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So, we all know the famous 12 balls 1 scale riddle. It has been here repeated many times, however, the provided solutions always start with splitting the twelve balls into three groups of four. This made me wonder whether it would also be possible to solve the riddle in another starting combination, e.g. four groups of three.

Just in case someone doesn't know the riddle, let me quickly reiterate:

You are given twelve balls and a scale. Eleven of the balls weigh exactly the same, but one of them weighs slightly more or less (you don't know which one of these). You are allowed to use up to three weighings on the scale to determine which one of the balls is different.

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    $\begingroup$ The original riddle also requires to determine if the different ball is lighter or heavier. Is this not required here? $\endgroup$ – EKons Aug 23 at 18:29
  • $\begingroup$ I have been working with this version, but it would most certainly be interesting to see solutions to both versions! $\endgroup$ – hmhmmm Aug 23 at 18:36
  • $\begingroup$ By breaking it into four groups rather than 3 you have necessarily required one extra weighting in the worst case situations, so no. $\endgroup$ – Octopus Aug 23 at 18:46
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No, it's not possible:

If you start by weighing 3 vs 3 and it comes out equal, you only have two weighings to find the fake coin (of unknown weight) among six.

  • In weighing 2, if you have more than three potential fakes on the scale, and it comes out unbalanced, then the remaining weighing (with only three potential results) will not allow you to pick the fake out of more than three options.

  • In weighing 2, if you have less than three potential fakes on the scale, and it comes out balanced, the remaining weighing will again not allow you to pick the fake out of more than three options.

  • So you must have exactly three potential fakes involved weighing 2 (plus some known-good coins to balance the scales). If your weighing there comes out equal, then you must weigh at least two of your coins from your remaining fakes - but if that comes out unequal, then you don't know which of the two was the fake.

If you start by weighing 6 vs 6, you will only have nine potential results from your remaining two weighings (left-left, left-center, left-right, ..., right-right). But there are still twelve potential fake coin options: it can be a light fake in the lighter group, or a heavy fake in the heavier group. Nine results are not enough to let you distinguish among twelve options.

So your first weighing cannot be divided into groups of 3.

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  • $\begingroup$ As sad as this is sad to hear, I believe it answers my question. Thank you for digging into this, I was really curious about it! $\endgroup$ – hmhmmm Aug 23 at 18:45
  • $\begingroup$ @hmhmmm You can avoid sadness by carrying some (e.g. three) known-good reference weights with you at all times; then you can start e.g. by weighing 5 against 4 (+ 1 good). If the scales tilt, next round is 3 light-suspects + 3 heavy-suspects against 6 good. $\endgroup$ – Bass Aug 23 at 23:07
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I'm not entirely sure how to single out the exact ball here as the description given seems to imply that I have four groups of three balls and only three times to use the scale in order to determine the ball.

Based on that, I can use standard logic to figure out which group the different ball is in. Let's say that our groups are labeled A, B, C, and D, and let's also go ahead and declare that the different ball is in group A for simplicity and that the weight for all of the same balls is 0.5kg.

With only three attempts I would:

Weigh groups A and B together.
Weigh groups C and D together.

I now know which coupled set contains the ball since the weight will differ slightly.

From here, I split the set that was different so A and B.
I now weigh A.

Closing in on the end of the tunnel, I know that C and D together was 1kg. Since I now know the weight of group A, I can perform basic math to split the weight appropriately from the measurement of A and B together.

Finally, I subtract the weight of A from the weight of C and D, if the result is equal to A then B contains the different ball, otherwise A contains the different ball.

Again, I'm not 100% sure this is the answer you're looking for as I don't believe I have enough context on what the correct answer should state. However, this does tell you what group the different ball is in.

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  • $\begingroup$ Thanks for the attempt! The point is to single out the one ball that is heavier/slighter, not the group it belongs to. For more context, you can check out this excellent answer for the riddle, which starts with three groups of four balls. I'd like to see a solution that starts with a different number of groups. $\endgroup$ – hmhmmm Aug 23 at 18:42

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