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Jimmy has two red disks and one blue disk, and he places them in a row such that the blue disk is in the center. Every move, Jimmy switches one of the outer disks with the center disk. Find the probability that the blue disk is in the center after $n$ moves.

I let $E_n$ be the probability that the blue disk is in the center after $n$ moves. Then, I had \begin{align*} E_1 &= 0 \\ E_2 &= \frac{1}{2} 1 + \frac{1}{2} E_1 = \frac{1}{2}. \end{align*} However, I wasn't sure how to proceed on the case for $E_3.$

questionasker
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  • The aim of this exercise is that you learn to construct a recursion relation between the probabilities after $n$ turns in terms of the probabilities after $n-1$ terms. You can tem use the initial state (Blue in the middle) to derive the required solution. – M. Wind Jun 19 '21 at 17:35
  • $E$ usually means expectation. Better to use $P$ for probabilities. – TonyK Jun 19 '21 at 19:13

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After $n-1$ turns, let $L(n-1)$, $M(n-1)$, $R(n-1)$ be the probabilities that the blue disk is in the Left, Middle or Right position. We now perform the next turn and compute the new probabilities. We obtain the following recursion relations:

$$L(n) = (1/2)L(n-1) + (1/2)M(n-1)$$ $$M(n) = (1/2)L(n-1) + (1/2)R(n-1)$$ $$R(n) = (1/2)R(n-1) + (1/2)M(n-1)$$

Summing the three equations we see that the sum of probabilities is conserved, as expected. Since $L+ M + R = 1$ holds, we can use this to rewrite the equation for $M$ as follows:

$$M(n) = 1/2 - (1/2)M(n-1)$$

We can easily verify that $M = 1/3$ is a possible solution. Indeed, for large $n$ the $M(n)$ converge to this value. The trend towards this constant value is oscillatory. By plugging in the initial values for M we get numbers that confirm the oscillatory behaviour. The exact solution is found to be:

$$M(n) = 1/3 + (2/3) (-1/2)^n$$

M. Wind
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