Question

Consider a game in which a red die and a blue die are rolled. Let \(x_{R}\) denote the value showing on the uppermost face of the red die, and define \(x_{B}\) similarly for the blue die. a. The probability distribution of \(x_{R}\) is $$ \begin{array}{lrrrrrr} x_{R} & 1 & 2 & 3 & 4 & 5 & 6 \\ p\left(x_{R}\right) & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 \end{array} $$ Find the mean, variance, and standard deviation of \(x_{R}\). b. What are the values of the mean, variance, and standard deviation of \(x_{B} ?\) (You should be able to answer this question without doing any additional calculations.) c. Suppose that you are offered a choice of the following two games: Game 1: Costs $$\$ 7$$ to play, and you win \(y_{1}\) dollars, where \(y_{1}=x_{R}+x_{B}\) Game 2: Doesn't cost anything to play initially, but you "win" \(3 y_{2}\) dollars, where \(y_{2}=x_{R}-x_{B}\). If \(y_{2}\) is negative, you must pay that amount; if it is positive, you receive that amount. For Game 1, the net amount won in a game is \(w_{1}=\) \(y_{1}-7=x_{R}+x_{B}-7 .\) What are the mean and standard deviation of \(w_{1}\) ? d. For Game 2, the net amount won in a game is \(w_{2}=\) \(3 y_{2}=3\left(x_{R}-x_{B}\right) .\) What are the mean and standard deviation of \(w_{2}\) ? e. Based on your answers to Parts (c) and (d), if you had to play, which game would you choose and why?

Short answer

The mean of the results of both games is 0. However, Game 1 would be the best to play as it has a smaller standard deviation (2.41) compared to Game 2 (7.25), implying less variation or risk in the outcomes.

Step-by-step solution

Calculating the mean, variance, and standard deviation of \(x_{R}\)

We know that for a probability distribution, the mean or expected value \(E(x)\) is given by \(\sum{x*p(x)}\), the variance \(Var(x)\) is \(\sum{(x-E(x))^2*p(x)}\) and the standard deviation \(SD(x)\) is the square root of the variance. Calculating these for \(x_{R}\) we get: \[E(x_{R})=\sum_{i=1}^{6}{i*\frac{1}{6}}=3.5\] \[Var(x_{R})=\sum_{i=1}^{6}{(i-3.5)^2*\frac{1}{6}}=2.92\] \(SD(x_{R})=\sqrt{2.92}\approx 1.71\].

Mean, Variance, and Standard Deviation of \(x_{B}\)

Since we have symmetric and identical distribution for both dies, the mean, variance and standard deviation of \(x_{B}\) occupy the same values as \(x_{R}\) thus \(E(x_{B})=3.5\), \(Var(x_{B})=2.92\), and \(SD(x_{B})=1.71\).

Mean and Standard Deviation of \(w_{1}\)

For Game 1, we need to calculate the mean and standard deviation for \(w_{1}\) where \(w_{1}=x_{R}+x_{B}-7\). Using the properties of means and variances we know that: \[E(w_{1})=E(x_{R}+x_{B}-7)=E(x_{R})+E(x_{B})-7=3.5+3.5-7=0\] For the standard deviation, because die rolls are independent, variance of \(w_{1}\) is: \(Var(w_{1})=Var(x_{R}+x_{B})=Var(x_{R})+Var(x_{B})=2.92+2.92=5.83\). Thus, \(SD(w_{1})=\sqrt{5.83}\approx2.41\).

Mean and Standard Deviation of \(w_{2}\)

For Game 2, we compute the mean and standard deviation for \(w_{2}\) given by \(w_{2}=3(x_{R}-x_{B})\). Now, \(E(w_{2})=E(3x_{R}-3x_{B})=3E(x_{R}-x_{B})=3(E(x_{R})-E(x_{B}))=3(3.5-3.5)=0\). Again, since the die rolls are independent, \(Var(w_{2})=9Var(x_{R}+x_{B})=9(2.92+2.92)=52.56\). Hence, \(SD(w_{2})=\sqrt{52.56}\approx7.25\).

Choosing the Game to Play

Based on the calculations, both games have the same mean of 0 but Game 1 has a smaller standard deviation of 2.41 compared to Game 2 which has a standard deviation of 7.25. A lower standard deviation means less risk or variability, so if one had to play, Game 1 would be a more prudent choice if one aims to minimize the risk.