Solution:
For n∈Z, let an be the fraction of the time Mr. Fat spends at n. By symmetry, an=a−n for all n.
For n>0, we have
an=52an−1+52an+1,
or
an+1=25an−an−1.
This Fibonacci-like recurrence can be solved explicitly to obtain
an=α⋅2∣n∣+β⋅2−∣n∣
for all n∈Z. Now we also have
n∈Z∑an=1,
so we better have α=0, so that a0=β and a±1=2β.
Now we also have
a0=52a−1+52a1+51,
so β=31.
This matches perfectly with ∑n∈Zan=1.