History Main / TwelveCoinsPuzzle

* ''Manga/TheKindaichiCaseFiles'':. Kindaichi solves a ten-bags-of-coins variant almost immediately, necessitating him to explain the solution to a bewildered Miyako. The person who had posed the problem [[spoiler: turns out to be the murder suspect, and she bemoans that she should have realized Kindaichi would see through her tricks given how casually he blew through the question]].

* ''Series/BrooklynNineNine'': Captain Holt asks the detectives a version of this in the "[[Recap/BrooklynNineNineS2E18CaptainPeralta Captain Peralta]]" episode with twelve men on an island instead of coins and offers tickets to a Music/{{Beyonce}} concert for anyone who can solve it. [[spoiler: Turns out that his own mentor asked him the same riddle and he couldn't figure it out so he set the whole thing up so the detectives would figure out the answer and he could impress his mentor by pretending he figured it out.]]

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* ''Literature/WithATangledSkein'': TheHeroine has to solve this puzzle to find a certain soul disguised as a demon, out of the 12 demons. She only knows that it is one disguised soul amongst demons, so initially does not realize how weight plays a factor. After some internal panic, she remembers that sin weighs souls, so a good soul will be lighter to drift to heaven. It [[spoiler:doesn't work due to Satan, but]] things resolve regardless.

* ''Manga/TheKindaichiCaseFiles'' uses the ten-bags-of-coins variant. Kindaichi solves it almost immediately, necessitating him to explain the solution to a bewildered Miyako. The person who had posed the problem [[spoiler: turns out to be the murder suspect, and she bemoans that she should have realized Kindaichi would see through her tricks given how casually he blew through the question]].

* Used in ''Literature/WithATangledSkein'' by Creator/PiersAnthony. The heroine had to use this puzzle to find a certain soul disguised as a demon, out of the 12 demons. It [[spoiler:doesn't work due to Satan, but]] things resolve regardless.
** She also only knows that it is one disguised soul amongst demons, so initially does not realize how weight plays a factor. After some internal panic, she remembers that sin weighs souls, so a good soul will be lighter to drift to heaven.

The usual version is 12 coins (or bags, or whatever) with three weighings, which has several solutions, some more intuitive than others. It's easier if you know whether the counterfeits are heavier or lighter than the genuine coins, so in trickier versions of the puzzle you must figure this out as well. A third variation may give you 13 coins (but you know whether the counterfeits are heavier or lighter) - some of the 12-coins-unknown-weight solutions also work for this (simply leave the extra coin out, and then if it was the counterfeit, you'll figure that out in the three weighings). 13-coins-and-unknown-weight is not possible in the standard three weighings.

If the more complicated version is used, and whether the bad coin is heavier or lighter is not known, the procedure is a little different and more complicated.
* Put four coins on each side. There are two possibilities:
** 1. One side is heavier than the other. If this is the case, remove three coins from the heavier side, move three coins from the lighter side to the heavier side, and place three coins that were not weighed the first time on the lighter side. (Remember which coins are which.) There are three possibilities:
*** 1.a) The same side that was heavier the first time is still heavier. This means that either the coin that stayed there is heavier or that the coin that stayed on the lighter side is lighter. Balancing one of these against one of the other ten coins will reveal which of these is true, thus solving the puzzle.
*** 1.b) The side that was heavier the first time is lighter the second time. This means that one of the three coins that went from the lighter side to the heavier side is lighter. For the third attempt, weigh two of these coins against each other. If one is lighter, it is the unique coin; if they balance, the third coin is the light one.
*** 1.c) Both sides are even. This means that one of the three coins that was removed from the heavier side is heavier. For the third attempt, weigh two of these coins against each other: if one is heavier, it is the unique coin; if they balance, the third coin is the heavy one.
** 2. Both sides are even. If this is the case, all eight coins are identical and can be set aside. Take the four remaining coins and place three on one side of the balance. Place 3 of the 8 identical coins on the other side. There are three possibilities:
*** 2.a) The three remaining coins are lighter. In this case you now know that one of those three coins is the odd one out and that it is lighter. Take two of those three coins and weigh them against each other. If the balance tips then the lighter coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is lighter.
*** 2.b) The three remaining coins are heavier. In this case you now know that one of those three coins is the odd one out and that it is heavier. Take two of those three coins and weigh them against each other. If the balance tips then the heavier coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is heavier.
*** 2.c) The three remaining coins balance. In this case you know that the unweighed coin is the odd one out. Weigh the remaining coin against one of the other 11 coins and this will tell you whether it is heavier or lighter.

There are other solutions, some of them with spiffy mnemonics, e.g. [[http://rec-puzzles.org/index.php/Weighing%20Solution here]] and [[http://demonstrations.wolfram.com/The12CoinWeighingProblem/ here]].

Interestingly, if you know whether the odd coin is heavier or lighter, you can handle up to 27 coins in three weighings (weigh 9 against 9, then 3 against 3, then 1 against 1).[[note]]Why not 28? Each weighing has only 3 possible results - left heavier, right heavier, equal - so the whole process has only 3*3*3 = 27 possible sets of results, no matter which coins you weigh each time. And a process is only guaranteed to work if each choice of odd coin leads to a different set of results. Each additional weighing triples the limit.[[/note]]

Another similar puzzle has you given a number of bags of coins and a mass measurement scale that you can use once, and being told that one bag's coins all weigh X and the rest all weigh Y. The solution for this one involves taking one coin from bag A, two coins from bag B, etc., and figuring out how much the resulting measurement differs from what you would get if all the coins weighed the same.

* It appears using a spell that detects the power level of cubes in the Creator/{{Infocom}} game ''[[VideoGame/{{Enchanter}} Spellbreaker]]''; naturally, the spell can be used only three times. And you don't know whether the correct cube will glow ''more'' or ''less'' than the fakes.

* Mentioned in ''VideoGame/FireEmblemTheSacredStones'', where Ewan presents this puzzle to Ross in a support conversation. However, in a slight twist, he doesn't ask how to do it in three weighings: he gives the more generic challenge of asking what is the ''minimum number'' of weighings required (and makes it 25 pebbles, not 12 coins). Ross immediately answers 24 (i.e., [[CuttingTheKnot weighing the pebbles one by one]]), then after some encouragement from Ewan, answers 4. Close, but no cigar - the answer is still 3.
* Appears in a few different variants, usually involving weights rather than coins, in the ''VideoGame/ProfessorLayton'' series of games.
* There is a similar puzzle in ''VideoGame/{{Drakensang}} 2 The River Of Time''. In order to open a chest, you're given several iron and lead weights and you must put the right number of them in the four boxes around the altar. If you can't find the solution, you can ask the druid Megalesios to solve it for you, but you won't gain any experience.
* Appears in the Creator/JulesVerne-inspired adventure game ''VideoGame/{{Voyage}}'', where you have to isolate the lightest of twelve 'isotopes' in the fewest number of steps, with varying points available depending on which of the above solutions you use.
* Subverted in ''VisualNovel/VirtuesLastReward'', where there are four coins to weigh and a scale that gives you five weighings before resetting itself. As the weights of the coins are constant, and there are only four of them, the resetting is just a nuisance.
* One of the puzzles in ''VideoGame/{{Safecracker}} (1996)'' involves a key-weighing machine that can only be used in two moves, in which the objective is to find one out of eight keys that's heavier than the others. Choosing the wrong key or making more than two moves scrambles the keys and resets the puzzle.

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* ''Website/{{TED}}'' has the [[https://www.youtube.com/watch?v=tE2dZLDJSjA&list=PLJicmE8fK0EiFRt1Hm5a_7SJFaikIFW30&index=13 Counterfeit Coin Riddle]], where you are an imprisoned mathematician who must earn freedom by determining which of twelve coins is counterfeit, using only a marker and three weighings on a scale.

!!Examples

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* One ''TabletopGame/PerplexCity'' card has a variant, where you must figure out how to identify pills of a different weight from five jars in one weighing.

* ''Website/{{TED}}'' has the [[https://www.youtube.com/watchv=tE2dZLDJSjA&list=PLJicmE8fK0EiFRt1Hm5a_7SJFaikIFW30&index=13 Counterfeit Coin Riddle]], where you are an imprisoned mathematician who must earn freedom by determining which of twelve coins is counterfeit, using only a marker and three weighings on a scale.

* Appears in the Jules Verne-inspired adventure game ''Voyage'', where you have to isolate the lightest of twelve 'isotopes' in the fewest number of steps, with varying points available depending on which of the above solutions you use.

* Subverted in ''VideoGames/VirtuesLastReward'', where there are several coins to weigh and a scale that gives you limited weighings before apparently resetting itself. The joke, however, is that the weights of the coins don't change when it resets, so the apparent resetting is just a nuisance.

* One of the puzzles in ''VideoGame/{{Safecracker}} (1996)'' involves a key-weighing machine that can only be used twice per turn, in which the objective is to find one out of eight keys that's heavier than the others. Choosing the wrong key randomizes the entire group, requiring you to start again.

* ''VideoGame/{{Safecracker}} (1997)'' has a machine in the Crabb Manor's basement containing 8 keys - with an instruction manual on how to use it. You have to move two groups of these keys to each side of the scale and weigh them to see which contains the heavier key, then push a button to guess which one is the odd one out. The machine dispenses the heavier key if you guess right, which unlocks a safe elsewhere, but guessing wrong randomizes which key is heavier.

* One of the puzzles in ''VideoGames/{{Safecracker}} (1996)'' involves a key-weighing machine that can only be used twice, in which the objective is to find one out of eight keys that's heavier than the others.

* One ''PerplexCity'' card has a variant, where you must figure out how to identify pills of a different weight from five jars in one weighing.

* It appears using a spell that detects the power level of cubes in the Creator/{{Infocom}} game ''VideoGame/{{Spellbreaker}}''; naturally, the spell can be used only three times. And you don't know whether the correct cube will glow ''more'' or ''less'' than the fakes.

* Appears in a few different variants, usually involving weights rather than coins, in the ''ProfessorLayton'' series of games.

* It appears using a spell that detects the power level of cubes in the Creator/{{Infocom}} game ''VideoGame/{{Spellbreaker}}''; naturally, the spell can be used only three times.

** The problem is that there's no reason to think that the real cube would be ''less'' magical than the forgeries, and several walkthroughs even make the assumption that the real coin is heavier, so to speak. You would think the easier solution would work half the time, but it doesn't--after three weighings [[SchrodingersGun the game determines which coins might be the fake given the information you have, and if there's more than one possibility the game will switch it so that you have the wrong coin]]. Infocom was great.

* Used in ''With a Tangled Skein'', in Creator/PiersAnthony's ''Literature/{{Incarnations of Immortality}}'' series. The heroine had to use this puzzle to find a certain soul disguised as a demon, out of the 12 demons.
** It [[spoiler:doesn't work due to Satan, but]] things resolve regardless.

* One of the puzzles in ''Safecracker (1996)'' involves a key-weighing machine that can only be used twice, in which the objective is to find one out of eight keys that's heavier than the others.

* Used in ''With a Tangled Skein'', in PiersAnthony's ''{{Incarnations of Immortality}}'' series. The heroine had to use this puzzle to find a certain soul disguised as a demon, out of the 12 demons.

* ''TheKindaichiCaseFiles'' uses the ten-bags-of-coins variant. Kindaichi solves it almost immediately, necessitating him to explain the solution to a bewildered Miyako. The person who had posed the problem [[spoiler: turns out to be the murder suspect, and she bemoans that she should have realized Kindaichi would see through her tricks given how casually he blew through the question]].

* ''TheKindaichiCaseFiles'' uses the ten-bags-of-coins variant.

** The problem is that there's no reason to think that the real cube would be ''less'' magical than the forgeries, and several walkthroughs even make the assumption that the real coin is heavier, so to speak. You would think the easier solution would work half the time, but it doesn't--after three weighings the game determines which coins might be the fake given the information you have, and if there's more than one possibility the game will switch it so that you have the wrong coin. Infocom was great.

Interestingly, if you know whether the odd coin is heavier or lighter, you can handle up to 27 coins in three weighings (weigh 9 against 9, then 3 against 3, then 1 against 1).[[hottip:*:Why not 28? Each weighing has only 3 possible results - left heavier, right heavier, equal - so the whole process has only 3*3*3 = 27 possible sets of results, no matter which coins you weigh each time. And a process is only guaranteed to work if each choice of odd coin leads to a different set of results. Each additional weighing triples the limit.]]