Problem source

A very generous shop-owner is hiding small diamonds in 
chocolate bars. Each diamond is hidden independently of any other diamond, 
and on average there is one diamond per kilogram of chocolate.
\begin{questionparts}
\item I go to the shop and roll a fair six-sided die once. 
I decide  that if I roll a score of $N$, I will buy $100N$ grams 
of chocolate. 
Show that the probability that I will have no diamonds is
\[
\frac{\e^{-0.1}}{ 6} \l \frac{1 - \e^{-0.6}  }{  1 - \e^{-0.1}} \r
\]
Show also that the expected number of diamonds I find is 0.35.
\item Instead, I decide to roll a fair six-sided 
die repeatedly until I score a 6. If I roll my first 6 on my $T$th throw, 
I will buy $100T$ grams of chocolate. 
Show that the probability that I will have no diamonds is 
\[
\frac{\e^{-0.1}}{ 6 - 5\e^{-0.1}}
\]
Calculate also the expected number of diamonds that I find.
(You may find it useful to consider the
the binomial 
expansion of $\l 1 - x \r^{-2}$.)
\end{questionparts}

Solution source

Not that the number of diamonds per kilogram is $1$ so we are assuming it is $Po(M)$ where $M$ is the mass in kg. In particular $\E[X] = M$ and $\mathbb{P}(X = 0) = e^{-M}$

\begin{questionparts}
\item $\,$ \begin{align*}
&& \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^6\mathbb{P}(\text{no diamonds and roll }n)  \\
&&&= \sum_{n=1}^6 \tfrac16 e^{-\frac{n}{10}} \\
&&&= \frac{e^{-0.1}}6 \left ( \frac{1-e^{-0.6}}{1-e^{-0.1}}\right) \\
&& \E[\text{diamonds}] &=  \sum_{n=1}^6 \E(\text{diamonds}|N = n)\mathbb{P}(N = n) \\
&&&=  \sum_{n=1}^6 0.1n \cdot \frac16 \\
&&&= 0.1 \cdot \frac{7}{2} = 0.35
\end{align*}

\item $\mathbb{P}(T = k) = \left ( \frac56 \right)^{t-1} \frac16$, so

\begin{align*}
&& \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^\infty\mathbb{P}(\text{no diamonds and  }T=n)  \\
&&&= \sum_{n=1}^\infty e^{-0.1n} \left ( \frac56 \right)^{n-1} \frac16 \\
&&&= \frac{e^{-0.1}}{6} \frac1{1- \frac56 e^{-0.1}} \\
&&&= \frac{e^{-0.1}}{6 - 5e^{-0.1}} \\
\\
&& \E[\text{diamonds}] &=  \sum_{n=1}^\infty \E(\text{diamonds}|T = n)\mathbb{P}(T = n) \\
&&&= \sum_{n=1}^\infty 0.1n \cdot  \left ( \frac56 \right)^{n-1} \frac16 \\
&&&= \frac{0.1}{6}  \sum_{n=1}^\infty n \cdot  \left ( \frac56 \right)^{n-1} \\
&&&= \frac{1}{60} \frac{1}{(1- \tfrac56)^2} \\
&&&= \frac{6}{10} = \frac35
\end{align*}

\end{questionparts}