Question

1. if a person rolls doubles when he tosses two dice, he wins $5. the cost to play the game is $1. is this game fair?

Explanation:

Step1: Calculate the probability of rolling doubles

When rolling two dice, there are a total of \(n(S)=6\times6 = 36\) possible outcomes.
The number of ways to roll doubles is \(n(E)=6\) (i.e., \((1,1)\), \((2,2)\), \((3,3)\), \((4,4)\), \((5,5)\), \((6,6)\)).
The probability \(P(\text{doubles})=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}\).
The probability of not rolling doubles \(P(\text{not doubles}) = 1-\frac{1}{6}=\frac{5}{6}\).

Step2: Calculate the expected value

Let \(X\) be the random variable representing the net gain.
If a person rolls doubles, the net gain \(x_1=5 - 1=4\) (since they pay \(1\) to play and win \(5\)).
If a person does not roll doubles, the net gain \(x_2=- 1\) (they just lose the \(1\) they paid to play).
The expected value \(E(X)=x_1P(x_1)+x_2P(x_2)\).
Substitute \(x_1 = 4\), \(P(x_1)=\frac{1}{6}\), \(x_2=-1\), \(P(x_2)=\frac{5}{6}\) into the formula:
\(E(X)=4\times\frac{1}{6}+(-1)\times\frac{5}{6}=\frac{4 - 5}{6}=-\frac{1}{6}\approx - 0.17\).

Answer:

Since the expected value \(E(X)=-\frac{1}{6}\approx - 0.17
eq0\), the game is not fair.