permalink

E and 2.