Cheese before the cat?

Question

A mouse starts at cell S of a maze. From $S$ it goes to cell $C_1$ or cell $C_2$ , each with probability $0.5$ . From $C_1$ it reaches the cheese with probability $0.4$ or moves to $C_2$ with probability $0.6$ . From $C_2$ it meets the cat with probability $0.5$ or moves to $C_1$ with probability $0.5$ . Cheese and cat end the walk.

Starting from $S$ , what is the probability the mouse reaches the cheese before the cat?

Accepted Answer

Let f(X) be the probability of reaching cheese before cat from cell X, with f(	ext{cheese}) = 1 and f(	ext{cat}) = 0. First-step analysis gives one equation per interior cell. f(C2) = 0.5 \cdot 0 + 0.5\,f(C1) = 0.5\,f(C1), f(C1) = 0.4 \cdot 1 + 0.6\,f(C2) = 0.4 + 0.6\,f(C2). Substitute the first into the second: f(C1) = 0.4 + 0.6\,(0.5\,f(C1)) = 0.4 + 0.3\,f(C1) \implies 0.7\,f(C1) = 0.4 \implies f(C1) = \frac{4}{7}. Then f(C2) = 0.5 	imes 	frac47 = 	frac27. Finally, from the start: f(S) = 0.5\,f(C1) + 0.5\,f(C2) = 0.5 \cdot 	frac47 + 0.5 \cdot 	frac27 = 	frac27 + 	frac17 = \frac{3}{…

Cheese before the cat?

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