General

@Aura0fDeath: Ah okay, and when I said flaming, I meant don't flame me for what I think a step consists of, not that I thought you were flaming me xD. The original riddle sounds hard, and I'm positive you can find the real answer somewhere by googling, just fyi.

@DragonHaQ: not flaming. and btw, just fyi, the original riddle (if i remember correctly) allows a TOTAL of 3 measurements, not 3 steps. So if a step takes 2 measurements, that counts for your 2/3. Which makes it much much harder =/

@Aura0fDeath: No, that basiler proposed 6 with 6, so it is impossible to tell which group has the abnormal ball because one group will be heavier and one group will be lighter, and the abnormal ball can also be either heavier or lighter than the rest of the balls. In the first step, your allowed to way all three groups. (I did not create this riddle, someone other guy did, so please don't flame me).

[quote=DragonHaQ]@Aura0fDeath: A step consists of comparing two groups on the balance.

You can tell which side is abnormal using three groups. If two groups both comprised of 4 balls are equal, than the last group will either be heavier or lighter than the two other groups. In your example is null using my first step.

@Inyurface: 's method is incorrect. Her first step is all luck. If you use two groups, how will you know which side to choose from? One group will be heavier, one group will be lighter o.O The rest of her answer is null.

Obviously @Peter1001: had the right answer, I even cited where he got his information on page 3 of comments. I didn't consider him a winner because he googled. No one on basil possibly has the brain capacity to answer so thoroughly and so distinctly.

EDIT: @Kinghippo7: A Riddle is "a question or statement so framed as to exercise one's ingenuity in answering it or discovering its meaning; conundrum," as according to dictionary.com... If your ingenuity was not exercised I can understand why this isn't a riddle to you.

@xxxPogo: I'm concerned on why you had to check o.O[/quote]

so ar eyou saying...in the first step, you can weigh all three groups of 4 against each other to determine which are equal, and if the abnormal one is lighter or heaver?

if you're only allowed one, then it's the same idea that @Inyurface proposed =/

@Aura0fDeath: A step consists of comparing two groups on the balance.

You can tell which side is abnormal using three groups. If two groups both comprised of 4 balls are equal, than the last group will either be heavier or lighter than the two other groups. In your example is null using my first step.

@Inyurface: 's method is incorrect. Her first step is all luck. If you use two groups, how will you know which side to choose from? One group will be heavier, one group will be lighter o.O The rest of her answer is null.

Obviously @Peter1001: had the right answer, I even cited where he got his information on page 3 of comments. I didn't consider him a winner because he googled. No one on basil possibly has the brain capacity to answer so thoroughly and so distinctly.

EDIT: @Kinghippo7: A Riddle is "a question or statement so framed as to exercise one's ingenuity in answering it or discovering its meaning; conundrum," as according to dictionary.com... If your ingenuity was not exercised I can understand why this isn't a riddle to you.

@xxxPogo: I'm concerned on why you had to check o.O

@Lillianaire: That would be another of the 3 weightings though.

@Aura0fDeath: That's what I've been trying to get at the whole time. I had to look it up too.

@DragonHaQ: I've heard this somewhere before, but it was you're only allowed to make 3 measurements....not just 3 steps. I'm not sure if in yours you're allowing multiple measurements per step.

Secondly, there's an issue with the solution, specifically steps 1 and 2. One side will always be heavier than the other whether the abnormal ball is lighter or heavier. How can you tell which side contains the abnormal one. For example, let's say the ball is lighter; that means that the abnormal side is actually lighter; if you take the heavier side and weigh those 2 v 2 or 1 v 1, it'll still be equal because the abnormal ball was in the lighter pile, and you've used up your 3 measurements/steps. I think the original riddle specifically stated that the ball was heavier, to rule out this issue, but I'm not 100% sure.

Lastly, the solution posted by @Inyurface also works;

1) seperate into two sets of 6. Pick the side that has the abnormal weight...let's say the ball is given that it is heavier. We then pick the set of heavier balls.
2) Seeperate the heavier set of 6 into two groups of 3. Again, take the heavier set.
3) Lastly, choose two at random and weigh them against each other. If they are equal, then the abnormal one is the one not weighed. if one is heavier, well, you got your answer.

Lastly, if the original riddle DOESN'T state whether the abnormal ball is heavier or lighter, then neither of these solutions will technically work. I'll google it since i'm interested now.

EDIT: Actually, @peter1001 has the correct answer...>_< his method can determine whichever is the abnormal ball, regardles if it's heavier or lighter.

I needa do-do

No sir, I only have 2 balls.

Omg your title made me laugh

Step one consists of two steps in itself... obviously, that basiler did not guess correctly within three steps.
EDIT: step two also has two steps in it also. So it takes a total of 5 steps.

...I'm on vacation, I don't want to think >.<

@zerosabor: I explained what a step consists of on Page 2 of comments.

[quote=elmospetrock]Not assumed knowledge haha. If the abnormal group from step 1 was lighter, then in the last step you would have to find the lighter ball of the two. Likewise for if the abnormal group was heavier. I guess you can just say that this problem has two solutions.[/quote]
No, you aren't getting what I'm saying. If the two 4's you weigh are equal weight, then [i]you don't know if the abnormal group from step 1 is heavy or lighter then the other two[/i] just that it must be different. You have no solution for the case of unknown weight.

[quote=Crescere]Am I the only one who had a "o___o" expression the first glance at the title?[/quote]

Hi there

[quote=LowWillpower]Yes, but if the two groups you weigh in the first step are not abnormal, you know the abnormal group, but you do not know if it is supposed to be lighter or heavier, which is something you use as assumed knowledge in the next step.[/quote]

Not assumed knowledge haha. If the abnormal group from step 1 was lighter, then in the last step you would have to find the lighter ball of the two. Likewise for if the abnormal group was heavier. I guess you can just say that this problem has two solutions.

[quote=elmospetrock]From step one, since only one of the balls is heavier or lighter than the rest, that means only one group out of three will be heavier or lighter than the other two. If the abnormal group was lighter, you would need to find the lighter groups for the rest of the equation. Same for if the abnormal group was heavier.[/quote]
Yes, but if the two groups you weigh in the first step are not abnormal, you know the abnormal group, but you do not know if it is supposed to be lighter or heavier, which is something you use as assumed knowledge in the next step.

@elmospetrock: Ah, in that case, congratulations again xD. The first step is probably the trickiest haha.

[quote=DragonHaQ]@peter1001: I never said it was incorrect. I just showed you how to citate your work using an URL (not a MLA citation)
@elmospetrock: Did you use the internet to solve this?[/quote]

I didn't use the internet haha. At first I thought you would have to split the 12 balls into groups of 6 and 6, but then I realized that you didn't know if the ball was heavier or lighter to begin with. I used 3 groups of 4 to improvise.

@peter1001: I never said it was incorrect. I just showed you how to citate your work using an URL (not a MLA citation)
@elmospetrock: Did you use the internet to solve this?

[quote=LowWillpower]How do you know if it is heavier or lighter. This seems kinda like the solution, but your first weigh could be 2 groups of the same weight, then you do not know if the last is heavier or lighter, just that it is odd.[/quote]

From step one, since only one of the balls is heavier or lighter than the rest, that means only one group out of three will be heavier or lighter than the other two. If the abnormal group was lighter, you would need to find the lighter groups for the rest of the equation. Same for if the abnormal group was heavier.

@peter1001: http://www.mathsisfun.com/pool_balls_solution.html
@vaelietta: Haha, good one...
@LowWillpower: Using three groups, two will be equal and one will be abnormal. He then splits the abnormal one.

[quote=elmospetrock]1)Make three groups with 4 balls in each group. One group will be heavier or lighter than the other two.
2)Split the abnormal group of 4 into 2 groups of 2. Now, going back to step one, if the abnormal group was heavier, you need to find the heavier group of 2. If the abnormal group was lighter, you need to find the lighter group of 2.
3)Weight each ball from the group of 2 separately.[/quote]
How do you know if it is heavier or lighter. This seems kinda like the solution, but your first weigh could be 2 groups of the same weight, then you do not know if the last is heavier or lighter, just that it is odd.

My IQ is only 159. I'm sorry, I can't solve it.

@ExSaber: No.
@CrazyChaosZ0: It step consists of moving ball/balls.
@NotUoid: Yeah, most likely...
@elmospetrock: Good job. We have a winner xD, though I hope you didn't use the internet.

[quote=Jimmy985]Lets say we are finding a heavier ball.

1.) Place 6 balls on each side, one side will be heavier. dispose of the 6 balls on the lighter side.
2.) Place the remaining 6 balls on the scale with 3 balls each side. Dispose of the 3 balls on the lighter side.
3.) Place 1 of the remaining 3 balls on one side of the scale and another ball on the opposing side. If they balance the remaining ball is the heaviest if one side is greater that is the heavier ball.[/quote]
You can't say whether you are finding a heavier or lighter ball. Weighing 6v6 tells you nothing. I looked up the solution online, and that isn't how you do it.

Throw each ball at a different member from a group of 12 identical kids' face and see which ball breaks more/less bones.

Lets say we are finding a heavier ball.

1.) Place 6 balls on each side, one side will be heavier. dispose of the 6 balls on the lighter side.
2.) Place the remaining 6 balls on the scale with 3 balls each side. Dispose of the 3 balls on the lighter side.
3.) Place 1 of the remaining 3 balls on one side of the scale and another ball on the opposing side. If they balance the remaining ball is the heaviest if one side is greater that is the heavier ball.

[quote=Inyurface]Step 1: Place half the balls on one side and half the balls on the other side. Whichever side is heavier/lighter has the odd ball.
Step 2: Now that you have 6 balls, place 3 on one side and 3 on the other. The heavier/lighter one has the odd ball.
Step 3: Place 2 of the 3 balls on the balance to find out which one is heavier or lighter. And then start concluding from there~[/quote]

This is exactly what I was going to say, but you don't know if you're looking for the heavier or lighter half. I think someone else on the first page got it though.

1)Make three groups with 4 balls in each group. One group will be heavier or lighter than the other two.
2)Split the abnormal group of 4 into 2 groups of 2. Now, going back to step one, if the abnormal group was heavier, you need to find the heavier group of 2. If the abnormal group was lighter, you need to find the lighter group of 2.
3)Weight each ball from the group of 2 separately.

What is defined as a "step"? Lifting them one by one could be considered one step, unless we're considering each individual movement but that'd be absurd because then you'd have to include picking them up, setting them down, etc. o.o

[quote=Inyurface]Step 1: Place half the balls on one side and half the balls on the other side. Whichever side is heavier/lighter has the odd ball.
Step 2: Now that you have 6 balls, place 3 on one side and 3 on the other. The heavier/lighter one has the odd ball.
Step 3: Place 2 of the 3 balls on the balance to find out which one is heavier or lighter. And then start concluding from there~[/quote]

Did he get it?

@DragonHaQ: Bleh

[quote=qtprincessxoxo]I'd use 7 of them to summon shenlong, I'd bow down, and ask him to find the abnormal one for me[/quote]
Dragon ball reference ftw.

@Deadly1337: No, there's enough information.
@Inyurface: But how can you tell which side is < or > if both sides are not equal?

What in the worldddddddddddddd

You balance 3 balls on one side, then you balance 3 balls on the other.
If they're of equal weight, you know that none of those 6 are the heavier/lighter one.
You'd have eliminated 6 balls at that point; if the group of 6 you used were equal, discard them.

There will be 6 balls remaining. From here, you repeat this process, and there you will get 3... no, never mind. You'd be left with the question of is the ball lighter or heavier at the end.

Am I the only one who had a "o___o" expression the first glance at the title?

Step 1: Place half the balls on one side and half the balls on the other side. Whichever side is heavier/lighter has the odd ball.
Step 2: Now that you have 6 balls, place 3 on one side and 3 on the other. The heavier/lighter one has the odd ball.
Step 3: Place 2 of the 3 balls on the balance to find out which one is heavier or lighter. And then start concluding from there~

you don't supply us with enough information to attempt it, we have no visuals, nothing.

@Msjazz: But you can only solve this in 3 steps.

lift them one by one

It's the middle of July.