Question

A player rolls a die and receives the number of dollars equal to the number on the die EXCEPT when the die shows a 6. If a 6 is rolled, the player loses $6. If the game is to be fair, what should be the cost to play?

Answer 1

We expect to win k dollars with probability 1/6 if k is between 1 and 5, and we lose 6 dollars with probability 1/6. So, the expected value for this random variable is

$1\cdot \dfrac{1}{6}+2\cdot \dfrac{1}{6}+3\cdot \dfrac{1}{6}+4\cdot \dfrac{1}{6}+5\cdot \dfrac{1}{6}-6\cdot \dfrac{1}{6} = \dfrac{1}{6}(1+2+3+4+5-6)=\dfrac{3}{2}$

So, you win on average 1.5 dollars, and if that's the cost to play, the game is fair.

Answer 2

Using the expected value formula, the cost of playing a fair game should be $1.5

• The probability of obtaining any single value on a die roll is = 1/6

X ____ 1 __ 2 ___ 3 ___ 4 ____ 5 ___ - 6

P(X)__1/6 _ 1/6 __ 1/6__ 1/6 ___ 1/6 ___ 1/6

For a fair game ;

• E(X) ≥ 0

• E(X) = Σ[(X × P(X)]

(1 × 1/6) + (2 × 1/6) + (3 × 1/6) + (4 × 1/6) + (5 × 1/6) - (6 × 1/6) = $1.5

Since the expected value is $1.5 and greater than 0 ; then the player can expect to win $1.5 in the long run and not lose.

Therefore, the player should be willing to pay $1.5

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