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Problem Text

A game in a casino is played with a fair coin and an unbiased  cubical die whose faces are labelled $1, 1, 1, 2, 2$ and $3.$  In each  round of the game, the die is rolled once and the coin is tossed  once. The outcome of the round is a random variable $X$. The value,  $x$, of $X$ is determined as follows. If the result of the toss is  heads then $x= \vert ks -1\vert$, and if the result of the toss is  tails then $x=\vert k-s\vert$, where $s$ is the number on the die  and $k$ is a given number.  Show that $\mathbb{E}(X^2) = k +13(k-1)^2 /6$.  Given that both $\mathbb{E}(X^2)$ and $\mathbb{E}(X)$ are positive integers, and
  that $k$ is a single-digit positive integer, determine the value  of $k$, and write down the probability distribution of $X$.
  A gambler pays  $\pounds 1$ to play the game, which  consists of two rounds. The gambler is paid:
\begin{enumerate}
 \item $\pounds w$, where $w$ is an integer, if the sum of the outcomes of the two rounds  exceeds $25$;
\item
 $\pounds 1$ if the sum of the outcomes equals $25$;
\item 
nothing if the sum of the outcomes is less that $25$.
\end{itemize}
Find, in terms of $w$, an expression  for the amount the gambler expects to be paid in a game, and deduce  the maximum possible value of $w$, given that the casino's owners choose $w$ so that the game is in their favour.

Solution (Optional)

\begin{align*}
&& \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right)  \right) \\
&&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) +  \frac16 \left ( 6k^2-20k+20\right) \right) \\
&&&= \frac1{12} \left (26k^2-40k+ 26\right) \\
&&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\
&&&= \frac{13}{6}(k-1)^2+k
\end{align*}

Since $k$ a single digit positive number and $\mathbb{E}(X^2)$ is an integer, $6 \mid k-1 \Rightarrow k = 1, 7$.

\begin{align*}
\mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right)  \right) = \frac23 \not \in \mathbb{Z}\\
\mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right)  \right) = 8
\end{align*}
Therefore $k = 7$

The probability distribution is

\begin{align*}
&& \mathbb{P}(X=4) = \frac1{12} \\
&& \mathbb{P}(X=5) = \frac1{6} \\
&& \mathbb{P}(X=6) = \frac12 \\
&& \mathbb{P}(X=13) = \frac1{6} \\
&& \mathbb{P}(X=20)= \frac1{12} \\
\end{align*}

The only ways to score more than $25$ are:
$20+6, 20+13, 20+20, 13+13$

The only ways to score exactly $25$ are $20+5$

\begin{align*}
\mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\
&= \frac{7}{48} \\
\mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\
\\
\mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144}
\end{align*}

The casino needs $\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}$

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