Two Jars End Up Equally Contaminated, and Conservation Proves It

Two jars, six cups each. One holds a blue liquid, the other a pink one. Pour two cups of pink into the blue jar, stir it thoroughly, then pour two cups of the mixture back. Which jar ends up more contaminated by the other colour?

Almost everyone picks one, and the reasoning is good: the two cups going back were already diluted, so less pink came home than went out. The premise is true. The conclusion does not follow, and the two jars end up exactly equally contaminated.

The argument that needs no fractions

Count what is in the room instead of what moved. There are six cups of blue liquid in total and six cups of pink, and nothing evaporated. Both jars finish holding six cups, because two cups left one jar and two came back.

Now suppose jar one finishes with $x$ cups of pink in it. Its volume is six, so it holds $6 - x$ cups of blue. But all six cups of blue are somewhere, so jar two must hold

cups of blue. The two intrusions are the same number, and the only two facts used were conservation and the equality of the final volumes. Nothing about mixing, nothing about the order of the pours, and no algebra beyond (1).

That is a stronger claim than the puzzle asks for, and it is worth being precise about which half of the answer it covers. The equality follows from conservation alone. The amount does not, and getting a number out requires knowing how well things mixed.

The number, for the record

With perfect stirring, pour $Q$ cups out of a jar of $V$. After the first pour, jar one holds $V + Q$ cups of which $Q$ are foreign, so the pour back removes the fraction $Q/(V+Q)$ of everything in it. The foreign liquid left behind is

and by the same computation jar one gave away $V - V^2/(V+Q) = VQ/(V+Q)$ of its own liquid, which is the same expression. Each jar therefore ends

For $V = 6$ and $Q = 2$ that is $12/8 = 1.5$ cups crossing in each direction, so each jar ends one quarter foreign and three quarters its own colour.

The slow route agrees with the fast one, which is the point worth making about it. Grinding through (2) is not wrong and it is not a trap; it is simply more work than (1), and it tells you less, because it answers the question for one particular schedule instead of all of them.

Stir badly and the equality survives

This is the part that convinced me the exchange argument is the real explanation rather than a shortcut. Suppose the stirring is terrible, so the two cups poured back are much pinker or much bluer than the jar's true proportions. Equation (2) is now false and the number $1.5$ is gone. Equation (1) is untouched, because it never mentioned proportions.

Twenty thousand random pour schedules with the transferred mixture deliberately skewed by up to eighty percent away from the jar's true composition all kept the two intrusions exactly equal, while the amount itself wandered away from $1.5$. That is the correct split between what conservation gives you and what mixing gives you.

The hypothesis that isload-bearing is the equal final volumes. Stop after the first pour and jar one holds eight cups against jar two's four: jar one has two cups of pink in it and jar two has no blue at all. Or make the pour back a different size from the pour out, and the equality breaks in exactly the same way. It is the levels that have to match, not the stirring.

Do it again, and again

Nothing stops you repeating the round trip, and the general behaviour is a small surprise. Let $c_k$ be the share of jar one that is still its own liquid after $k$ round trips, with perfect stirring throughout. Tracking one pour out and one pour back gives

an affine map whose fixed point is $c^{*} = 1/2$, independent of both $V$ and $Q$. Subtracting the fixed point turns (4) into a clean contraction, and with $c_0 = 1$ it solves to

For the six-and-two jars the ratio is exactly $1/2$, so the purities run $3/4$, $5/8$, $9/16$, $17/32$, each round trip halving whatever advantage the jar had left. Setting $k = 1$ in (5) returns the $3/4$ from (3), which is the check that (4) was written down correctly.

The limit needs no computation either. After enough exchanges both jars hold the same mixture, and there are equal totals of the two liquids, so that common mixture is half of each. The arithmetic in (5) only tells you how fast, and the answer is geometrically, at rate $(V-Q)/(V+Q)$. A tiny pour barely mixes anything; a pour with $Q = V$ finishes in one step.

Where the setup stops making sense

The pour size cannot exceed the jar's contents, so $Q \le V$. At the boundary $Q = V$ the first pour empties jar two entirely and hands jar one twelve cups of a half-and-half mixture; the pour back returns six of them, and both jars end exactly half and half. The equality holds, trivially, and (3) gives $V/(V+Q) = 1/2$ as it should.

One word in the question does real damage if left alone. "More contaminated" can mean a larger amount of foreign liquid or a higher concentration of it, and those are different questions. They happen to have the same answer here only because both jars finish at the same volume, which is the same hypothesis that made (1) work. If the volumes differed, the amounts could match while the concentrations did not.

And the claim is about the foreign liquid in each jar, not about the jars looking alike. They do not look alike. One is mostly blue and the other mostly pink, and the thing that matches is the size of each one's intrusion.

Sources and further reading

• The puzzle in its traditional form — Wine and water mixing problem
• The principle behind (1) — Conserved quantity
• Why (4) converges, and how fast — Fixed-point iteration

Every number above was checked before publication in exact rational arithmetic, never floating point: symbolically for general $V$ and $Q$, then over every $V$ from one to twenty-four with $Q$ in quarter-cup steps, then over twenty thousand random pour schedules with perfect stirring and twenty thousand more with the stirring deliberately sabotaged. A negative control confirmed the hypothesis is load-bearing: stop the sequence with the levels unequal and the two intrusions come apart.