Problem source

In a lottery, each of the $N$ participants pays $\pounds c$ to the organiser and picks a number from $1$ to $N$. The organiser picks at random the winning number from $1$ to $N$  and all those participants who picked this number receive an equal share of the prize, $\pounds J$.
\begin{questionparts}
\item The participants pick their numbers independently and with equal probability. Obtain an expression for the probability that no participant picks the winning number, and hence determine the organiser's expected profit. Use the approximation
\[
\left( 1 - \frac{a}{N} \right)^N \approx \e^{-a}
\tag{$*$}
\]
 to show that if $2Nc = J$ then the organiser will expect to make a loss.
\textbf{Note}: $\e > 2$.  
 
\item Instead of the numbers being equally popular, a fraction $\gamma$ of the numbers are popular and the rest are unpopular. For each participant,  the probability of picking any given popular number is $\dfrac{a}{N}$ and the probability of picking any given unpopular number is $\dfrac{b}{N}\,$. Find a relationship between $a$, $b$ and $\gamma$. Show that, using the approximation $(*)$, the organiser's expected profit can be expressed in the form
\[
A\e^{-a} + B\e^{-b} +C
\,,
\]
where $A$, $B$ and $C$ can be written in terms of  $J$, $c$, $N$ and $\gamma$. 
In the case $\gamma = \frac18$ and $a=9b$, find $a$ and $b$. Show that, if $2Nc = J$, then the organiser will expect to make a profit.
\textbf{Note}: $\e < 3$.  
\end{questionparts}

Solution source

\begin{questionparts}
\item The probability no-one picks the winning number is $\left ( 1 - \frac{1}{N}\right)^N \approx \frac1e$. 

\begin{align*}
&& \mathbb{E}(\text{profit}) &= Nc - (1-e^{-1})J \\
&&& < Nc -(1- \tfrac12 )J \\
&&& < Nc - \frac12 J \\
&&&= \frac{2Nc-J}{2}
\end{align*}

Therefore if $J = 2Nc$ the expected profit is negative. 

\item $\,$ \begin{align*}
&& 1 &= \sum_{\text{all numbers}} \mathbb{P}(\text{pick }i) \\
&&&= \sum_{\text{popular numbers}} \mathbb{P}(\text{pick }i)+\sum_{\text{unpopular numbers}} \mathbb{P}(\text{pick }i) \\
&&&=\gamma N \frac{a}{N} + (1-\gamma)N \frac{b}{N} \\
&&&= \gamma a + (1-\gamma)b
\end{align*}

\begin{align*}
&& \mathbb{P}(\text{no-one picks winning number}) &= \mathbb{P}(\text{no-one picks winning number} | \text{winning number is popular})\mathbb{P})(\text{winning number is popular}) + \\
&&&\quad + \mathbb{P}(\text{no-one picks} | \text{unpopular})\mathbb{P}(\text{unpopular}) \\
&&&= \left (1 - \frac{a}{N} \right)^N \gamma + \left (1 - \frac{b}{N} \right)^N (1-\gamma) \\
&&&\approx \gamma e^{-a} + (1-\gamma)e^{-b} \\
\\
&& \mathbb{E}(\text{profit}) &= Nc - (1-\gamma e^{-a} - (1-\gamma)e^{-b})J \\
&&&= Nc-J+J\gamma e^{-a} +J(1-\gamma)e^{-b}
\end{align*}

If $\gamma = \frac18$ and $a=9b$, then $1=\frac18 a + \frac78b = 2b \Rightarrow b = \frac12, a = \frac92$ and

\begin{align*}
&& \mathbb{E}(\text{profit}) &= Nc-J +J\tfrac18e^{-9/2}+J\tfrac78e^{-1/2} \\
&&&= Nc-J+\tfrac18Je^{-1/2}(e^{-4}+7)
\end{align*}

If we can show $e^{-1/2}\frac{e^{-4}+7}{8} > \frac12$ we'd be done, so

\begin{align*}
&& e^{-1/2}\frac{e^{-4}+7}{8} &> \frac12 \\
\Leftrightarrow && e^{-4}+7 &>4e^{1/2} \\
\Leftrightarrow && 49+14e^{-4}+e^{-8} &>16e \\
\end{align*}

But clearly the LHS $>49$ and the RHS $<48$ so we're done

\end{questionparts}