Markov Dispenser

Solution

Answer: 361.22

Let \(X_n\) be the \(n\)-th number dispensed, where \(X_n\in\{5,10\}\). We want \[ \mathbb{E}[X_1+X_2+\cdots+X_{50}] = \sum_{n=1}^{50}\mathbb{E}[X_n]. \] So it is enough to find the expected value of each term. Let \[ p_n=\Pr(X_n=5). \] Since the first number is always \(10\), \[ p_1=0. \] Now build a recurrence for \(p_n\). If the previous number was \(5\), then the next number is \(5\) with probability \(0.7\). If the previous number was \(10\), then the next number is \(5\) with probability \(0.4\). So \[ p_{n+1}=0.7p_n+0.4(1-p_n). \] \[ p_{n+1}=0.4+0.3p_n. \] This recurrence settles toward an equilibrium value \(p\) satisfying \[ p=0.4+0.3p. \] So: \[ 0.7p=0.4 \qquad\Rightarrow\qquad p=\frac{4}{7}. \] The general solution is therefore \[ p_n=\frac{4}{7}+\left(p_1-\frac{4}{7}\right)(0.3)^{\,n-1}. \] Since \(p_1=0\), \[ p_n=\frac{4}{7}\left(1-(0.3)^{\,n-1}\right). \] Now compute the expected value of the \(n\)-th number: \[ \mathbb{E}[X_n] = 5p_n+10(1-p_n) = 10-5p_n. \] \[ \mathbb{E}[X_n] = 10-5\cdot \frac{4}{7}\left(1-(0.3)^{\,n-1}\right). \] So \[ \mathbb{E}[X_n] = 10-\frac{20}{7}+\frac{20}{7}(0.3)^{\,n-1} = \frac{50}{7}+\frac{20}{7}(0.3)^{\,n-1}. \] Now sum from \(n=1\) to \(50\): \[ \mathbb{E}[X_1+\cdots+X_{50}] = \sum_{n=1}^{50}\left(\frac{50}{7}+\frac{20}{7}(0.3)^{\,n-1}\right). \] \[ = 50\cdot \frac{50}{7} + \frac{20}{7}\sum_{n=1}^{50}(0.3)^{\,n-1}. \] The geometric series is \[ \sum_{n=1}^{50}(0.3)^{\,n-1} = \sum_{k=0}^{49}(0.3)^k = \frac{1-(0.3)^{50}}{1-0.3} = \frac{1-(0.3)^{50}}{0.7}. \] So \[ \mathbb{E}[X_1+\cdots+X_{50}] = \frac{2500}{7} + \frac{20}{7}\cdot \frac{1-(0.3)^{50}}{0.7}. \] Since \(0.7=\frac{7}{10}\), \[ \frac{20}{7}\cdot \frac{1}{0.7} = \frac{20}{7}\cdot \frac{10}{7} = \frac{200}{49}. \] Hence \[ \mathbb{E}[X_1+\cdots+X_{50}] = \frac{2500}{7} + \frac{200}{49}\left(1-(0.3)^{50}\right) \approx 361.22 \] The expected sum is 361.22.