Three mislabeled jars, one draw to fix them.
Three jars labeled apples, oranges, and mixed, and every label is wrong. Prove you can relabel all three after drawing a single fruit, provided you draw it from the right jar. A tiny puzzle about squeezing every drop of information from a constraint.
Three sealed jars sit on a table. One contains only apples, one contains only oranges, and one contains a mix of both. The jars carry the labels APPLES, ORANGES, and MIXED, one label each.
You are told a single hard fact: every jar is labeled incorrectly. Not "possibly wrong", not "at least one wrong". All three labels are wrong.
You may reach into exactly one jar of your choosing, without looking inside, and pull out exactly one fruit. Looking at that one fruit, you must deduce the true contents of all three jars.
Which jar do you draw from, and how does the rest follow?
Two questions to sit with before the hints. First: how much information does one fruit carry? It is one of two kinds, so a single draw yields at most one bit, yet there are six ways to assign three contents to three jars. Second: how much work is the "all labels wrong" constraint doing? Of those six assignments, how many does it eliminate before you touch anything? The puzzle is only solvable because the constraint does most of the work in advance, and choosing where to draw is really choosing where one bit settles everything that remains.
Stuck? Open hints one at a time
Each hint gives away one more layer. Stop the moment something clicks.
Hint 1 a nudge
Count the assignments that survive "all labels wrong". Of the six ways to fill three jars, only two leave every label incorrect (these are the derangements of three items). One bit is exactly enough to split two cases, if you draw from the right jar.
Hint 2 the key move
Draw from the jar labeled MIXED. Its label is wrong, so it is pure: whatever fruit you pull, the whole jar is that fruit. Drawing from a jar that might be mixed teaches you almost nothing; drawing from one that cannot be mixed converts one fruit into full knowledge of that jar.
The solution
Reveal the full solution
Draw from the jar labeled MIXED. Suppose you pull an apple.
The MIXED label is wrong, so that jar is pure, and the apple tells you it is pure apples. Now look at the jar labeled ORANGES. It cannot be oranges (its label is wrong) and it cannot be apples (the mixed-labeled jar is apples), so it must be the mixed jar. The jar labeled APPLES takes the only content left: oranges. Pulling an orange runs the mirror image of the same argument. Either way, one fruit relabels all three jars with certainty.
Why must the draw come from the MIXED-labeled jar? Try the jar labeled APPLES instead. It is not apples, so it is oranges or mixed, and both of those jars contain oranges. Pull an orange and you have learned nothing that separates the two surviving worlds; pull an apple and you know that jar is mixed, but the draw only sometimes resolves things, which fails the "with certainty" bar. Only the MIXED-labeled jar guarantees that a single fruit identifies its jar completely, and the wrong-label constraint then forces the other two by elimination.
What this puzzle is really testing
Constraint propagation. The solver's actual work happened before the draw: "all labels wrong" collapses six possible worlds to two, and the draw is designed so its one bit splits exactly those two. This is the smallest clean example of a pattern that scales: prune the state space with the constraints you were given for free, then spend your expensive observations only on what remains. The same economy drives search-strategy choices in graphs, where the question is always which probe collapses the most possibilities, and it is a gentler cousin of the elimination bookkeeping in the 25-horses puzzle.
Follow-ups to expect
Four jars (apples, oranges, mixed, and empty, all mislabeled): how many draws now? What if only at least one label is wrong, where the constraint stops doing your work and no single draw can suffice? And the general version: with n jars all mislabeled, the surviving worlds are the derangements of n, which for n = 4 number nine, so count how many bits of drawing you need. For the habit of turning constraints into eliminated states, the graphs module is where it pays off in code.