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We have a fair coin, and start at the 12 o'clock marker on a clock. At each step, flip the coin. If heads, move clockwise, if tails move counter-clockwise. As you land on a number, mark that number as visited.

Which number(s) on the clock has the highest probability of being the last one to be visited? Which one has the lowest? Alternatively, what are the probabilities of each number being the last one to be visited?

I ran some simulations in python. The probabilities seem evenly distributed across 2-10 and 12, but half as big for 1 and 11. My intuition tells me that 6 might be the reasonable choice for "most likely to be last", but intuition isn't very good for probability.

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Can you show the simulation code? The results you got are highly counter-intuitive (but the intuition, you know...) – user58697 Mar 13 '14 at 4:38
Do you stop after a fixed number of moves? – awkward Mar 13 '14 at 11:22
You stop once everything has been visited. – Coin_problem_arm Mar 13 '14 at 13:20
@user58697: See my answer ;-) – joriki Apr 18 at 20:46

This problem is treated at Random walk on $n$-cycle, with the slight difference that the starting number is also marked. In that case, all numbers other than the starting number have the same probability to be the last one visited. You are effectively giving equal probability $\frac12$ for $1$ or $11$ to be the starting number in that problem, so all other numbers have the same probability to be the last, and these two have half of that probability because they can only be the last number if the other one is selected as the starting number, with probability $\frac12$. So your simulation was right and user58697's intuition was wrong.

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