Twelve Marbles, Three Weighings, and Twenty-Seven Ways to Land

Twelve marbles look identical. One of them has the wrong weight, and you are not told whether it is heavy or light. With three weighings on a two-pan balance, find it and say which way it is off.

It can be done, always, and the reason it can be done is a counting argument you can check in ten seconds. The reason the obvious first move fails is the same counting argument run backwards.

Count the answers, count the outcomes

There are twelve candidates for the odd marble and two directions it can be off in, so

Each weighing returns one of three things: left pan down, right pan down, or level. Three weighings therefore produce at most

so nothing forbids it. Note what equation (2) does and does not say. It says three weighings might be enough. It does not say they are, because a bound on capacity is not a construction. Both halves of the problem need doing.

The same count kills two weighings outright: $3^2 = 9 < 24$, so no scheme however clever gets there.

Splitting six against six is not slow, it is impossible

The instinct from searching a sorted list is to halve. Put six marbles against six. That weighing always tips, because the odd marble is on one side or the other and nothing can offset it. So the level outcome never occurs, and a three-answer instrument has been used as a two-answer one.

Worse, neither branch narrows things much. If the left pan goes down, the odd marble is one of the six on the left being heavy, or one of the six on the right being light. Twelve answers survive, and two weighings offer nine sequences:

The conclusion is not that halving is a slower route to the answer. It is that no continuation exists. Equation (3) is a statement about outcome sequences rather than about any particular plan, so it rules out every possible pair of follow-up weighings at once.

The right first move splits into three

The ideal split of 24 answers into three groups is $24/3 = 8$ each, and four against four achieves it exactly. If the pans tip, the odd marble is one of the four on the heavy side being heavy or one of the four on the light side being light, which is eight answers. If they balance, the odd marble is one of the four set aside, in either direction, which is also eight.

That is as tight as it gets, and it is why the problem has exactly twelve marbles in it. There is no slack to waste on a bad second weighing either.

A schedule that works

Number the marbles 1 to 12 and weigh $\{1,2,3,4\}$ against $\{5,6,7,8\}$.

If they balance, the odd one is among 9, 10, 11, 12 and marbles 1 to 8 are known normal. Weigh $\{9,10,11\}$ against $\{1,2,3\}$. If that tips, you learn both which trio contains the culprit and its direction, and 9 against 10 finishes: whichever misbehaves is it, and if they balance it is 11. If the second weighing also balances, marble 12 is the odd one, and 12 against 1 gives the direction.

If the left pan goes down, then either one of 1 to 4 is heavy or one of 5 to 8 is light. Weigh $\{1,2,5\}$ against $\{3,4,6\}$. This is the step worth studying, because it mixes marbles from both original pans on purpose. Three cases:

In each case the third weighing separates three possibilities using its three outcomes, except the last, where only two are left and the balance cannot come out level. The case where the right pan goes down first is the mirror image, with the roles of the two original pans exchanged.

Enumerating all 24 answers through this schedule gives 24 distinct outcome sequences and the correct marble and direction in every case. The three leftover sequences in figure 2 are not wasted design; they are the three occasions where the remaining ambiguity is a pair rather than a triple, and a balance holding two unequal things cannot report level.

Why the impossibility argument covers clever plans too

The usual objection to a counting proof is that it seems to assume the weighings are chosen in advance. It does not, and the reason is worth spelling out.

A strategy, adaptive or not, induces a map from the 24 answers to the outcome sequence that answer would produce. Two answers mapping to the same sequence are indistinguishable, because the sequence is literally everything you observe. So a strategy succeeds if and only if that map is injective, and injectivity into a set of 9 sequences is impossible from a domain of 12. How the later weighings are chosen never enters the argument.

Why twelve, and not thirteen

Equation (2) permits $2n \le 27$, so $n \le 13$. Fourteen is dead on arrival: $2 \times 14 = 28 > 27$.

Thirteen is the interesting boundary. It fails as stated, and it succeeds under two small relaxations. If you have one extra marble known to be normal, thirteen unknowns can be resolved with the direction in three weighings, because a reference marble lets you build weighings that would otherwise be unbalanced. And if you are asked only to name the odd marble without saying which way it is off, thirteen works with no reference at all. The general pattern, for $k$ weighings with direction required and no reference marble, is

so each extra weighing roughly triples what you can handle. The puzzle chose twelve because it is the largest number for which the answer is a clean yes, and because the tightness of equation (4) is what makes the solution feel forced rather than lucky.

Sources and further reading

• The problem and its variants — Balance puzzle
• The argument in equations (2) and (3) — Pigeonhole principle
• The framework figure 2 draws — Decision tree model

The schedule was checked by exhaustive enumeration rather than by sampling: all 24 answers were pushed through it, the deduced marble and direction compared against the truth as a pair, and the 24 outcome sequences confirmed pairwise distinct with no branch exceeding three weighings. The six-against-six split was enumerated too, and reports twelve surviving answers in each tipping branch and an empty level branch.