Please help me solve the following two questions. I know the answers, but there was no help on how to get them.
- If I paid a dollar per point (333 pays 9, 444 12, 666 18, 321 6), what's the maximum you're willing to pay? Now, how much value does option for the second roll add?
- If you win 2 dollars for each dice you roll 4 or higher assuming you can roll 3 dice, how much would you be willing to pay for this game?
2.1 How much would you pay if you have the option to roll again (voiding results of first roll) but the second roll is with only 2 dice?
$Q1$. Let $X_i, i=1,2,3,$ be the number of points with the $i^{th}$ die, and let $X$ be the total number of points. So $X = X_1 + X_2 + X_3$. Then,
\begin{eqnarray*} E\left(X\right) &=& E\left(X_1\right) + E\left(X_2\right) + E\left(X_3\right) \\ &=& 3 E\left(X_1\right) \\ &=& 3 \times \dfrac{7}{2} \\ &=& \$10.50 \end{eqnarray*}
So to have an advantage you would be prepared to pay $\$10.49$.
The second part of this question is answered at the end.
$Q2$. Here, let $X_i, i=1,2,3,$ be the amount returned from the $i^{th}$ die, and let $X$ be the total return. So again, $X = X_1 + X_2 + X_3$.
\begin{eqnarray*} E\left(X\right) &=& E\left(X_1\right) + E\left(X_2\right) + E\left(X_3\right) \\ &=& 3 E\left(X_1\right) \\ &=& 3 \times 2 \times \dfrac{1}{2} \\ &=& \$3.00 \end{eqnarray*}
So to have an advantage you would be prepared to pay $\$2.99$.
$Q2.1$. If we rolled only two dice, then by a similar process to $Q2$ we see the expected return is $2 \times 2 \times \frac{1}{2} = \$2.00$.
Therefore, it is beneficial to use the option of a second roll only when the return from the first roll is less than $\$2.00$, which means it is $\$0.00$.
Let $X$ be the return from the first roll, and $Y$ the return from the second roll, and $Z$ the total return. So $Z = X + Y$.
The probability distribution for $X$ and $Y$:
\begin{eqnarray*} P\left(X = 0\right) &=& \frac{1}{8} \\ P\left(X = 2\right) &=& \frac{3}{8} \\ P\left(X = 4\right) &=& \frac{3}{8} \\ P\left(X = 6\right) &=& \frac{1}{8} \\ && \\ P\left(Y = 0\right) &=& \frac{7}{8} + \frac{1}{8} \times \frac{1}{4} = \frac{29}{32}\\ P\left(Y = 2\right) &=& \frac{1}{8} \times \frac{1}{2} = \frac{2}{32} \\ P\left(Y = 4\right) &=& \frac{1}{8} \times \frac{1}{4} = \frac{1}{32} \end{eqnarray*}
Thus, \begin{eqnarray*} E\left(X\right) &=& 2 \times \frac{3}{8} + 4 \times \frac{3}{8} + 6 \times \frac{1}{8} = 3 \\ E\left(Y\right) &=& 2 \times \frac{2}{32} + 4 \times \frac{1}{32} = 0.25 && \\ E\left(Z\right) &=& E\left(X\right) + E\left(Y\right) \\ &=& \$3.25. \end{eqnarray*}
The option increases the expected return by $\$0.25$. So to have an advantage you would be prepared to pay $\$0.24$ for the option of a second roll.
The second part of $Q1$ I left till now because I'm not sure of your meaning for "option of a second roll". I'll assume it's similar to that in $Q2.1$: specifically, that you can pay (upfront) for an option to roll the $3$ dice again if you don't like the outcome of the first roll of the $3$ dice. Let me know if you mean something different.
It's clear that you would choose to roll again only if the return of the first roll was less than $\$10.50$.
We proceed in a similar way to $Q2.1$. Let $W$ be the return from the first roll, let $V$ be the return from the second roll, and let $U$ be the total return. So $U = W + V$. The probability distribution for $W$ has ($216 = 6^3$):
\begin{eqnarray*} P\left(W = 11\right) &=& 27/216 \\ P\left(W = 12\right) &=& 25/216 \\ P\left(W = 13\right) &=& 21/216 \\ P\left(W = 14\right) &=& 15/216 \\ P\left(W = 15\right) &=& 10/216 \\ P\left(W = 16\right) &=& 6/216 \\ P\left(W = 17\right) &=& 3/216 \\ P\left(W = 18\right) &=& 1/216 \end{eqnarray*}
Thus, \begin{eqnarray*} E\left(W\right) &=& \frac{1}{216}\left(11\times 27 + 12\times 25 + 13\times 21 + 14\times 15 + 15\times 10 + 16\times 6 + 17\times 3 + 18 \right) \\ &=& 1395 / 216 \\ &\approx& \$6.46 \end{eqnarray*}
And, \begin{eqnarray*} E\left(V\right) &=& \frac{1}{2} E\left(X \right) \qquad\mbox{(since we re-roll in exactly half the cases)} \\ &=& \frac{1}{2} \times 10.50 \\ &=& \$5.25 \end{eqnarray*}
Thus,
\begin{eqnarray*} E\left(U\right) &=& E\left(W\right) + E\left(V \right) \\ &\approx& 6.46 + 5.25 \\ &=& \$11.71 \end{eqnarray*}
The option increases the expected return by $\$1.21$. So to have an advantage you would be prepared to pay $\$1.20$ for the option of a second roll.