More Difficult Than The Einstiens?

It is possible to find out, but it's all down to your luck, because the most pennies that you can weigh in three turns is 6.

No luck involved here!! Just pure genius and intellect.And it is possible to solve this in more than one way!

Well, as far as I can determine, the information to confidently guess which penny is what can be obtained, but a definite answer cannot.

If such a problem does require so much genius and intellect, and you have it, cure cancer or something damnit. Don't waste time on stuff like this. Any how, I am privileged enough to be able to waste my time here. good day.

Calm down, it's only a puzzle. I'll have a go at it...

Separate the 12 pennies into 3 groups of 4. Balance 2 of the groups against each other. If they are the same weight, the bad penny isn't in either group, so you know it is in the one you didn't weigh. You can then balance 1 penny against another (from the 4 you didn't weigh). If they weigh the same, the bad penny is neither, so it must be one of the other two. Weigh these against each other, and one side will go up, my problem is I don't know whether it's heavier or lighter, so which penny?

If they weren't the same weight (going back to my first balancing now), I know one of the groups contains the bad penny, but again, I don't know if the bad penny is heavier or lighter, so which group?

I was trying to do it with a recursive method that you use to narrow it down, lowering the size of the group each time. Was I close? Can anyone see how to finish my method? I'll have another think about it...

You were very close indeed. But there is a way to determine the exact penny. I am sure if you think again then you should be able to crack it. If you need a hint then I will post it tomorrow.

Here goes... Separate the 12 pennies into 3 groups of 4. Balance 2 of the groups against each other. If:

- they are the same weight: the bad penny isn't in either group, so you know it is in the one you didn't weigh. Balance 3 from one of the groups you now know doesn't contain the bad penny against 3 from the group you now know does contain the bad penny. If:

- - they are the same weight: the bad penny must be the 1 penny left that you didn't weigh from the group you knew it was in.

- - the side you know doesn't have the bad penny goes down: the bad penny is lighter, and one of the 3 on the light side of the scales. Weigh 2 of the 3 against each other. If:

- - - they balance, the 1 you didn't weigh is the bad light penny!

- - - the left goes down, the right penny is the bad light penny, and vice versa.

- - the side you know the penny could be in goes down: the bad penny is heavier, and one of the 3 on the heavy side of the scales. Weigh 2 of the 3 against each other. If:

- - - they balance, the 1 you didn't weigh is the bad heavy penny!

- - - the left goes down, the left penny is the bad heavy penny, and vice versa.

- the left side goes down: I'm still not sure...

- the right side goes down: I'm still not sure...

Within striking distance...

Fleogan. You are basing it all that you are excluding the group with the penny. That is just a 33% chance. If you find you have chosen that pile you need to do one more weighing in order to establish which pile.

Wasn't it supposed to be a method that works everytime and not just if you randomly choose the correct piles/coins?