Marilyn Vos Savant’s website:
http://www.marilynvossavant.com/
Oh, I decided to check out Marilyn’s website and she still has a link up to this problem and a sampling of the responses she received. It’s a funny read:
http://www.marilynvossavant.com/articles/gameshow.html
A little logic puzzle that created a miniature sensation in the world of logic puzzles.
Really, what this problem does is highlight the limitations of logic because the two most common incorrect solutions people come up with are logical as far as they go but it is the precepts of the reasoning that leads to incorrect conclusions.
That’s why I thought I might want to make a second video explaining why the incorrect reasonings are incorrect but that doesn’t seem like it would be interesting at all, at least not to any other than someone like me. 8^)
Duration : 0:8:9
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this vidoe is awsome
this vidoe is awsome
Dammit. Back to the …
Dammit. Back to the drawing board. 8/
Coin: 1 2 3 4 5 6 7 …
Coin: 1 2 3 4 5 6 7 8 9 10 11 12
Weighing no.1: 123 vs 456 =no difference
This means that the false coin is hidden among the rest.
Weighing no.2: 789 vs xxx (x=coins you know are real)= no difference
Then, you know that the false one is either 10, 11 or 12, but you don’t know if it’s heavier or to light.
To find the false coin, you must weigh minimum 2 times more. (Thereby 4 weighings in total) Also, this is not right method. Nice try, though.
If weighing the …
If weighing the first 3 vs 3 coins shows no difference that mean the fake coin is in one of the other two stacks of three. That’s one weighing. Take those 6 coins (one of which is the fake) and weigh 3 of them against 3 coins that you know are real. ( Second weighing). Now you know which stack of 3 has the fake coin AND you know if the fake coin is heavier or lighter than the real coins. Now weigh two of the last three against each other. (Third weighing).
think about this: …
think about this: what if the first weighing doesn’t show any difference, how many weighings would you do in total to solve this problem? my calculation shows more than 3.
Now with the last …
Now with the last three coins, take one out and set it aside and weigh the last two coins and weight them against each other. If they weigh the same, they’re both real and the coin you set aside is the fake. If the two coins weigh different and you already know that the fake coin is heavier(or lighter) then you know which is the fake coin.
This is the …
This is the solution I came up with:
Make 4 piles of coins, 3 in each. Weigh two piles against each other. If they weigh the same, they’re all real coins. If not, one of the 6 coins is the fake. Now break the 6 coins with the fake into two stacks with 3 coins each. Weigh one of those stacks against three real coins to find out which 3-coin stack has the fake. Now you’re down to three coins and you know if the fake coin is heavier or lighter.
the puzzle goes …
the puzzle goes like this: you’ve got 12 coins, cointaing one false one. They’re all equal in appearance, but the false one differ in the weight: that is, it’s either too heavy or to light. You’ve got an weight scale (similar to the starsign libra). Within three measurements, where you weigh the coins against each other, you got to find the false one and determine whether it’s to heavy or to light. It’s as simple as that. gl.
what is the twelve …
what is the twelve coin puzzle?
randomly removing …
randomly removing one card from the second pool lowers the remain card in that pools probability to to 1/3 rd
i actually have one …
i actually have one hard one, if you’re interested. it’s called: the twelve coin puzzle”
yes. quite easy, …
yes. quite easy, huh?
If I understand …
If I understand this problem, the answer is 50% because there are only two possible results: Alan hangs, Boris and Charles do not OR Boris hangs, Alan and Charles do not.
This is not similar to the puzzle in the video because all three cards are in play as possible kings. In your puzzle, Charles is out of play.
let me ask you a …
let me ask you a similar question: the same question, but with two participants. Alan, Boris and Charles are in jail, and one of them is going to be hanged to death. They can’t talk to each other, neither can they know who is being hanged. Thus, Alan ask the guard who’s NOT being hanged, and the guard tells him Charles is not. When Alan further ask the guard what he told Boris on the same question, he once again mention Char. So what’s Alan’s probability to be hanged? And pls. excuse my enligsh
he never said that …
he never said that it was 3/3, but the king constitute 1/3 of the 3 cards, and thereby it’s 1/3 to find the card if it’s shuffeled randomly. So this is a probability problem.
I’m sorry but the …
I’m sorry but the problem is direct and the solution I give is correct. There is no other answer. If you come up with something else, you are making a mistake.
yes you should …
yes you should switch we did this in maths class
Why, she doesn’t …
Why, she doesn’t have the highest IQ if that is what you meant.
I adore Marilyn vos …
I adore Marilyn vos Savant.
the problem is full …
the problem is full of holes… it lets you ume that it’s a probability problem, but it’s really not. It’s more an intuition problem than a logical one.
there are three cards, not 3/3
similarly, if there were 9 cards and 3/9 of them were shown to be false, only then would it be a probability problem
The problem allows only ONE card to be shown false, not 1/3 of the 3/3.
yes you should …
yes you should switch, prob of winning if you switch is 2/3..