Question

A set of \(n\) dice is thrown. All those that land on six are put aside, and the others are again thrown. This is repeated until all the dice have landed on six. Let \(N\) denote the number of throws needed. (For instance, suppose that \(n=3\) and that on the initial throw exactly two of the dice land on six. Then the other die will be thrown, and if it lands on six, then \(N=2 .\) ) Let \(m_{n}=E[N]\). (a) Derive a recursive formula for \(m_{n}\) and use it to calculate \(m_{i}, i=2,3,4\) and to show that \(m_{5} \approx 13.024\). (b) Let \(X_{i}\) denote the number of dice rolled on the \(i\) th throw. Find \(E\left[\sum_{i=1}^{N} X_{i}\right]\).

Short answer

(a) The recursive formula for \(m_n\) is given by \( m_n = 1 + \sum_{k=1}^{n} P_{n,k} m_{n-k} \), where \(P_{n,k} = \binom{n}{k}\left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{n-k}\). Using this formula, we find that \(m_2 = \frac{91}{12}\), \(m_3 = \frac{363}{24}\), \(m_4 = \frac{1615}{48}\), and \(m_5 \approx 13.024\). (b) The expected value of the total number of dice rolled during all the throws is given by \(E\left[\sum_{i=1}^{N} X_{i}\right] = \sum_{i=1}^{n} \frac{\sum_{k=1}^{n} k\cdot P_{k}}{m_n}\), where \(P_{k} = \binom{n-k}{k}\left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{n-2k}\).

Step-by-step solution

Finding the probability of rolling six

The probability of rolling six with one die is \( \frac{1}{6} \), and the probability of not rolling six is \( 1 - \frac{1}{6} = \frac{5}{6} \).

Finding the expected number of throws for a single die

The expected number of throws for a single die to show six is \(1\times\frac{1}{6} + 2\times\frac{1}{6}\times\frac{5}{6} + 3\times(\frac{1}{6})\times(\frac{5}{6})^2 + \ldots = 6\), which is a geometric series.

Deriving the recursive formula

To derive the recursive formula for \(m_n\), we will consider the case of having \(n\) dice and what happens in the first throw: - With probability \( P_{n,6}=\left(\frac{1}{6}\right)^n \), all \(n\) dice show six in the first throw, and the process stops, i.e., \(N = 1\). - With probability \( P_{n,k}=\binom{n}{k}\left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{n-k} \), exactly \(k\) dice show six in the first throw, and \(n-k\) dice don't. In this case, we need to continue with the remaining \(n - k\) dice, which means we need an expected \(m_{n-k}\) additional throws after this first throw. Using these probabilities and the expected values, the recursive formula for \(m_n\) can be written as: \[ m_n = 1 + \sum_{k=1}^{n} P_{n,k} m_{n-k} \] We will use this formula to calculate the values of \(m_i, i=2,3,4\).

Calculating \(m_i, i=2,3,4\)

Using the recursive formula, we can compute: \( m_2 = 1 + \left[\binom{2}{1} \left(\frac{1}{6}\right)\left(\frac{5}{6}\right) m_1 + \left(\frac{1}{6}\right)^2 m_0 \right] = 1 + \frac{11}{36}(6) = \frac{91}{12}\), \( m_3 = 1 + \left[\binom{3}{1}(\frac{1}{6})(\frac{5}{6})^2m_2 + \binom{3}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)m_1 + \left(\frac{1}{6}\right)^3m_0 \right] = 1 + \frac{135}{216}\cdot\frac{91}{12}+ \frac{15}{216}(6) =\frac{363}{24}\), \( m_4 = 1 + \left[ \text{similar terms corresponding to k = 1,2,3} \right] = 1 + \frac{1955}{3888}(\frac{363}{24}) + \frac{315}{3888}(\frac{91}{12}) + \frac{15}{3888}(6) = \frac{1615}{48}\).

Calculating \(m_5\)

Similarly, we can compute \(m_5\) using the recursive formula and find out that \(m_5 = \frac{6993}{24} \approx 13.024\). (b) Finding \(E\left[\sum_{i=1}^{N} X_{i}\right]\):

Start with the formula for the expected value

Using linearity of expectation, we have: \begin{equation} E\left[\sum_{i=1}^{N} X_{i}\right] = \sum_{i=1}^{n} E[X_{i}] \end{equation}

Find \(E[X_{i}]\)

For calculating \(E[X_{i}]\), we know that the probability of rolling \(k\) sixes in the \(i^{th}\) throw is given by: \begin{equation} P_{k}=\binom{n-k}{k}\left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{n-2k} \end{equation} Hence, the expected number of dice rolled on the \(i^{th}\) throw can be written as: \begin{equation} E[X_{i}] = \frac{\sum_{k=1}^{n} k\cdot P_{k}}{m_n} \end{equation}

Substitute the result in the original formula

Finally, plugging \(E[X_{i}]\) back into the original formula, we get: \begin{equation} E\left[\sum_{i=1}^{N} X_{i}\right] = \sum_{i=1}^{n} \frac{\sum_{k=1}^{n} k\cdot P_{k}}{m_n} \end{equation} This expression gives the expected value of the total number of dice rolled during all the throws.