# Can you solve the Pinocchio "green hats" logic puzzle?

Originally published at: Can you solve the Pinocchio "green hats" logic puzzle? | Boing Boing

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(A)

If Pinocchio has no hats, then the statement “all my hats are green” is (vacuously) true. Therefore, he must have at least one hat.

It’s a bit misleading, because the strongest conclusion that can be made from the two statements is that Pinocchio has at least one hat that is not green. But (A) definitely follows from the two statements, and the other four do not.

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I’m gonna go ahead and tick off the math folks and declare that a “vacuously true” statement is not true. It may be mathematically meaningless, but in the real world, if a statement is not entirely true, then it isn’t true. So even though A and C cannot be true simultaneously, Pinocchio’s statements do not allow us to narrow things further.

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As the video explains, this is a problem from a math contest. In that context, (C) is not false in the case Pinocchio has no hats.

In the “real world”, you might say (C) is misleading, but that’s only if you choose to read more into the statement than is actually there.

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Gotta love symbolic logic (seriously, logic proofs are like a better crossword), however… it feels a little unfair to call it a general ‘logic puzzle’ when a specific tool is needed to solve it.

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All we can conclude from “Pinocchio always lies” is that he is not standing or sitting.

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Feels like the answer should be “Pinocchio has at least one hat that is not green”, but it isn’t among the list of choices.
If the question was to select the answer that was MOST true, then I would say A, but that’s not the question.

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“A” is the correct choice only if one of the answers must be true. But that’s not a stated condition. As it stands, you can’t conclude anything from the two premises. None of the answers is provably correct.

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Isn’t one proper interpretation of “always lies” to negate “All my hats are green” to “Some of my hats, if I have any, are not green?” If so, I don’t see this ‘solution’ among the available options:

A) Pinocchio has at least one hat.
(B) Pinocchio has only one green hat.
(C) Pinocchio has no hats.
(D) Pinocchio has at least one green hat.
(E) Pinocchio has no green hats.

granted some of those are possible (including the empty set “(C)”) but none seem to fully describe the “always lies” as interpreted above [shrug] (the answer lies in set theory somewhere a bit over there)

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Good article on Pinocchio in the NYer - The Transformations of Pinocchio | The New Yorker

I know this one. After thirty days Pinocchio will leave the island.

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Yes, this seems a little poorly worded. Answer A as written doesn’t rule out him having hats that are only green.

I haven’t watched the video, but at the moment, we can only surmise that if Pinocchio always lies, that if he owns any hats, at least one of them is not green. But considering how people lie, it could be that it was a lie about owning any hats at all. Technically misleading, an attempt not to lie, but the intent makes it a lie. It all depends upon context.

“But all my hats are green, and not any other colour!”

“Oh, I have hats, and all my hats are green!”

Damnable liars, though, in reality they can’t be trusted to reliably lie all the time.

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I resorted to what I remembered of set theory: “All my hats are green” is (I think) equivalent to “The set of hats that are mine is a subset of the set of hats that are green.” But the empty set is a subset of every other set: therefore Pinocchio must have at least one hat.

Unless I am missing something, if you negate the universal statement, then you get an existential one:
¬∀x P(x) ⇔ ∃x ¬P(x)
So you need at least one hat that exists and is not green.

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I think the important fact we are missing is: “Did Pinocchio take Symbolic Logic in University?” Or the ever popular logic puzzle caveat that the individual in the problem will follow perfect logic.

I always found the negation of “all are” to “some are not” strange when the examples are common language, and the speaker would probably also consider having no hats as a possible negation.

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The best defence of this question would be to say that if Pinnochio is lying, then his statements must be testably false.

Put another way, in order to lie, you have to contradict something you know to be true. If Pinnochio has no hats, then he has no facts about his hat collection to misrepresent. So (A) is true because if it were false, he would not be lying, even though he also wouldn’t be telling the truth.

But yeah, posing this as a right-vs-wrong question hinges on the assumption that you will interpret it in an orthodox way. I feel like it rewards the homeworkly kind of intelligence over the kind that takes things apart from first principles.

It at least is an impressive demonstration of how many erroneous conclusions can be drawn from two sentences.

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I came here to say this tasted like a formal symbolic logic problem more than a math or philosophy problem. Man, I loved that class in college.

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“All my hats are green” is making two statements and claiming both are true. A clearer way to say it would be “I have at least two hats and each one is green.” To negate an “and” statement, either or both parts must be false, so if we know Pinocchio’s statement is a lie, we know one of a number of things are true:

I have one hat and it is green. Or,
I have one hat and it is not green. Or,
I have no hats and they are green (the “vacuously true” statement). Or,
I have no hats and they are not green.

With the information given, we can’t say anything more than that, so we cannot know that any of the conclusions offered are true.

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