A Blank Slot Is Evidence: 1/4 Against 1/3 on a Ring of Six

Six slots sit in a ring, numbered $1$ through $6$ in cyclic order. Two of them are marked, and the two marks are next to each other. You are dropped on the ring at a uniformly random slot, and the slot you land on turns out to be blank. One step remains: either advance to the next slot around the ring, or re-draw a fresh uniform slot and take that one instead. Which is safer?

The problem takes fifteen seconds to state and it separates two kinds of candidate almost perfectly. One reads the blank landing as a fact about the world and folds it into the arithmetic. The other treats it as flavour text.

Where the instant answer comes from

Two marks out of six is a third, and that fraction is sitting right there in the statement. So the reflex is that both moves are worth

and the choice is a coin flip you may as well decide on aesthetics. What makes this trap durable is that equation (1) is true. It is the correct value of one of the two options. Re-drawing really does put all six slots back on the table, including the two marked ones, so re-drawing really is a one in three. The error is in copying that number onto the other option as well.

The blank rules out two slots, not one

Here is the sentence most people get wrong when they try to say what the blank taught them: it did not eliminate one possibility. It eliminated both marked slots at once. Whatever your position is, it is one of the four blanks, and every one of those four is equally likely because the original draw was uniform.

Write $B$ for the event that you landed blank and $A$ for the event that the slot one step ahead of you is marked. What the problem asks for is $P(A \mid B)$, and the definition of a conditional probability turns that into two counts:

The denominator is easy. Four of six slots are blank, so $P(B) = 4/6$. The numerator is where the geometry enters, and it is the only step in the whole problem that requires thought.

One dangerous slot in four

Each mark has exactly one slot immediately behind it, so across the two marks there are two slots that would step onto trouble. But the two marks are touching. The slot immediately behind the second mark is the first mark, and you already know you are not standing on a mark, so that candidate is gone. Exactly one blank slot is dangerous: the one immediately behind the leading mark.

So $P(A \cap B) = 1/6$, and equation (2) closes:

Advancing is a one in four. Re-drawing is a one in three. The difference is

which is small in absolute terms and large as a ratio: re-drawing raises your risk by a third of itself. You advance. You never re-draw.

Adjacency is doing all the work

This is the part the video had no room for, and it is the part that decides whether you have understood the problem or memorised its answer. Pull the two marks apart and the conclusion reverses.

Suppose the marks sit at cyclic distance $2$ instead of $1$. There are still four blanks. But now neither mark has a mark behind it, so both of the slots immediately behind a mark are blanks you could be standing on. The step risk doubles to $2/4 = 1/2$, which is worse than re-drawing, and the correct advice flips.

On a ring of six there are only three shapes a pair of marks can take, and two of them punish you for advancing. The general rule is short: the number of dangerous blanks is two minus one for every place where a mark is followed by another mark. A block of touching marks hides its own interior behind itself.

The most surprising member of the family is the smallest. With a single mark on a ring of $n$ slots, advancing is $1/(n-1)$ and re-drawing is $1/n$, so re-drawing wins for every $n$. One mark and you should always re-draw. Two touching marks and you should never re-draw. Nothing about the two situations looks different from the outside, and a candidate who has only memorised the answer to the second will confidently give it for the first.

The general ring

Take $n$ slots and a single block of $k$ touching marks, with $1 \le k < n$. There are $n-k$ blanks, and exactly one of them sits behind the block, so

Advancing wins when $1/(n-k) < k/n$, which rearranges to $n(k-1) > k^2$. For $k = 1$ the left side is zero and advancing never wins, as we just saw. For $k \ge 2$ the condition reads

At $k = 2$ that threshold is $4$, so advancing wins on every ring with five slots or more, and the six-slot version of the puzzle sits comfortably past the boundary. The specialisation of (5) at $k=2$ is worth keeping, since the gap is what tells you how much the choice is worth:

The gap is positive for $n > 4$, zero at $n = 4$ where the two moves are genuinely interchangeable, and negative below that. It also decays like $1/n$, so on a long ring the advantage of advancing shrinks toward nothing even though it never disappears. At $n = 6$ equation (7) returns $2/24 = 1/12$, which is equation (4) again.

Why the argument is really about information

There is a cleaner way to see the whole thing, and it is the version worth saying out loud under pressure. Re-drawing throws away everything you learned. Advancing keeps it. The blank landing is a genuine observation with a genuine likelihood attached, and a move that re-randomises your state deletes that observation from the record.

Stated that way it sounds like it should always favour advancing, which is exactly the trap the single-mark case sets. Information is worth having, and it is still possible for the informed move to be the worse bet, because the geometry of what you learned matters as much as the fact that you learned something. Four blanks with one killer among them beats a fresh draw. Five blanks with one killer among them does not.

Sources and further reading

• Conditional probability and the ratio in equation (2) — Conditional probability, and Bayes' theorem for the same computation in its usual dress
• Uniform draws on a finite set, which is what both moves are — Discrete uniform distribution
• Cyclic order, and what "the next slot" means on a ring — Cyclic order

Every number above was checked twice before publication: in exact rational arithmetic, and by walking all six landings against all six rotations of the mark pair and counting. The adjacency claim was checked a third way, against a run built to break it, where the marks are separated and advancing becomes the worse move.