Balance Scale Puzzle

Hi to All:

Since the logic puzzle seemed popular, I thought that I might another challenge.

You have 12 balls, 11 of which are identical and one is identical in appearance but not in weight. In addition you have a simple balance scale. Now find an algorithm to determine which ball is the odd one and determine whether it is lighter or heavier than the others. The only constraint is that you can not use the balance scale more than 3 times.

Seems pretty simple.

Put six balls on one side and six on the other. Take the six balls from the heavier side and divide them into two groups of three. Record the weight difference. Weigh the two groups of three. Record the weight difference. From the three in the heavier group, place two on one side and one on the other. Record the weight difference.

In three tries the fake ball should be eliminated. The missing weight in all three tries combined should be the weight of the fake ball? If each time the scale was used there was a gram difference, then the fake ball weighs three grams more than a real ball?

I dunno. Maybe not.


Do I know before hand what the weight of a real ball is? If I did, then the weight difference could be determined in the first try of weighing the two groups of six balls.

Hi détrop:

You’ve got the right idea. However, in the last weighing you don’t know if the single ball is heavy because two lighter balls may weigh more than one heavy ball. (Basically you have to weigh the same number of balls on both sides of the scale). Even more fundamentally, after your first weighing, you don’t know if the odd ball is on the light side.

PS You don’t know the actual weight.

Yes! Dètrop’s idea is the lift off. The above is key. But there’s more to that statement than meets the eye.

Ed, this is a fun puzzle. Thanks.

You could split the 12 balls into 3 groups of 4…


You pick up each ball in turn. The one that is a different weight to the others is the odd one. Put it and one of the 11 balls on either side of the balance to work out whether it is heavier or lighter.

Not an algorithm, but what I’d do if given this puzzle in real life…

SIATD had it right, although I think part of the rule is that you cannot weight them without using the scale, and only three times at that. How simple it would be to merely pick one up with your hand. Clever, SIATD.

I had a feeling that my solution was incomplete, Ed3. Thanks for providing clues…these kinds of puzzles help me get a better understanding of logic. Stick around and walk me through it.

Alright, weigh the two groups of six. One side weighs four ounces and the other three. So far there is a missing ounce. Simple enough, but it gets tricky.

We don’t know if it is because the fake ball is heavier than the rest and is on the side that weighs four ounces, or if it is lighter and is on the side that weighs three.

It seems like we should continue and break the groups of six into groups of three and weigh them, but that would leave us with one ball on one side and two on the other in our final weighing.

This is heavy man. I have a feeling it involves some complex calculus and/or algebra, which I suck at. In fact, I’m not sure I even spelled them right.

This algorithm you speak of is somehow determined in the three weighings by some sort of probability rule which works in the ‘chances of elimination’ of the fake ball in the three weighings. Which is to say, a recurring pattern in weight difference, combined with the chances of one ball out of twelve being on one side of the scale as opposed to the other produces the solution.

Am I on the right track, Ed3? Help me out, but don’t solve it.

(any second now some fifteen year-old is going to come in here and make us all look like asses. I love it when that happens.)

Speak up, geeks and nerds alike. Those lonely isolated days in the library studying complex mathematics needs to pay-off, so now is your chance.

You love this idea of the loners being secretly smarter than all of us, detrop. Can’t help but wonder what implications this has for your life…

As a general rule the conditions which promote intellectual development in society are often negative circumstances which although result in positive things, like being smart, they occur for the wrong reasons. Often enough the cliche of the smart guy being a loner is very true- and there is a reason for that.

The tragedy is that people view intellectual development as an alternative to social skill, while in reality there should be no distinction. In the modern world there is a basic malaise surrounding education and intellectualism- only a few, in proportion to the many, are intelligent enough to stand-out, and as such in a contrast to the many, they are considered the oddity. The atmosphere today is not conducive to vigorous education, but instead it makes simple minded consumers, and it does this my auditing the material of study…by dumbing down the curriculum.

Those exceptions often find themselves having a natural tendency to want to learn more, and the capacity to do so, but at the expense of various social conventions and normalities.

Thus is the pun of the nerd or geek, this stereotyped personality that is over intelligent when the irony is that everyone else is so stupid.

My comment was made in jest, although there is some truth to it.

Weighting 1: split them equally into 3 groups, and then weight any 2 groups. If the result is balance, then it means that the different ball must be in the other group that’s not weighted, this implies that we have 2 chances left to weight the 4 balls in this group. The implication is final. However, if the result is imbalance, then proceed.

Weighting 2: make a note of which of the 2 weighted groups is heavier, and which one is lighter. Pick 2 balls out of each group respectively and weight these newly formed 2 equal groups with a total of 4 balls.

Weighting 3: if the result of the previous weighting is balance, then the different ball must be among the unselected 4 balls. If the result is imbalance, then the different ball must lie in 1 of our new groups. The imbalance result must be correspondent to the note, which we made before, in other words, the group that’s now heavier must come from the group that was heavier before. The following lines a, b, and c describe the situation that is ultimately the same for us irrespective of the outcome of the previous weighting, provided that you involve the different ball, or rather the group that contains the different ball, into the next step.

a - Heavy group H: ball H1, ball H2;

b - Lighte group L: ball L1, ball L2;

c - Auxiliary group A (which comes from innocent balls that we already rationally dismissed from previous weightings): ball A1, ball A2.

Mix A1 and A2 with, say L2. So now we have A1, A2, and L2 on one side of the libra, H1, H2, and L1 on the other side. Make the final weighting. The result can’t possibly be balance, it can only be imbalance. This is the kind of result that I wanted to achieve. I’m stuck at this stage. Now I’m inclined to regard this as a paradox instead as a puzzle. I’ve tried to assign possible values 1, 2, 3 to the balls, but it didn’t help here. But I still have faith in analysing the following situation by way of multiple attempts at grouping, so that eventually the different ball will be somehow logically singled out:

H1 L2 A1
H2 L2 A2

Bravo, Uniqor!

I hope you are right so far.

Are you saying that it is a real paradox, or do you believe there is a solution?

Ed3, I hope this gig is real. Nah, Ed wouldn’t do that because he’s laid back and studious.

Hi to All:

Someoneisatthedoor was right, and as Detrope noted, I should have specified that all the weighing must be done on the balance scales.

The most progress by far is in the current work of Uniqor.

Detrope is pretty much right about my personality. The problem is logically solvable, and does not require mathematics or statistics.

Hi Uniqor:

Assuming that the scales balanced on the first weighing how would you determine which ball is heavy or light on the next 2 weighings? This part is fairly simple, but I do not know exactly how you would do it.

The problem is in fact more difficult if the first weighing is unbalanced, but you have had some cleaver insights.

Thanks to all Ed

Nope. I definitely cannot do this puzzle. Uniqor, I am cheering for you. And when you’re done, let’s celebrate with a bottle of…[ fill in the blank ]. Anything you like, name it.

You’d have two more weighings and only four balls to weigh. The only alternative to Uniqor’s scenario is this:

Take the four balls from the unweighed group and divided them into two groups. Weigh those.

There must be a difference of weight noticed in this second weighing. All the other balls have been excluded.

Record this weight difference.

Now, in the final weighing we have to determine whether the fake ball is lighter or heavier (but I don’t see how). We’re going to split those groups of two balls into one individual ball and weigh those together.

Here are two new problems- when we make these two new single ball groups, we have a one in four chance of picking the fake ball in the final weighing of the two individual balls. Also, even if we do pick the fake ball, we still wouldn’t know if it was heavier or lighter. We would only be able to know it is on the scales by the imbalance. Which side is a mystery.

This sucks, Ed3. My brain hurts. You said there wouldn’t be any math but I don’t see how.

I see how its possible to find the fake ball, but I don’t see how one could ever know the real weight of either the fake ball or the original balls unless they either knowingly picked the fake ball, or had a fourth weighing with one real ball and a counter weight, to determine the true weight without comparing ball weight. Right? If one always has that chance of picking the fake ball in any one of the three weighings, then they could never be certain which side of the scales it was on.

I don’t get it. Look, Ed3, if one cannot know the weight of the real ball, then regardless of how many weighings one has, one can never know which ball is fake.

Let’s say I lucked out and picked two balls to be weighed individually, one of which was the fake ball. Still, how could I know what it was supposed to weigh if I can’t know the weight of the real ball. You follow me?

Alright, I gotta idea. Throw the scales away…you don’t need em’. Line the twelves balls up beside on another on the table, and lift one end of the table slightly. Which ever ball rolls faster down the plane is the heavier ball, and thus the fake one, or, which ever ball is the slowest rolling ball is lighter, and thus the fake one. Gravity.

Just kidding.

Uniqor, take over. I’m stumped.

Affirmative, sir. Promote me up a rank if mission successful.

I offer this new and improved solution, I hope it’s also a final solution, if so, I’m indebted to all your encouragement.

Weighting 1:

Split into 3 equal groups and weight any 2 groups.

If the result is balance, then the odd ball must be in the unweighted group. Pick 2 balls out of that group and weight them.

If the result is balance, then the odd ball must be among the other 2 balls. Weight those 2 balls and we’re done.

However, if the result is inbalance, then the odd ball must be among the 2 balls. Weight any of these 2 balls against any of the other 2 balls and we’re done.

Weighting 2:

Proceed into this stage if the result of the very first weighting is inbalance.

The table below describes the current nasty situation.


Swope H3 and H4 with A3 and A4, thus we formed Newauxiliarygroup containing A1, A2, H3 and H4. Weight Lightgroup against Newauxilarygroup.

a. The result of this weighting can be balanced. If so, it implies that H3 and H4 are innocent balls. It follows that the odd ball must be among H1 and H2, also, it must be heavy, otherwise Heavygroup wouldn’t be heavier than Lightgroup in the first place. So weight H1 and H2, vola.

b. The result of this weighting can also be Lightgroup < Newauxiliarygroup.

c. The result of this weighting can’t possibly be Lightgroup > Newauxiliarygroup.

So follow situation b, which would imply that the odd ball is among H3 and H4, also, it is heavier. Weight H3 and H4 and fill in arendt’s blank.

Now tell us, Ed3, shall we screw the cork out?

One last thing, a thing which could potentially send Congressional Medal Of Honor to my mother. If you choose to form your Newauxiliarygroup by selecting any 2 balls from Lightgroup, then repeat the steps with the same logics mutatis mutandis.

How are you meant to pick up the balls and put them on the scales without noticing that one of them is a different weight to all the others?

Uniqor, though ingenious, your solution does not work, there’s two mistakes in it.

  1. If the result of weighing one is balance it will be impossible to determine whether the ball is lighter/heavier 25% of the time:

i.e. In aux group ABCD, compare A vs. B, balance. Weigh A vs. C, balance. Is D heavier or lighter? The other solution, weigh AB vs. CD results in a worse scenario.

  1. For your solution in the case of initial imbalance, if the reweigh throws up situation b, that L1-4 are lighter again, the ball is not, as you said, obviously heavier, there’s still a chance it’s lighter and is one of L1-L4.

I’m fairly confident that this puzzle is actually impossible.

Actually, scrap fairly, I’m 100% that this puzzle is impossible (hope I don’t have to eat my words now). :laughing:

No cigar, SIATD. You place the balls on the scales with a robotic arm, insensitive to gravitational forces, which you control with a joystick. Quit trying the easy way out.

I think it is you who should do the promoting, sir.

It is you who needs to pat me on the head here. Just who do you think you’re kidding, busta?

Anyway, you gotta deal with Matt now. Apparantly your theory is flawed.

[ detrop gets a bag of popcorn and a soda ]