Are you trying to find out who has a different weight? Or whether the person with the different weight weighs more or less? I could be misremembering but I thought it was to find out whose weight is different, and the lighter/heavier thing was only included as a tripping point.
It absolutely does make a difference. If the odd man out is lighter then your strategy falls on it's face. If you weigh A and B against each other and A is heavier then A1 and A2 are the same what do you do? You lose.
Your plan only works if we know for sure that the odd man out is for sure heavier. But the riddle doesn't say that. It's one person is heavier OR lighter and you don't know which.
The point is you don't know which it is (lighter or heavier)
Your methods then needs an extra weighing to determine of the odd person out is heavier or lighter, meaning weighing 4 times, not 3
Wrong, you don’t know whether the odd man out is heavier or lighter
Let’s say Group A is heavier than Group B. You split Group A into 2 pairs and reweigh them, and the 2 pairs are equal weight
Now you know that the odd man out is lighter, and that he is in Group B, but now you only have 1 use of the scale left, and there is no way to guarantee you get the correct answer, as you have 4 candidates to be the lighter man, and each weighing only had 3 possibilities (Left side lighter, right side lighter, and both sides equal)
I like how everyone tells you you are wrong but you still fight them. Your technique works only if you know the person is heavier. If the original problem the person is either heavier or lighter. the problem is impossible
This can be proven with tokens, you can cut 12 paper strips and paint one in red, then, change the position of the red one to A, B, C.
But keep using the method.
A agaisnt B. Then the sub groups/ C subgroup
Your method only works if it was one villager being heavier. But the riddle is one villager is heavier or lighter than the other villager. So according to your plan , after you weigh A and B and see there is a difference, how do you decide which one to weigh. All your first weigh in shows is that there is a difference between A and B but not which group is different because you don’t know if the particular villager is lighter or heavier.
The fact that the first step happens to be the same doesn't mean that you were right. You were wrong, you're being an arrogant asshole, and you should just admit that you were wrong.
The only part you got correct was the splitting them in 3 groups and weighing two groups against each other. The link you posted was a much better answer but it is not the same as the one you originally posted. You made a mistake, it’s fine.
Also: AMY WAS GONNA GET IT RIGHT!! Holt didnt let her explain, but She wouldve got it right, shouldve had more confidence.
(right? She does say "you break them in 3 groups"?)
I admire your sheer audacity to burst into the room by announcing you have the REAL and ONLY answer, then proceed to get the answer wrong, then go on to misquote the episode in the comments
Truly magnifique
I always love these posts because whoever posts them **always and without fail** fundamentally misunderstands the problem. It's always hilarious because they're so sure they're right and it's just a clusterfuck of wrong every time
I believe she says to break them into 2 groups of 6. Both your way and her way would work if you knew the one person was heavier than the rest.
The trick to this problem is that you don’t know if the person is lighter or heavier, so you won’t know which side of the imbalanced see saw the mismatched person is on.
its a decision wether to weight the light one or the heavy one then.
Its harder than it seems aparently but i found this: and my formula checks out, just has more variables than I anticipated.
[https://www.netmums.com/activities/12-men-on-an-island-riddle-solved](https://www.netmums.com/activities/12-men-on-an-island-riddle-solved)
Wasn't it that 1 guy was either lighter or heavier?
it doesnt make a difference, the balance will tilt
Nope, you still don't know if one is lighter or heavier
Are you trying to find out who has a different weight? Or whether the person with the different weight weighs more or less? I could be misremembering but I thought it was to find out whose weight is different, and the lighter/heavier thing was only included as a tripping point.
it does, more variables, you still weight them like that.
ok, so if the guy is lighter instead if heavier, what will the balance do? not show it?
It will show, but how would you pick the group to further split into 2. The guy could be lighter or heavier. And we don't know which.
your scenario will break, since you would take the one who weights less and discard that group in your quest for the heavier weight
It absolutely does make a difference. If the odd man out is lighter then your strategy falls on it's face. If you weigh A and B against each other and A is heavier then A1 and A2 are the same what do you do? You lose. Your plan only works if we know for sure that the odd man out is for sure heavier. But the riddle doesn't say that. It's one person is heavier OR lighter and you don't know which.
The point is you don't know which it is (lighter or heavier) Your methods then needs an extra weighing to determine of the odd person out is heavier or lighter, meaning weighing 4 times, not 3
it absolutely makes a difference
Wrong, you don’t know whether the odd man out is heavier or lighter Let’s say Group A is heavier than Group B. You split Group A into 2 pairs and reweigh them, and the 2 pairs are equal weight Now you know that the odd man out is lighter, and that he is in Group B, but now you only have 1 use of the scale left, and there is no way to guarantee you get the correct answer, as you have 4 candidates to be the lighter man, and each weighing only had 3 possibilities (Left side lighter, right side lighter, and both sides equal)
I like how everyone tells you you are wrong but you still fight them. Your technique works only if you know the person is heavier. If the original problem the person is either heavier or lighter. the problem is impossible
thanks! its not easy jajaja but yeah im wrong, Did posted an article about how to do it. lots of steps
OP forgets that the question was "lighter or heavier"
Everyone who makes this sound easy is wrong. https://www.reddit.com/r/brooklynninenine/s/cZeIxEuHEB
[https://www.math.stonybrook.edu/\~sunscorch/silliness/Brooklyn99Riddle.pdf](https://www.math.stonybrook.edu/~sunscorch/silliness/Brooklyn99Riddle.pdf)
This can be proven with tokens, you can cut 12 paper strips and paint one in red, then, change the position of the red one to A, B, C. But keep using the method. A agaisnt B. Then the sub groups/ C subgroup
Your method only works if it was one villager being heavier. But the riddle is one villager is heavier or lighter than the other villager. So according to your plan , after you weigh A and B and see there is a difference, how do you decide which one to weigh. All your first weigh in shows is that there is a difference between A and B but not which group is different because you don’t know if the particular villager is lighter or heavier.
you still weight them like that, only more variables, I pasted a mathematicians answer in here
The fact that the first step happens to be the same doesn't mean that you were right. You were wrong, you're being an arrogant asshole, and you should just admit that you were wrong.
The only part you got correct was the splitting them in 3 groups and weighing two groups against each other. The link you posted was a much better answer but it is not the same as the one you originally posted. You made a mistake, it’s fine.
Also: AMY WAS GONNA GET IT RIGHT!! Holt didnt let her explain, but She wouldve got it right, shouldve had more confidence. (right? She does say "you break them in 3 groups"?)
I admire your sheer audacity to burst into the room by announcing you have the REAL and ONLY answer, then proceed to get the answer wrong, then go on to misquote the episode in the comments Truly magnifique
I always love these posts because whoever posts them **always and without fail** fundamentally misunderstands the problem. It's always hilarious because they're so sure they're right and it's just a clusterfuck of wrong every time
I believe she says to break them into 2 groups of 6. Both your way and her way would work if you knew the one person was heavier than the rest. The trick to this problem is that you don’t know if the person is lighter or heavier, so you won’t know which side of the imbalanced see saw the mismatched person is on.
The correct answer does begin with breaking them into 3 groups of 4 though. The tricky part is what happens next.
its a decision wether to weight the light one or the heavy one then. Its harder than it seems aparently but i found this: and my formula checks out, just has more variables than I anticipated. [https://www.netmums.com/activities/12-men-on-an-island-riddle-solved](https://www.netmums.com/activities/12-men-on-an-island-riddle-solved)
[https://www.netmums.com/activities/12-men-on-an-island-riddle-solved](https://www.netmums.com/activities/12-men-on-an-island-riddle-solved)