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The Wason Card Problem. I'd be interested to know whether you pick the right cards, and if you agree with how the solution is reasoned out.
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With the proviso that I didn't realize that those four cards are the only cards, yes and yes.
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The Wason Card Problem.
I'd be interested to know whether you pick the right cards, and if you agree with how the solution is reasoned out.
Ohmygod, gimmie an "a" for god's sake, gimme an "a".
Don't you, Hydra, understand semantics? 
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just so we're all on the same page, the correct answer is A and 7.
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On a different (contrary) page, the correct answer is conditional. Depending on the results, flipping either the A or the 7 card could prove the statement false. 
ÅΓª╥┐↕§
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if you agree with how the solution is reasoned out. Yes; I noted, however, that were the number of vowels only two, then either number would have done.
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Nice syntax, Jackie!
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if you agree with how the solution is reasoned out. Yes; I noted, however, that were the number of vowels only two, then either number would have done.
Of which we have no way of knowing it without turning over both numbered cards.
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Thanks, Anna, though I confess I don't get what warrants the compliment...
Thanks, Faldage--I had a feeling I was missing something somewhere, and re-reading the article helped me realize what it was: as many logic puzzles as I've done, I still fell into the trap: just because four does not have a vowel on the other side, it doesn't automatically mean that seven does. [smacking forehead e] (Now, if we'd been told that there were a total of two consonants and two vowels on those cards...)
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Thanks, Anna, though I confess I don't get what warrants the compliment...
... your succinct use of the subjunctive.
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This is the bit from the text which explains the correct answer. Quote:
I must look at the card with the vowel showing to find out what is on the other side because it could be an odd number and thus would show me that the statement is false. I must also look at the card with the odd number to find out what is on the other side because it could be a vowel and thus would show me that the statement is false. I don't need to look at the card with the consonant because the statement I am testing has nothing to do with consonants. Nor do I need to look at the card with the even number showing because whether the other side has a vowel or a consonant will not help me determine whether the statement is false.
In other words: A and 7.
I think this is a pretty tough problem! The solution depends on a very literal-minded interpretation of "Which card(s) must you turn over to determine whether the following statement is *false*?" which statement, to answer the problem correctly, should not be equated with proving that the statement is true.
In other words, it is not a "true or false?" problem; but a "prove this false" problem. Considering this fact, that "whether" in the statement is very unkind. It tempts you into thinking "whether or not" where "not" means "true" and therefore resolve the statement in your mind into the "true or false" dichotomy you are meant to avoid to solve the ******* problem.
Apparently, the problem is made difficult by the "confirmation bias": The fact that the brain prefers to try to prove something is true (and therefore not false) than false (and therefore not true).
Anway, it got me.
(Well done Faldage)
Last edited by Hydra; 10/26/06 02:30 PM.
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But again, [not to sound pompous or campy,] the answer should be A or 7, or possibly both, depending on the interim result. Disproving the statement may be possible with a single card. 
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The instructions are "Which card(s) must you turn over to determine whether the following statement is false?". Your solution is to "Which card(s) MIGHT you turn over...".
Turning over A and 7 must determine the outcome. Turning over either A or 7 might determine the outcome.
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But even turning over A and 7 may not prove the statement false, depending on what is revealed. For that matter it was not even stipulated that this was the complete card set vice a sampling from a larger pack [not an original observation at this point]. Given the latter possibility, 'might' is all there is to work with. So "A or 7" is correct if one or both of those cards falsifies the statement. The answer does not suggest being committed to selecting a particular card; only that one of them (if any) will be sufficient, despite any undue melodrama, to dispatch the villain.
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Four cards are presented: A, D, 4, and 7. There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false? "If a card has a vowel on one side, then it has an even number on the other side."
question: why would you interpret this to suggest that there are more than four cards?
suggested answer: to further muddy the situation, and to further muddy the situation only.
Last edited by tsuwm; 10/27/06 07:19 PM.
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Four cards are presented: A, D, 4, and 7. There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false? "If a card has a vowel on one side, then it has an even number on the other side."
question: why would you interpret this to suggest that there are more than four cards?
suggested answer: to further muddy the situation, and to further muddy the situation only.
Why would you assume that there are only four cards?
But irregardless of whether or not there are or aren't only four cards or not, only by the turning over of A and 7 do you have a hope of disproving the hypothesis.
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it's a classic philosophical/psychological test which seems to always have four "cards", as here and here .
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Quote:
it's a classic philosophical/psychological test which seems to always have four "cards", as here and here .
Or here?
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The whole thing is a long-winded approach to seeing if the examinee understands that the assumption "if a then b" implies the statement "if not b then then not a." And, the red herring "not a" (which does not prove "not b") is there to tempt the examinee.
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And, the red herring "not a" (which does not prove "not b") is there to tempt the examinee.
Not to mention "b" (which does not prove "a").
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" Four cards are presented: A, B, 4, and 7. There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false? "If a card has a vowel on one side, then it has an even number on the other side."OptionsA+ odd... yes the statement is false. A+ even...no you can't prove the statement false. B+ odd...not germane B+ even...not germane 4+ vowel...not germane 4+ consonant...not germane 7+ vowel... yes, the statement is false7+ consonant...not germane. Only one of the four cards MUST be turned over to prove the statement false. And the answer is "A". Not "A" and then maybe "7". Only the "A" has to be turned over; even if turning over the "7" is needed for the proof. Geez! I'd like to sell you guys a used car and a garage in the swamp.
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Or either "7" and then, maybe, "A", one. You got no call to claim that "A" is any way more special than "7".
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Oh yes I do, Faldage, but I'm in my cups tonight so I'll explain it to you monyanna*.  * " phonetically speaking."
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On second thought I might forget my incisive logic in the fog of the grey dawn of morning so I'll regale you with it now. as follows... Listen closely..."IF" (the man says) "the vowel is ..." Get it? The proposition (and question) is based upon that contention. Really Faldage, you MUST stop making up your own rules.
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And the logical equivalent* is "If the number is not ..."
*The logical contrapositive, to be specific.
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Listen closely..."IF" (the man says) "the vowel is ..."
Get it? The proposition (and question) is based upon that contention.
Really Faldage, you MUST stop making up your own rules.
The proposition is "If a card has a vowel on one side, then it has an even number on the other side." (BTW, that is a direct copy and paste. The phrase "the vowel is" does not occur even in the discussion much less the statement of the problem.)
"On one side" means if the card has a vowel on either side, not just the side that's showing. It does matter what's on the other side of the 7.
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Don't be like Faldage who would rather drink muddy water than admit that he is wrong. Instead listen carefully this one last time and you will become a more knowledgable man even if you are a woman. [Four cards are presented: A, B, 4, and 7. There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false? If a card has a vowel on one side, then it has an even number on the other side.]Ok, last chance... The question requires that you find a card to turn over that has a vowel on one side. Only card "A" has a vowel showing of the four. Card "4" (an even number) can sport any letter and still be consistant with the statement in question. Turning over card "4" can't prove or disprove the statement unless card "A" has been turned over first. Therefore you MUST turn over card "A" first in order to prove or disprove the statement. As in... Which card(s) must you turn over to determine whether the following statement is false?See that wasn't so hard, now was it?
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Oh, I don't mind admitting that I'm wrong WHEN I'M WRONG.
Assuming only these four cards you just might could turn card A over and find an even number. You haven't proven the hypothesis false. If you then turn over card 7 and find a vowel, then and only then have you proven the hypothesis false. And in this case turning over just the 7 card will be sufficient to prove it false and you can take your A card and paint it green for all I care.
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If a card has a vowel on one side, then it has an even number on the other side
If the 7 card has A, E, I, O, or U on the side that you can't see then there is a card which has a vowel on one side and an odd number on the other side which makes the statement false.
If the problem stated that these are the fronts of the cards and the statement said "If a card has a vowel on the front, then it has an even number on the back" you would be right, but it doesn't say that. Both sides are equally sides of the card
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There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false? "If a card has a vowel on one side, then it has an even number on the other side." _____________________________________________________________________ Options (1) and (2) (1) Has a 80% chance of proving the statement false. A + even number = cannot determine A + odd number = statement false (2) Has a 20% chance of proving the statement false 7 + vowel = statement false 7 + consonant = cannot determine Unless you are a mindreader you MUST choose "A" first in your effort to prove that the statement in question is false. So the answer is "A", and that's that. 
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Quote:
There is a letter on one side of each card and a number on the other side. Which card(s) must you turn over to determine whether the following statement is false?
"If a card has a vowel on one side, then it has an even number on the other side."
_____________________________________________________________________
Options (1) and (2)
(1) Has a 80% chance of proving the statement false.
A + even number = cannot determine
A + odd number = statement false
(2) Has a 20% chance of proving the statement false
7 + vowel = statement false
7 + consonant = cannot determine
Unless you are a mindreader you MUST choose "A" first in your effort to prove that the statement in question is false.
So the answer is "A", and that's that.
Careful there Milo, Thanksgiving is coming up. Someone might mistake you for a Thanksgiving Turkey instead of a Christmas one.
You've had it explained to you several times so I won't bother doing it again.
Meanwhile, behind door two is a bag of used dog food. Wanna change your mind about which door you choose?
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"You've had it explained to you several times so I won't bother doing it again." said Faldage.
__________________________________________________________________
Aw shoot, you got me there, Faldage, and I promise I won't bother you by asking you to explain it to me again. But one thing...
What part of 80% consonants and 20% vowels (not counting the "Y") is it that you don't understand?
Take your time...
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there you go again. there are exactly four(4) cards in evidence, no more.. no less -- of these there is one vowel, one consonant, one even number, one odd number on display.
you (re)do the math.
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Oh nooooooo! Not you too, Joe Friday? This is terrible.
Worse than being wrong is being the only one who is right.
Can any swinging chick or Dick, Tom or Harry, in this building help me out?
Look, this ain't rocket science, it's probability theory. "Basic" probability theory. Just enough to keep your own shirt for five minutes in a fast game of Mississippi Tont.
Yes Joe, I know that there are only four cards in the hole, but I also know that the odds are four to one against turning up a vowel in the flip.
Naturally the odds are even on in turning up an odd number or even number, ad infinitum.
Now take a deep breath and apply this information to the question at hand.
Please?
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>but I also know that the odds are four to one against turning up a vowel in the flip.
this is where you've totally, irrevocably screwed the pooch. the odds of turning up a vowel "in the flip" depend completely on the (perverted?) nature of the person who selected the card in the first instance.
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Quote:
>but I also know that the odds are four to one against turning up a vowel in the flip.
this is where you've totally, irrevocably screwed the pooch. the odds of turning up a vowel "in the flip" depend completely on the (perverted?) nature of the person who selected the card in the first instance.
That's my boy, Joe, you are completely right. But since we know not of the nature of his "perversion" we must employ the "laws of probability" in order to make the best guess.
And the best guess is to choose card "A" first in order to avoid having to guess a vowel at one in four odds.
Whew, now I feel better knowing that Joe is on my side.
Last edited by themilum; 11/02/06 01:34 AM.
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So, Milo. Now that you know door two has a bag of used dog food behind it, are you or are you not going to change your pick?
Besides, the only letters shown are A and B. As far as you know they're not even Latin alphabet. Might be Greek or Cyrillic. What does that do to your "probabilities"?
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Besides, the only letters shown are A and B.
The problem doesn't even require that there are only letters, numbers, or that there is only one thing on the back. For example, the 7 card could have a picture of some dog food, a blood stain and a star, "SEVEN", ... It could even be blank.
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calling Mr. Schroedinger.
formerly known as etaoin...
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The odds are irrelevant. The quest is to prove the statement false. So finding a vowel on the back of the 7 card 'cuts the mustard'. Even flipping the 7 to find 'DUMMY' on the other side would render the statement bogus, since there would indeed be a vowel there. What is "used dog food" anyway?  Would it be a sack of canine chow that has been opened, or some of its eaten contents heaved back up, or perhaps 'fully digested'? 
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Used is used. In one end and out the other.
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Quote:
The odds are irrelevant. The quest is to prove the statement false. So finding a vowel on the back of the 7 card 'cuts the mustard'. Even flipping the 7 to find 'DUMMY' on the other side would render the statement bogus, since there would indeed be a vowel there.
What is "used dog food" anyway? Would it be a sack of canine chow that has been opened, or some of its eaten contents heaved back up, or perhaps 'fully digested'?
Wrongo, Aramis11, the quest is to find...
Which card(s) must you turn over to determine whether the following statement is false?
As in the real world when we attempt to operate within a set of parameters we must accept the dictates of those parameters if we wish to participate; and the mechanics of this question envolves probabilities.
Trust me.
But don't trust Faldage. Faldage get's confused because (believe it or not) he has experienced so much during his long tour of this planet so that sometimes the past and present tend to blend together.
Take dogfood for example. Faldage is referring to a puzzle of serval years back, one that I had thought that he and Wordwind had resolved by experimentation in a sleazy bar to my satifaction.
It is a good puzzle, but one that Faldage is wrongly analoging to the unresolved puzzle of the moment.
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It is possible the determination cannot be made. If this 'real world' is limited to only the four cards, the statement could be shown not to be false. But one shot can sink the ship. If the first does, the second need not be assayed. In that case, "must you turn over" does not apply. Probabilites do not dictate the outcome. This is a conditional arrangement. It could have said, "Which cards should be selected to ensure the veracity of the statement can be determined?"
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If the universe is limited to these four cards it will always be possible to prove or disprove the hypothesis by turning over the A card and the 7 card. If these are but a subset of the cards available, it will be possible to disprove it but not to disprove it. If these are the full set, turning over either the A card xor [sic] the 7 card may disprove it but cannot prove it.
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I would choose to turn over the 'A' card, only.
Let's walk through the reasoning:
If I turn over the 'A' and there is an even number then the statement is NOT false.
If I turn over the 'A' and there is an odd number then the statement is false.
If I turn over the 'A' and there is another letter then the statement is false.
However, the same rule(s) apply to the '4' card (with respect to a vowel on the other side).
I believe (as I think has been implied in earlier posts) the puzzle is not succinctly described and there cannot be a definitive answer.
------
"I am certain there is too much certainty in the world" -Michael Crichton
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Yeah, ParkinT, but like in life we must play with the cards that we are dealt. So, rather, lets try this imperfect construct to the cruel semantics and see if it floats... A card: even (1 in 2) or odd (1 in 2) Odd number proves statenent wrong. An even number continues the search. 7 card: consonant (20 chances) vowel (5 chances) A vowel proves the statement wrong. A consonant continues the search. 4 card: a vowel proves the statement right because a "b" card has no bearing on the results. So you see, you MUST turn over the A card first unless you have time to waste which I guess we here all do or else we would be out building bridges or chasing girls or something instead of fretting over puzzles designed to be ambiguous.
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Quote:
A Game Show Puzzle
Imagine that you are on a game show, a curtain rises and the host shows you four doors behind it. The first two doors are closed, and have a letter on the front of the door. The first door shows the letter "A" and the second the letter "B". The third and fourth doors are open and you can see what's inside: inside the third door is a beautiful model holding a goat by a leash; inside the fourth door is another beautiful model, sensuously stroking a new car. Because the last two doors are open, you can't see what letter is on the outside of the door; don't assume that they are "C" and "D" simply because the other two start the alphabet!
The host of the game show says: "We have a rule on this show that if a door has a vowel on it then there is always a goat within. The question I have for you is whether the four doors that you see violate this rule. Of course, we could check the rule by opening every closed door and checking the front of every open door, but we would like to test the rule in the easiest way possible. If you can tell me the way to test the rule that checks the fewest doors, you will win the brand new car you see behind the last door!"
What will you say to the host? Recall that checking the first two doors means opening them to see what's behind them, and checking the second two doors means closing them to see what letter is on the front of the door. What is the smallest number of doors that you need to check, and which doors are they?
Source (and solution): The Fallacy Files.
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Quote:
Quote:
A Game Show Puzzle
Imagine that you are on a game show, a curtain rises and the host shows you four doors behind it. The first two doors are closed, and have a letter on the front of the door. The first door shows the letter "A" and the second the letter "B". The third and fourth doors are open and you can see what's inside: inside the third door is a beautiful model holding a goat by a leash; inside the fourth door is another beautiful model, sensuously stroking a new car. Because the last two doors are open, you can't see what letter is on the outside of the door; don't assume that they are "C" and "D" simply because the other two start the alphabet!
The host of the game show says: "We have a rule on this show thatif a door has a vowel on it then there is always a goat within. The question I have for you is whether the four doors that you see violate this rule. Of course, we could check the rule by opening every closed door and checking the front of every open door, but we would like to test the rule in the easiest way possible. If you can tell me the way to test the rule that checks the fewest doors, you will win the brand new car you see behind the last door!"
What will you say to the host? Recall that checking the first two doors means opening them to see what's behind them, and checking the second two doors means closing them to see what letter is on the front of the door. What is the smallest number of doors that you need to check, and which doors are they?
Source (and solution): The Fallacy Files.
What, Hydra? Are you illustrating the reasoning behind the topical Watson Puzzle with this modified hybrid example?
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