Does anyone know how to answer this Q worth 9MKS:

A friend bets you $100 on a game involving two six-sided dice, one red and one green.

You choose the number of times the pair of dice will be rolled. You win if the number of

times a red 6 is rolled is at most 2 and the number of times a green 6 is rolled is at least 2.

a) How many times should the dice be rolled to maximise your chance of winning?

b) With that number of rolls, what are your expected winnings?

Thanks in advance

This is quite an interesting question.

**Part a**First, it's important to realise that the results of each dice are independent of each other,

**and** that the probability of rolling a 6 on any given roll on either dice is constant.

Let \(n\) be the number of times the pair of dice are rolled, where \(n\geq 2\).

Let \(R\) be the number of observed \(6\)'s on the

**red** dice from \(n\) rolls.

Let \(G\) be the number of observed \(6\)'s on the

**green** dice from \(n\) rolls.

Note: \(R\) and \(G\) have the same distribution, so you could just define a single variable \(X\), but I chose to separate them for clarity.That is, \[R\sim \text{Bi}(n,\ 1/6)\quad\text{and}\quad G\sim\text{Bi}(n,\ 1/6).\] We are told that you win the game so long as \(R\leq 2\)

**and** \(G\geq 2\). Hence, \[\text{Pr}(\text{win})=\text{Pr}(R\leq 2)\times \text{Pr}(G\geq 2)\] From here, you could using your CAS define a function say \[b(n)=\texttt{binomCdf}(n,\,1/6,\,0,\,2)\cdot \texttt{binomCdf}(n,\,1/6,\,2,\,n)\] and then use trial and error to find the value of \(n\) that gives you the highest probability of winning. Or, you could continue as follows: \begin{align*}\text{Pr}(\text{win})&=\text{Pr}(R\leq 2)\times\Big[1-\text{Pr}(G\leq 1)\Big]\\

&=\left[\sum_{k=0}^2\binom{n}{k}\left(\frac16\right)^k\left(\frac56\right)^{n-k}\right]\left[1-\sum_{k=0}^1\binom{n}{k}\left(\frac16\right)^k\left(\frac56\right)^{n-k}\right]\\

&\qquad \vdots\\

&=\frac{1}{250}\left(\frac{5}{36}\right)^n(n^2+9n+50)\big[5\!\times\!6^n-5^n(n+5)\big] \end{align*}

Using a graph of \(\text{Pr}(\text{win})\) against \(n\), it is quite easy to see that \(\boxed{n=12\,}\) rolls will maximise the chance of winning.

**Part b**When \(n=12\), \[\text{Pr}(\text{win})\approx 0.419101\quad\text{(6DP)}\] and so your expected

**earnings** from the game is \[E=\$100 \times 0.419101=\$41.91\quad\text{(nearest cent)}.\] Assuming you also bet \(\$100\) to play the game, you expected

**winnings** is \[W=\$41.91-\$100=\boxed{-\$58.09\,}\quad \text{(nearest cent)}\]