Question

Use Scenario 6-5.Consider the following game.You pay me an entry fee of x dollars; then I roll a fair die. If the die shows a number less than 3 I pay you nothing; if the die shows a 3 or 4, I give you back your entry fee of x dollars; if the die shows a 5, I will pay you $1; and if the die shows a 6, I pay you $3. What value of x makes the game fair (in terms of expected value) for both of us?
A.$2
B.$4
C.$1
D.$0.75
E.$0.5

Answer 1

Answer:

C.$1

Step-by-step explanation:

Since, when a die is rolled,

Then the total possible outcomes = 6 ( i.e. 1, 2, 3, 4, 5 or 6 )

Outcomes of getting number less than 3 = 2 ( i.e. 1 or 2 )

Outcomes of getting  3 or 4 = 2

Outcomes of getting 5 = 1,

Outcomes of getting 6 = 1,

$\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

Thus, the probability of getting number less than 3 = $\frac{2}{6}=\frac{1}{3}$

The probability of getting 3 or 4 = $\frac{2}{6}=\frac{1}{3}$

The probability of getting 5 = $\frac{1}{6}$

The probability of getting 6 = $\frac{1}{6}$

∵ If the die shows a number less than 3 then nothing will get; if the die shows a 3 or 4, we will get x dollars; if the die shows a 5, we will get $1; and if the die shows a 6, we will get $3

So, the expected value = $x\times \frac{1}{3}+0\times \frac{1}{3}+1\times \frac{1}{6}+3\times \frac{1}{6}$

$=\frac{x}{3}+\frac{1}{6}+\frac{1}{2}$

$=\frac{2x+1+3}{6}$

$=\frac{2x+4}{6}$

The game is fair if,

$\frac{2x+4}{6}=x$

$2x + 4 = 6x$

$4 = 4x$

$\implies x = 1$

Hence, the value of x would be $ 1.