Steps for a bug on a triangle to fall in the hole

Question

A bug sits at corner X of a triangle with corners X, Y, Z. Each step it hops to one of the other two corners, each with probability 1/2. Corner Z has a hole: once the bug reaches Z it falls in and stops. Starting at X, on average how many hops until the bug falls in the hole?

Accepted Answer

Let eX and eY be the expected hops to reach Z from X and from Y; Z has value 0. By the symmetry of X and Y (each connects to the other non-hole corner and to Z), eX = eY =: e. From X, one hop goes to Y (value e) or to Z (value 0), each with probability 1/2: e = 1 + 	frac12 e + 	frac12 \cdot 0 \;\Longrightarrow\; 	frac12 e = 1 \;\Longrightarrow\; e = 2. Takeaway. With symmetry you can collapse equal-by-symmetry states into one unknown; the bug reaches the hole in 2 hops on average.

Steps for a bug on a triangle to fall in the hole

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