In this Jane Street puzzle, its solution says the mean stroll length is $20$. The arugment is kind of plausible but not clear. I would like to see a rigorous proof utilizing the symmetry.
2026-04-02 05:21:17.1775107277
The mean random walk on a spherical lattice excursion length
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For any irreducible finite Markov chain, the expected return time to the initial state $x_0$ is $1/\pi(x_0)$, where $\pi$ is the stationary distribution. See [1] or [2], Prop. 1.19 page 13. For simple random walk on a regular graph, the stationary distribution is uniform (see example 1.21 in [2]) so $1/\pi(x_0)$ is just the number of states of the chain, i.e. the number of white hexagons on the soccer ball.
[1] James Robert Norris. Markov chains. Cambridge university press, 1998.
[2] https://www.yuval-peres-books.com/markov-chains-and-mixing-times/ https://pages.uoregon.edu/dlevin/MARKOV/mcmt2e.pdf