Problem source

\begin{questionparts} \item $X_{1},X_{2},\ldots,X_{n}$ are independent identically distributed random variables drawn from a uniform distribution on $[0,1].$ The random variables $A$ and $B$ are defined by 
\[
A=\min(X_{1},\ldots,X_{n}),\qquad B=\max(X_{1},\ldots,X_{n}).
\]
For any fixed $k$, such that $0< k< \frac{1}{2},$ let 
\[
p_{n}=\mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k).
\]
What happens to $p_{n}$ as $n\rightarrow\infty$? Comment briefly on this result. 
\item Lord Copper, the celebrated and imperious newspaper proprietor, has decided to run a lottery in which each of the $4,000,000$ readers of his newspaper will have an equal probability $p$ of winning $\pounds 1,000,000$ and their changes of winning will be independent. He has fixed all the details leaving to you, his subordinate, only the task of choosing $p$. If nobody wins $\pounds 1,000,000$, you will be sacked, and if more than two readers win $\pounds 1,000,000,$ you will also be sacked. Explaining your reasoning, show that however you choose $p,$ you will have less than a 60\% change of keeping your job. \end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& p_n &= \mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k) \\
&&&= \mathrm{P}(A\leqslant k) +\P(B\geqslant1-k) - \mathrm{P}(A\leqslant k\mbox{ or }B\geqslant1-k)\\
&&&= 1-\mathrm{P}(A\geq k) +1-\P(B \leq 1-k) - \l 1- \mathrm{P}(A\geq k\mbox{ and }B\leq 1-k)\r\\
&&&= 1 - \P(X_i \geq k) - \P(X_i \leq 1-k) + \P(k \leq X_i \leq 1-k) \\
&&&= 1 - k^n - (1-k)^n + (1-2k)^n
\end{align*}

Therefore as $n \to \infty$ $p_n \to 1$, since $k, (1-k), (1-2k)$ are all between $0$ and $1$ and so their powers will tend to $0$.

\item Let $N = 4\,000\,000$. The probability exactly one person wins is $Np(1-p)^{N-1}$. The probability exactly two people win is $\binom{N}{2} p^2 (1-p)^{N-2}$. We wish to maximise the sum of these probabilities. To find this maximum, differentiate wrt $p$.

\begin{align*}
\frac{\d}{\d p} : && \small N(1-p)^{N-1}-N(N-1)p(1-p)^{N-2} + N(N-1)p(1-p)^{N-2} - \frac12 N(N-1)(N-2)p^2(1-p)^{N-3} \\
&&= N(1-p)^{N-3} \l (1-p)^2 - \frac12(N-1)(N-2)p^2\r \\
\Rightarrow && \frac{(1-p)}{p} = \sqrt{\frac{(N-1)(N-2)}{2}} \\
\Rightarrow && p = \frac{1}{1+ \sqrt{\frac{(N-1)(N-2)}{2}}}
\end{align*}

This will be a maximum, since this is an increasing function at $p=0$ and decreasing at $p=1$ and there's only one stationary point.

Note that $p > \frac{\sqrt{2}}{(N-2)}$ and $p < \frac{\sqrt{2}}{N-1+\sqrt{2}} < \frac{\sqrt{2}}{N}$ and so:

\begin{align*}
Np(1-p)^{N-1} &< \sqrt{2}(1-\frac{\sqrt{2}}{N-2})^{N-1} \\
&\approx \sqrt{2} e^{-\sqrt{2}}
\end{align*}
\begin{align*}
\frac{N(N-1)}{2}p^2(1-p)^{N-2} &<(1-\frac{\sqrt{2}}{N-2})^{N-1} \\
&\approx e^{-\sqrt{2}}
\end{align*}

Alternatively, we can use a Poisson approximation. The number of winners is $B(N, p)$ where we are hoping $np$ is small but not zero. Therefore it's reasonable to approximation $B(N,p)$ by $Po(Np)$. (Call this value $\lambda$). Then we wish to maximise:

\begin{align*}
&& p &= e^{-\lambda} \l \lambda + \frac{\lambda^2}{2} \r \\
&&&=  e^{-\lambda} \lambda \l 1+ \frac{\lambda}{2} \r \\
\Rightarrow && \ln p &= -\lambda + \ln \lambda + \ln(1+\frac12 \lambda) \\
\frac{\d}{\d \lambda}: && \frac{p'}{p} &= -1 + \frac{1}{\lambda} + \frac{1}{2+\lambda} \\
&&&= \frac{-(2+\lambda)\lambda+2+2\lambda}{\lambda(2+\lambda)} \\
&&&= \frac{2-\lambda^2}{\lambda(2+\lambda)} \\
\Rightarrow && \lambda &= \sqrt{2}
\end{align*}
\begin{align*}
\frac{\sqrt{2}+1}{e^{\sqrt{2}}} &< \frac{\sqrt{2}+1}{1+\sqrt{2}+1+\frac{1}{3}\sqrt{2}+\frac{1}{6}} \\
&= \frac{30\sqrt{2}-18}{41} 
\end{align*}

Either way, we find we want to estimate $e^{-\sqrt{2}}(1+\sqrt{2})$

\end{questionparts}