2026 January Momath Mindbender
Published:
Puzzle from here. Consider a ladybug starting on the 12-hand of the clock. During each timestep, it will choose to walk either counterclockwise or clockwise, each with 50% probability. What is the probability that the last new number it visits is 6?
It will be useful to establish consistent notation. Let \(p_y\) be the probability that \(y\) is the last new number the ladybug visits. Let \(X_t\) be the position of the ladybug. Let \(T_y\) be the first time where \(X_t = y\).
I will also use \(P_x( ... )\) to denote the probability of an event given the starting point of a random walk of \(x\).
Gamblers Ruin
The first solution I came up with relies on the setup called “Gambler’s Ruin”. Consider a 1-D symmetric random walk starting at \(0 < x < d\). I claim that \(P_x(T_d < T_0) = \frac{x}{d}\).
First, observe that \(E(X_t) = x\) (this is called a Martingale process). This is because at each time step, we make no expected progress towards either endpoint. Therefore, at time \(T = \min( T_d, T_0)\), then we obtain the equation
\[x = E(X_0) = E(X_T) = 0 P(X_T = 0) + d P(X_T = d) = d P(T_d < T_0)\]which proves our claim.
Solution One
Now, to solve the puzzle, it remains to enumerate some simple cases. Consider the sequence in which the ladybug sees new numbers. Then, in order for 6 to be last, then it must either see 5 before 7 before 6, or 7 before 5 before 6. That is,
\[p_6 = P_{12}(T_5 < T_7 < T_6) + P_{12}(T_7 < T_5 < T_6)\]We can decompose
\[P_{12}(T_5 < T_7 < T_6) = P_{12}(T_5 < T_7) \cdot P_5(T_7 < T_6)\]and
\[P_{12}(T_7 < T_5 < T_6) = P_{12}(T_7 < T_5) \cdot P_7(T_5 < T_6)\]so that plugging in our expression
\[p_6 = (1/2) \cdot (1/11) + (1/2) \cdot (1/11) = 1/11\]Solution Two
Following this, I immediately noticed that \(p_1 = p_2 = \ldots = p_{11} = 1/11\). All the probabilities are uniform! This hints that there is a simpler explanation.
Indeed, in order for 6 to be the last new number visited, there must be an arc between 5 to 7, or from 7 to 5. All the steps before reaching either 5 or 7 are superfluous. Therefore, $p_x = P_5(T_7 < T_6)$.
In words, the probability of seeing 6 last is the probability of starting to one side of 6 and reaching the other side of 6 without hitting 6. This holds for any starting position.
Now, as there are 11 probabilities, the final step is to show that they sum to 1. One way of doing this (although there are likely many) is to shift a perspective: instead of varying the starting point, vary the end point. As there must be some number that is the last one visited, those probabilities must add to 1.