Conditional Reasoning, Skills in Deduction & Forming Proper Conclusions in this JFK Thing of Ours

[Steve Brodhay]

My understanding is to prove the statement is true/false. I say 2 cards. The E will show absolutely if the statement is false,but only support the statement may be true. The K can't prove it either way. The 7 needs to be turned to prove the odd numbers and vowels are true. The 4 can't prove it either way. Am I close?

[Steve Brodhay]

Sorry- I got the numbers mixed up. 4, not 7 has to be turned over. But it's still only 2 cards: E and 4.

[John Iacoletti]

This is an awesome puzzle. My answer is that you have to turn over everything but the 4. Some people are making the assumption that each card must have a letter on one side and a number on the other. That's not a given. The "K" could have a vowel on the other side. It's irrelevant what's on the other side of the 4 because if it's a vowel it confirms the proposition but if it's not it doesn't violate the proposition. You only need to look for falsifying evidence .

[Jon G. Tidd]

Glenn,

First, apologies for my weasling comment.

Second, my nature is to question the formulation of questions. Which brings me to:

Third, the formulation of your question. You write:

The condition is now established (true): 'If a card has a vowel on one side, then it has an even number on the other side.'

The problem is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true. How many and which card(s)?

Let's take each of your four cards. It turns out, each card must be turned over to see if the conditional statement is provisionally true for each of the four cards.

For example, the "4" card is turned over. The other side is "L". No conclusion can be reached.

For example, the "7" card is turned over. The other side is an "M". No conclusion can be reached.

For example. the "E" card is turned over. The other side is an "A". So we've got satisfaction of the conditional statement. But so what?

In the end, the conditional statement cannot be proved or disproved by your four-card example.

Edited August 11, 2015 by Jon G. Tidd

[Glenn Nall]

Glenn,

First, apologies for my weasling comment.

Second, my nature is to question the formulation of questions. Which brings me to:

Third, the formulation of your question. You write:

The condition is now established (true): 'If a card has a vowel on one side, then it has an even number on the other side.'

The problem is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true. How many and which card(s)?

Let's take each of your four cards. It turns out, each card must be turned over to see if the conditional statement is provisionally true for each of the four cards.

For example, the "4" card is turned over. The other side is "L". No conclusion can be reached.

For example, the "7" card is turned over. The other side is an "M". No conclusion can be reached.

For example. the "E" card is turned over. The other side is an "A". So we've got satisfaction of the conditional statement. But so what?

In the end, the conditional statement cannot be proved or disproved by your four-card example.

i'm back. first, i done forgot about that...

second - you're trying too hard (this is a famous study done decades ago by a couple of professionals. Their answer is correct, and it makes total sense once you hear it.).

read this again: The problem is SIMPLY to decide what are the minimum cards that need to be turned over to prove that the conditional statement is true.

another way it's been put is from this other version of the test:

You are shown a set of four cards placed on a table, each of which has a number on one side and a letter on the other side. The visible faces of the cards show 3, 8, M and O. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is a vowel?

[Glenn Nall]

good people - this is NOT MY test. I COPIED and PASTED the text into this thread, and I didn't change the answer in order to trick you.

From Wikipedia (I redacted the names so that no one cheats, and so that i get to use the word 'redacted' in a sentence):

"The [Doctor's Name Here] selection task (or four-card problem) is a logic puzzle devised by [Same Name Here] in 1966. It is one of the most famous tasks in the study of deductive reasoning. An example of the puzzle is:

You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red? (the problem stated in a different way)

A response that identifies a card that need not be inverted, or that fails to identify a card that needs to be inverted, is incorrect. The original task dealt with numbers (even, odd) and letters (vowels, consonants). <<< ...that's this one...

The importance of the experiment is not in justifying one answer of the ambiguous problem, but in demonstrating the inconsistency of applying the logical rules by the people when the problem is set in two different contexts but with very similar connection between the facts."

read more here:

https://en.wikipedia.org/wiki/Deductive_reasoning

[Glenn Nall]

This is an awesome puzzle. My answer is that you have to turn over everything but the 4. Some people are making the assumption that each card must have a letter on one side and a number on the other. That's not a given. The "K" could have a vowel on the other side. It's irrelevant what's on the other side of the 4 because if it's a vowel it confirms the proposition but if it's not it doesn't violate the proposition. You only need to look for falsifying evidence .

in fact, i think that it IS the case that each card contains both letter and number - in another version of this it so states. but this would not effect (affect?) the answer.

if you turn over K, you learn the same thing regardless what's on the other side - even or odd, letter or number, it neither proves nor disproves the postulate.

[Glenn Nall]

I am making an assumption that the cards have a letter on one side and a number on the reverse. This may not be true as it is not stipulated in the original premise.

Going on that assumption, you would have to turn over all cards except the K.

The E to prove there is an even number on the reverse, the 4 to prove there is a vowel on the reverse, and the 7 to prove there is a consonant on the reverse.

Of course, my original assumption may be in error.

Terry, I owe you an apology - you may be correct in that the cards each contain both a letter and a number. another version of this test says this is the case, but this version doesn't.

it doesn't matter, though. the answer is unaffected.

Edited August 11, 2015 by Glenn Nall

[Glenn Nall]

I don't mean to annoy anyone - i do feel that this is an appropriate (and fun) topic in each of us understanding more what's involved in the conclusion forming process which can either lead to errors in deduction or to progress in our pursuit of accuracy and truth in the solution.

one of the real reasons i'm into this thing so much is my passion for 'problem solving,' and i'm sure that's the case for many of ya'll. so these kinds of things are fun, and good for our brains (which we need to solve this thing!)

i'm just pasting in this little bit of text and this quick test i found (that I failed) without the answer. if any of you have seen it already, which is very likely, please don't publish the answer, or cheat.

so, check it out:

If...then...

Conditional reasoning is based on an 'if A then B' construct that posits B to be true if A is true.

Note that this leaves open the question of what happens when A is false, which means that in this case, B can logically be either true or false.

Conditional traps

A couple of definitions: In the statement 'If A then B', A is the antecedent and B is the consequent.

You can affirm or deny either the antecedent or consequent, which may lead to error.

Denying the consequent

Denying the consequent means going backwards, saying 'If B is false, then A must also be false.' Thus if you say 'If it is raining, I will get wet', then the trap is to assume that if I am not getting wet then it is not raining.

Denying the antecedent

Denying the antecedent is making assumptions about what will happen if A is false. Thus if you say 'If it is raining, I will get wet' and is not raining, I might assume that I will not get wet. But then I could fall in the lake.

Affirming the consequent

This is making assumptions about A if B is shown to be true. Thus if I make the statement 'If it is raining, I will get wet', then if I am getting wet it does not mean that it is raining.

The card trap

A classic trap was created some years ago;

Four cards are laid out as below:

The condition is now established (true): 'If a card has a vowel on one side, then it has an even number on the other side.'

The problem is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true. How many and which card(s)?

Discuss it among yourselves...

All four cards have to be turned over. The "K" could have an even number on the other side, and the "7" could have a vowel on the other side. To only turn over the "E" and the "4" would be to make an assumption about the other two cards.

Edit:

I changed my answer from four cards to two, the "E" and the "4", after looking more closely at the conditions. Our statement is only establishing a relationship between vowels and even numbers, and says nothing about consonants and odd numbers. In other words, consonants and odd numbers can have anything they want on the reverse side, and it will not affect the stated condition.

Edit:

On the other hand, if the "K" had an even number on the reverse, or the "7" had a vowel on the reverse, that would tend to invalidate the statement. I think I will go with four again.

almost, but not quite...

[John Dolva]

Glenn version one : "Four cards are laid out as below:
EK47.jpg
Given the condition: 'If a card has a vowel on one side, then it has an even number on the other side.'

version two : "Four cards are laid out as below:
EK47.jpg
The condition is now established (true): 'If a card has a vowel on one side, then it has an even number on the other side.'

"The problem is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true. How many and which card(s)?"



Mark ", the word "only" is not found in the statement we are dealing with.

The statement does not say that ONLY vowels will have even numbers on the reverse. The statement does not say that even numbers will ONLY have vowels on the reverse.

Those are ASSUMPTIONS that are not based upon our statement. They MAY or MAY NOT be true.

So turning over ANY card beyond the E would only tend to prove or disprove those ASSUMPTIONS, and not necessarily affect the statement we were given."



Nor is "all" or "any" mentioned anywhere.

"a card", "one side", "other side"

Both sides of "a card" are "one side" and both have a corresponding "other side". It doesn't matter if there is none or a hundred other cards. Each card is treated independently of each other card.

E and 4 are the only that can fulfill the condition. One or both *[or none] may be true.

2 cards, E and 4.

____________

The re statement to deal with cards with patches on them

"You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?"

requalifies (and thus recognises what I proposed) by saying "shows". Thta can be taken to mean the visible side. That is a restatement qualitively different from the original statement. *[in this case : one card, 8]

edit : typos, add *[ ]

Edited August 12, 2015 by John Dolva

[Paul Hailes]

John Dolve - but if you flip the 4 over and it has a vowel on the other side it only serves to confirm the original statement, not prove it. If it has a consonant on it, that does not invalidate the original statement because nowhere is it stated that consonant cards cannot have an even number on its opposite side.

[John Dolva]

Yes but the statement refers to "a card", not all or any cards with odd or even on one side. (I think this psych test shows I can be pedantic. But I already knew that.)

hmmmm..."The problem is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true. How many and which card(s)?"

minimum : one card, either card which has a vowel or an even number on a side, E or 4. , if that card does not prove the statement, then two cards, which may or may not prove it. which still would not invalidate it given another set.

[Robert Prudhomme]

Okay, I thought it over real hard, and it occurred to me we are only trying to prove one thing; whether or not a card with a vowel on it has an even number on the reverse side.

Therefore, only the "E" has to be turned over.

I thought at first the "4" would have to be turned over as well, to ascertain whether or not there was a vowel on the other side, but it also occurred to me there could be even numbers on cards outside of the vowel/ even number matchings.

Good trick, Glenn. One card.

P.S.

My wife looked at it for two seconds and had the answer. Beginner's luck?

Edited August 12, 2015 by Robert Prudhomme

[Mark Knight]

Robert, sometimes we condition ourselves to try to answer the question that wasn't asked.

It's a big part of being human. We all do it from time to time. Sometimes we need to slow down and ask ourselves, "Exactly what IS the question asking?" I know that defense attorneys prefer that their clients and their witnesses confine their answers to the questions that are specifically asked.

So sometimes the logic train jumps the rails when we try to answer more than we can possibly know from the data we have. [A certain Walker-did-it theory comes to mind here.]

[Paul Hailes]

John Dolva. There are 2 cards that can immediately DISprove the statement. The 2 are not the 2 you mention.

I think, respectfully, you are overthinking the original statement. I could be wrong, but it seems to me that Glenn has told us not to overthink, nor read too much into the problem.

Edited August 12, 2015 by Paul Hailes