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2015 Paper 1 Q13
D: 1500.0 B: 1501.1
probability negative binomial distribution geometric distribution fair die binomial expansion conditional probability infinite series

A fair die with faces numbered \(1, \ldots, 6\) is thrown repeatedly. The events \(A\), \(B\), \(C\), \(D\) and \(E\) are defined as follows. \begin{align*} A: && \text{the first 6 arises on the \(n\)th throw.}\\ B: && \text{at least one 5 arises before the first 6.} \\ C: && \text{at least one 4 arises before the first 6.}\\ D: && \text{exactly one 5 arises before the first 6.}\\ E: && \text{exactly one 4 arises before the first 6.} \end{align*} Evaluate the following probabilities:

  1. \(\P(A)\)
  2. \(\P(B)\)
  3. \(\P(B\cap C)\)
  4. \(\P(D)\)
  5. \(\P(D\cup E)\)
For some parts of this question, you may want to make use of the binomial expansion in the form: \[ (1-x)^{-n} = 1 +nx +\frac {n(n+1)}2 x^2 + \cdots + \frac {(n+r-1)!}{r! (n-1)!}x^r +\cdots\ .\]

Solution:

  1. \(\,\) \begin{align*} \mathbb{P}(A) &= \mathbb{P}(\text{the first 6 arises on the \(n\)th throw.}) \\ &= \mathbb{P}(\text{\(n-1\) not 6s, followed by a 6.})\\ &= \left ( \frac56\right)^{n-1} \cdot \frac16 = \frac{5^{n-1}}{6^n} \end{align*}
  2. There is nothing special about \(5\) or \(6\), so which comes first is \(50:50\), therefore this probability is \(\frac12\)
  3. There is nothing special about \(4\), \(5\) or \(6\) so this is the probability that \(6\) appears last out of these three numbers, hence \(\frac13\)
  4. \(\,\) \begin{align*} \mathbb{P}(D) &= \mathbb{P}(\text{exactly one 5 arises before the first 6.}) \\ &=\sum_{n=2}^{\infty} \mathbb{P}(\text{exactly one 5 arises before the first 6 which appears on the \(n\)th roll.}) \\ &= \sum_{n=2}^{\infty} \binom{n-1}{1} \left ( \frac46 \right)^{n-2} \frac16 \cdot \frac16 \\ &= \frac1{36} \sum_{n=2}^{\infty} (n-1) \left ( \frac23 \right)^{n-2} \\ &= \frac1{36} \sum_{n=1}^{\infty} n \left ( \frac23 \right)^{n-1} \\ &= \frac1{36} \frac{1}{\left ( 1- \frac23 \right)^2} = \frac14 \end{align*}
  5. \(\,\) \begin{align*} \mathbb{P}(D \cup E) &= \mathbb{P}(D) + \mathbb{P}(E) - \mathbb{P}(D \cap E) \\ &= \frac12 - \mathbb{P}(D \cap E) \\ &=\frac12 - \sum_{n=3}^{\infty} \mathbb{P}(\text{exactly one 5 and one 4 arises before the first 6 which appears on the \(n\)th roll.}) \\ &=\frac12 - \sum_{n=3}^{\infty} 2\binom{n-1}{2} \left ( \frac36 \right)^{n-3}\cdot \frac16 \cdot \frac16 \cdot \frac16 \\ &=\frac12 - \frac2{6^3}\sum_{n=3}^{\infty} \frac{(n-1)(n-2)}{2} \left ( \frac12 \right)^{n-3} \\ &=\frac12 - \frac2{6^3}\frac{1}{(1-\tfrac12)^3}\\ &= \frac12 - \frac{2}{27} \\ &= \frac{23}{54} \end{align*}

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1999 Paper 2 Q14
D: 1600.0 B: 1516.0
probability expected value optimisation strategy fair die sequential decision discrete probability distributions stopping problem

You play the following game. You throw a six-sided fair die repeatedly. You may choose to stop after any throw, except that you must stop if you throw a 1. Your score is the number obtained on your last throw. Determine the strategy that you should adopt in order to maximize your expected score, explaining your reasoning carefully.

Solution: Once you have thrown, all previous throws are irrelevant so the only thing which can affect your decision is the current throw. Therefore the strategy must consist of a list of states we re-throw from, and a list of states we stick on. It must also be the case that if we stick on \(k\) we stick on \(k+1\) (otherwise we can improve our strategy by switching those two values around). Therefore we can form a table of our expected score: \begin{array}{c|c|c} \text{stop on} & \text{possible outcomes} & \E[\text{score}] \\ \hline \geq 2 & \{1,2,3,4,5,6\} & \frac{21}{6} = 3.5 \\ \geq 3 & \{1,3,4,5,6\} & \frac{19}{5} = 3.8 \\ \geq 4 & \{1,4,5,6\} & \frac{16}{4} = 4 \\ \geq 5 & \{1,5,6\} & \frac{12}{3} = 4 \\ =6 & \{1,6\} & \frac{7}{2} = 3.5 \end{array} Therefore the optimal strategy is to stop on \(4\) or higher. If we cared about variance we might look at the variance of the two best strategies, \(4\) or higher has a variance of \(\frac{1+16+25+36}{4} - 16 = 3.5\) and \(5\) or higher has a variance of \(\frac{1+25+36}3 - 16 = \frac{14}3 > 3.5\) so \(4\) or higher is probably better in most scenarios.

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