The Riddler of this week(-end) has a simple riddle to propose, namely* given a random walk on the {1,2,…,N} torus with a ⅓ probability of death, what is the probability of death occurring at the starting point?*

The question is close to William Feller’s famous Chapter III on random walks. With his equally famous reflection principle. Conditioning on the time n of death, which as we all know is definitely absorbing (!), the event of interest is a passage at zero, or any multiple of N (omitting the torus cancellation), at time n-1 (since death occurs the next time). For a passage in zero, this does not happen if n is even (since n-1 is odd) and else it is a Binomial event with probability

For a passage in kN, with k different from zero, kN+n must be odd and the probability is then

which leads to a global probability of

i.e.

Since this formula is rather unwieldy I looked for another approach in a métro ride [to downtown Paris to enjoy a drink with Stephen Stiegler]. An easier one is to allocate to each point on the torus a probability p[i] to die at position 1 and to solve the system of equations that is associated with it. For instance, when N=3, the system of equations is reduced to

which leads to a probability of ½ to die at position 0 when leaving from 0. When letting N grows to infinity, the torus structure no longer matters and the probability of dying at position 0 implies returning in position 0, which is a special case of the above combinatoric formula, namely

which happens to be equal to

as can be [unnecessarily] checked by a direct R simulation. This √5 is actually the most surprising part of the exercise!

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