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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'?
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?
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.
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
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.
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.
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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