Year: 2018
Paper: 2
Question Number: 12
Course: LFM Stats And Pure
Section: Geometric Distribution
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In a game, I toss a coin repeatedly. The probability, $p$, that the coin shows Heads on any given toss is given by
\[ p= \frac N{N+1} \,, \]
where $N$ is a positive integer. The outcomes of any two tosses are independent.
The game has two versions. In each version, I can choose to stop playing after any number of tosses, in which case I win £$H$, where $H$ is the number of Heads I have tossed. However, the game may end before that, in which case I win nothing.
\begin{questionparts}
\item In version 1, the game ends when the coin first shows Tails (if I haven't stopped playing before that).
I decide from the start to toss the coin until a total of $h$ Heads have been shown, unless the game ends before then. Find, in terms of $h$ and $p$, an expression for my expected winnings and show that I can maximise my expected winnings by choosing $h=N$.
\item In version 2, the game ends when the coin shows Tails on two \textit{consecutive} tosses (if I haven't stopped playing before that).
I decide from the start to toss the coin until a total of $h$ Heads have been shown, unless the game ends before then. Show that my expected winnings are
\[ \frac{ hN^h (N+2)^h}{(N+1)^{2h}} \,.\]
In the case $N=2\,$, use the approximation $\log_3 2 \approx 0.63$ to show that the maximum value of my expected winnings is approximately £3.
\end{questionparts}\begin{questionparts}
\item Since we either win $h$ or $0$, to calculate the expected winnings we just need to calculate the probability that we get $h$ consecutive heads, therefore:
\begin{align*}
&& \mathbb{E}(\text{winnings}) &= E_h \\
&&&= h \cdot \left ( \frac{N}{N+1} \right)^h \\
&& \frac{E_{h+1}}{E_h} &= \frac{h+1}{h }\left ( \frac{N}{N+1} \right)
\end{align*}
Therefore $E_h$ is increasing if $h \leq N$, so we can maximise our winnings by taking $h = N$. (In fact, we could take $h = N$ or $h = N+1$, but arguably $h = N$ is better as we have the same expected value but lower variance).
\item We can have up to $h$ tails appearing (if we imagine slots for tails of the form $\underbrace{\_H\_H\_H\_\cdots\_H}_{h\text{ spaces and }h\, H}$ so, we have
\begin{align*}
&& \mathbb{P}(\text{wins}) &= \sum_{t = 0}^h \mathbb{P}(\text{wins and } t\text{ tails}) \\
&&&= \sum_{t = 0}^h\binom{h}{t} \left ( \frac{N}{N+1} \right)^h\left ( \frac{1}{N+1} \right)^t \\
&&&= \left ( \frac{N}{N+1} \right)^h \sum_{t = 0}^h\binom{h}{t}\left ( \frac{1}{N+1} \right)^t \cdot 1^{h-t} \\
&&&= \left ( \frac{N}{N+1} \right)^h \left ( 1 + \left ( \frac{1}{N+1} \right) \right)^h \\
&&&= \left ( \frac{N}{N+1} \right)^h \left ( \frac{N+2}{N+1}\right)^h \\
&&&= \frac{N^h(N+2)^h}{(N+1)^{2h}} \\
\Rightarrow && \E(\text{winnings}) &= h \cdot \frac{N^h(N+2)^h}{(N+1)^{2h}}
\end{align*}
If $N = 2$, we have
\begin{align*}
&& \E(\text{winnings}) &= E_h \\
&&&= h \cdot \frac{2^h\cdot2^{2h}}{3^{2h}}\\
&&&= h \cdot \frac{2^{3h}}{3^{2h}} \\
\Rightarrow && \frac{E_{h+1}}{E_h} &= \frac{h+1}{h} \frac{8}{9} \\
\end{align*}
Therefore to maximise the winnings we should take $h = 8$, and the expected winnings will be:
\begin{align*}
&& E_8 &= 8 \cdot \frac{2^{24}}{3^{16}} \\
\Rightarrow && \log_3 E_8 &= 27 \log_3 2 - 16 \\
&&&\approx 24 \cdot 0.63 - 16 \\
&&&\approx 17 - 16 \\
&&&\approx 1 \\
\Rightarrow && E_8 &\approx 3
\end{align*}
\end{questionparts}
This question was the most poorly attempted of all of the questions. While approximately one fifth of candidates attempted this question (more than questions 10 or 11), many of the solutions scored very low marks. The few candidates who were able to make progress on the question were however able to secure very good marks. There were a number of candidates who clearly did not understand the payoffs described in this question, thinking for example that the process continued until a tail was reached and was then related to the number of heads achieved. Many of the students attempted differentiation to maximise the expected winnings, but often did not progress beyond the non-integer value found to check the two possible integer values. In the second part many candidates struggled to find a useful method of counting successful outcomes, and therefore could not make much further progress on the question. In many cases the quality of explanation seen accompanying the method was not sufficiently detailed to demonstrate that a valid method was being attempted – candidates would be well advised to pay attention to the explanation of their method in questions such as this. In the final part, many candidates failed to recognise the similarity between the function to be maximised and that from the first part of the question and therefore attempted to work through the process again. The manipulation of logarithms for the final part of the question was generally well done.