The 12 Coin Weighing Puzzle, Fully Solved: All 3 Weighings, Step by Step
The 12-coin puzzle is famous for a reason: you have 12 coins that look identical, exactly one is fake, and a two-pan balance scale with no weights. The fake weighs slightly differently from the rest, but you are not told whether it's heavier or lighter. You get exactly three weighings to find the fake coin and say which way it's off. Most explanations online either solve an easier version (where you already know the fake is heavier) or give a real solution with a step missing or a branch that quietly breaks. Below is the complete tree: every weighing, every branch, and all 24 possible answers, checked case by case so nothing is left dangling.
Why three weighings is exactly enough — not one more
Start by counting what you actually need to distinguish. There are 12 coins, and each could be fake in one of 2 directions (heavier or lighter), so there are 24 possible truths. Each weighing has exactly 3 possible readings: left pan down, right pan down, or balanced. Three weighings in a row can therefore produce at most 3 × 3 × 3 = 27 distinct outcome sequences. Since 27 is just barely greater than 24, three weighings give you just enough room to tell every case apart — but only if every weighing is chosen carefully enough that no two different truths ever produce the same sequence of three readings.
This isn't a coincidence specific to 12 coins. Freeman Dyson proved in 1946 that the largest number of coins solvable in n weighings under these exact rules (one fake, direction unknown) is (3^n − 3) ÷ 2 — which works out to 12 for n = 3. Twelve isn't just "small enough to work," it's the maximum the counting argument allows. Try it with 13 coins and no strategy, however clever, can succeed in three weighings.
Weighing 1: split into three groups of four
Label the coins 1 through 12. Split them into three groups of four: A = {1,2,3,4}, B = {5,6,7,8}, and C = {9,10,11,12}. Weigh A against B, leaving C aside.
- A and B balance: coins 1-8 are all genuine. The fake is in C, and you still don't know which of the four or which direction.
- A is heavier than B: coins 9-12 are all genuine. Either one coin in A is the heavy fake, or one coin in B is the light fake — eight live candidates in total.
- A is lighter than B: the mirror image — either one coin in A is light, or one coin in B is heavy.
Branch 1 — the scale balances (fake is coin 9, 10, 11, or 12)
You now have four suspects and two genuine coins to use as reference weights (say, coins 1 and 2). Weigh {9, 10} against {11, 1}:
| Weighing 2 result | Weighing 3 | Conclusion |
|---|---|---|
| Balances | Weigh 12 vs coin 2 (genuine) | 12 heavier → 12 is the heavy fake. 12 lighter → 12 is the light fake. |
{9,10} heavier | Weigh 9 vs 10 | 9 > 10 → 9 is heavy. 9 < 10 → 10 is heavy. 9 = 10 → 11 is light. |
{9,10} lighter | Weigh 9 vs 10 | 9 > 10 → 10 is light. 9 < 10 → 9 is light. 9 = 10 → 11 is heavy. |
Notice the trick: mixing one suspect (11) with one known-genuine coin (1) on weighing 2 means a tip can only be explained by 9, 10, or 11 — never by the genuine filler — so every outcome still points at exactly one coin.
Branch 2 — the left side was heavier (fake is heavy in 1-4, or light in 5-8)
Eight candidates remain, each with a direction already fixed by weighing 1: 1, 2, 3, or 4 heavy, or 5, 6, 7, or 8 light. Weighing 2 reuses coins from both groups in one deliberately mixed weighing: {1, 2, 5} against {3, 4, 6} (coins 7 and 8 sit out).
| Weighing 2 result | Candidates left | Weighing 3 | Conclusion |
|---|---|---|---|
| Left heavier | 1 heavy, 2 heavy, or 6 light | Weigh 1 vs 2 | 1 > 2 → 1 heavy. 1 < 2 → 2 heavy. 1 = 2 → 6 light. |
| Right heavier | 3 heavy, 4 heavy, or 5 light | Weigh 3 vs 4 | 3 > 4 → 3 heavy. 3 < 4 → 4 heavy. 3 = 4 → 5 light. |
| Balances | 7 light or 8 light | Weigh 7 vs a genuine coin (9) | 7 < 9 → 7 light. 7 = 9 → 8 light. |
Read the logic of that weighing 2 grouping carefully, because it's the heart of the whole puzzle: coin 1 and 2 sit on the side that was already heavy, so if either is genuinely heavy it keeps that side down. Coin 5 moves to the heavy side carrying its (hypothesized) lightness, which would also tip that side down if 1 and 2 turn out innocent. Coin 6 moves to the other side carrying its lightness, which would lift that side up — read as "left heavier" again. Three different suspects, three different physical reasons, but they land on the same reading, which is exactly why weighing 3 needs to separate them.
Branch 3 — the left side was lighter (the mirror case)
Same two groups, same weighing 2 ({1,2,5} vs {3,4,6}), but every hypothesis flips direction: now 1-4 could be light and 5-8 could be heavy. Work through it the same way — left heavier on weighing 2 now means candidates 3-light, 4-light, or 5-heavy; right heavier means 1-light, 2-light, or 6-heavy; balanced means 7-heavy or 8-heavy — and the same pattern of weighing 3 (comparing the two same-side suspects, or checking the leftover coin against a genuine one) finishes it.
Every one of the 24 cases, checked
Running every possible fake coin and direction through this exact tree confirms it resolves all 24 cases with no collisions — only 3 of the 27 possible three-weighing outcome sequences go unused (one of which, "balanced every time," can't happen anyway, since the puzzle guarantees a fake exists). Here is which coins fall into which branch, by weighing 1's result:
| Weighing 1 result | Branch | Live candidates (8 cases each) |
|---|---|---|
| Balances | 1 | 9, 10, 11, or 12 — either direction |
| Left heavier | 2 | 1, 2, 3, or 4 heavy; or 5, 6, 7, or 8 light |
| Left lighter | 3 | 1, 2, 3, or 4 light; or 5, 6, 7, or 8 heavy |
Eight cases per branch, three branches, 24 cases total — and each one reaches a different, unambiguous final answer by the end of weighing 3, per the tables above.
Why the "obvious" strategy breaks down
If you already knew the fake was heavier, you could just keep halving: weigh 6 against 6, then 3 against 3, then 1 against 1, narrowing down every time. That shortcut fails here because an imbalance only tells you which side is heavier — it doesn't tell you whether the heavy side has the heavy fake or the light side has the light fake, and without that, halving groups repeatedly throws away the one piece of information (which direction) you still need at the end. That's exactly why weighing 2 in both unbalanced branches above deliberately moves coins between pans rather than just splitting the current suspect pool in half: the movement is what lets the same tip direction mean different things for different coins, and weighing 3 is there specifically to resolve that ambiguity.
Once the logic clicks, it's worth testing it against the easier version first — PassDrill's logic puzzles practice questions include the simpler 8-coin case where the fake is already known to be heavier (solvable in 2 weighings), plus this exact 12-coin, unknown-direction setup, so you can check your own first weighing against the reasoning above before working through the rest of the tree yourself.
Source: H. D. Grossman, "The Twelve-Coin Problem," Scripta Mathematica 11 (1945); F. J. Dyson, "The Problem of the Pennies," The Mathematical Gazette, Vol. 30, No. 291 (1946), pp. 231-234, for the (3^n-3)/2 bound on coins solvable in n weighings.