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here is the problem that I have been trying to do:

N+1 plates are laid out around a circular dining table, and a hot cake is passed between them in the manner of a symmetric random walk: each time it arrives on a plate, it is tossed to one of the two neighboring plates, each possibility having probability $\frac 12$. The game stops at the moment when the cake has visited every plate at least once. Show that, with the exception of the plate where the cake began, each plate has probability 1/N of being the last plate visited by the cake.

This is what I have so far,

Plates: 1, ..., N+1 I set k, 1 <= k <= N+1. Set the starting plate at plate #1.

Then, the event $A_k$ is the event that k is the last plate to be reached. So the probability of $A_k$ is?

Need to find the probability that the random walk reaches k and then multiply by 2 since it is a circular table? Not sure at this point.

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Pick any plate. Only way it can be last plate is cake must go to one its neighbors then reverse and go to its other neighbor before itself is reached. No matter what plate considered this is always the case. Now no matter what plate considered with probability 1 a neighbor will be reached before the plate in question. Hence all plates have same chance.

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  • $\begingroup$ Think about state of system when a neighbor plate is reached for first time. Then ask yourself how will plate in question be last plate. $\endgroup$
    – stevek
    Oct 1, 2014 at 20:42

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