Which one is more effective?

Enjoyed the question

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Solution:

First note that for each mouse:

P(A increases by 1) $= \frac{1}{10}$ ,

P(A decreases by 1) $= \frac{4}{10}$ ,

P(A stays the same) $= \frac{5}{10}$ .

So, ignoring mice where the results of A and B were the same:

P(A increases by 1) $= \frac{1}{5}$ , P(A decreases by 1) $= \frac{4}{5}$ .

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Hence:

For $i = 1,...,7, p_i = \frac{1}{5} p_{i+1} + \frac{4}{5} p_{i-1}$ . Looking for a solution of the form $p_n = \lambda^n$ , we get:

$5\lambda = \lambda^2 + 4$ , so $\lambda = 1, 4$ , and $p_i = A 4^i + B$ .

Boundary conditions:

$p_0 = 0 \implies B = -A$

$p_8 = 1 \implies A(4^8 - 1) = 1,$ so $A = \frac{1}{4^8 - 1}$ .

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$p_4 = \frac{4^4 - 1}{4^8 - 1} = \frac{1}{4^4 + 1} = \frac{a}{b}$ .

So $b - a = 4^4 = 2^8 = 256$ .

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Indeed any 1D random walk where the ratio between moving up 1 and down 1 is $a:b$ , and $p_0 = 0$ , is of the form $p_i = A(a^i - b^i)$ .

So the boundary condition on the top of $p_N = 1$ , gives us $p_i = \frac{a^i - b^i}{a^N - b^N}$ .