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2018 Paper 2 Q13
D: 1600.0 B: 1502.8
probability tree diagram recurrence relation Markov chain random walk circular motion symmetry pass the parcel

Four children, \(A\), \(B\), \(C\) and \(D\), are playing a version of the game `pass the parcel'. They stand in a circle, so that \(ABCDA\) is the clockwise order. Each time a whistle is blown, the child holding the parcel is supposed to pass the parcel immediately exactly one place clockwise. In fact each child, independently of any other past event, passes the parcel clockwise with probability \(\frac{1}{4}\), passes it anticlockwise with probability \(\frac{1}{4}\) and fails to pass it at all with probability \(\frac{1}{2}\). At the start of the game, child \(A\) is holding the parcel. The probability that child \(A\) is holding the parcel just after the whistle has been blown for the \(n\)th time is \(A_n\), and \(B_n\), \(C_n\) and \(D_n\) are defined similarly.

  1. Find \(A_1\), \(B_1\), \(C_1\) and \(D_1\). Find also \(A_2\), \(B_2\), \(C_2\) and \(D_2\).
  2. By first considering \(B_{n+1}+D_{n+1}\), or otherwise, find \(B_n\) and \(D_n\). Find also expressions for \(A_n\) and \(C_n\) in terms of \(n\).

Solution:

  1. \(\,\) \begin{align*} && A_1 &= \frac12 \\ && B_1 &= \frac14 \\ && C_1 &= 0 \\ && D_1 &= \frac14 \end{align*} \begin{align*} && A_2 &= \frac12 \cdot \frac12 + 2 \cdot \frac14 \cdot \frac14 = \frac38 \\ && B_2 &= \frac14 \cdot \frac12 + \frac12 \cdot \frac14 = \frac14 \\ && C_2 &=2 \cdot \frac14 \cdot \frac14 =\frac18 \\ && D_2 &= B_2 = \frac14 \end{align*}
  2. \begin{align*} && A_{n+1} &= \frac12 A_n+ \frac14(B_n + D_n) \\ && B_{n+1} &= \frac12 B_n+ \frac14(A_n + C_n) \\ && C_{n+1} &= \frac12 C_n+ \frac14(D_n +B_n) \\ && D_{n+1} &= \frac12 D_n+ \frac14(C_n +A_n) \\ \\ \Rightarrow && B_{n+1}+D_{n+1} &= \frac12 (B_n+D_n) + \frac12(A_n+C_n) \\ &&&= \frac12 \\ \Rightarrow && B_{n+1}&=D_{n+1} = \frac14 \\ \\ && C_{n+1} &= \frac12C_n + \frac14 \cdot \frac12 \\ &&&= \frac12 C_n + \frac18\\ &&&= \frac12 C_{n-1} + \frac1{8} + \frac1{16} \\ &&&= \frac1{8} + \frac{1}{16} + \cdots + \frac{1}{8 \cdot 2^{n-1}} \\ &&&= \frac18 \left (1 + \frac12 + \cdots + \frac1{2^{n-1}} \right) \\ &&&= \frac18\left ( \frac{1-\frac1{2^n}}{1-\frac12} \right) \\ &&&= \frac18 \left (2 - \frac{1}{2^{n-1}} \right) \\ &&&= \frac14 - \frac{1}{2^{n-1}} \\ \Rightarrow && A_n &= \frac14 + \frac1{2^{n-1}} \end{align*}

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