The last hour on the clock
A ladybug sits on the 12 of a clock face. At each step it moves to one of the two neighbouring hour marks with probability . It keeps going until every hour mark has been visited.
Find the probability that the 6 is the last hour mark to be visited.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
For the 6 to be last, the ladybug must first reach one of its neighbours, 5 or 7. What must happen next?
From 5, it must go all the way around the other way, to 7, before touching 6. That is a gambler's ruin problem.
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s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. The 6 is visited last exactly when, after first reaching a neighbour of 6, the ladybug walks all the way around the clock to the other neighbour before stepping on 6. By gambler's ruin that has the same probability whatever the mark, so all 11 marks are equally likely to be last.
Step 1: reach a neighbour. The ladybug will certainly reach 5 or 7 at some point; say it reaches 5 first. At that moment 6 and 7 have not been visited, and 6 is last if and only if the ladybug now reaches 7 before 6.
Step 2: gambler's ruin. Going from 5 to 7 without touching 6 means going the long way round: 10 steps counterclockwise. Unfold the clock into a line: the ladybug starts 1 step from 6 and 10 steps from 7 the other way. A fair walk reaches the far end first with probability
Step 3. The same happens if it reaches 7 first. So the probability that 6 is last is .
The surprise. The same argument works for any mark other than 12: every one of the 11 marks is last with probability . The 1, right next to the start, is as likely to be last as the 6 on the opposite side.