probability

[kay_wink08]

A game of chance requires the player to throw a 6-sided die and a 12-sided die.
(1,1) (1,2,3,4,5,6)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
(1,7)
(1,8)
(1,9)
(1,10)
(1,11)
(1,12)
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
(2,7)
(2,8)
(2,9)
(2,10)
(2,11)
(2,12)...and so on up to (6,12)

Above is the sample space.
The player pays $5 to play and wins $10 for any odd roll and wins nothing for any even roll. Calculate the expected winnings for the player and determine if this is a fair game or not.
I believe the game is fair because the number of possible odd numbers and even numbers are equal.
How would I calculate the winnings???
Any help would be very helpful.

 

[soroban]

Hello, kay_wink08!

A game of chance requires the player to throw a 6-sided die and a 12-sided die.
The player pays $5 to play and wins $10 for any odd roll and wins nothing for any even roll.
Calculate the expected winnings for the player and determine if this is a fair game or not.

I believe the game is fair because the number of possible odd numbers and even numbers are equal.
How would I calculate the winnings?

You're expected to be familiar with Expected Value.

The player wins $10 for an odd roll.
\(\displaystyle P(\text{win \$10}) \,=\,\tfrac{36}{72} \,=\,\tfrac{1}{2}\)

The player wins $0 for an even roll.
\(\displaystyle P(\text{win \$0}) \,=\,\tfrac{36}{72} \,=\,\tfrac{1}{2}\)

The player pays (loses) $5 on every roll.
\(\displaystyle P(\text{lose \$5}) \,=\,\tfrac{72}{72} \,=\,1\)

\(\displaystyle E \;=\;\tfrac{1}{2}(+10) + \tfrac{1}{2}(+0) + 1(-5) \;=\;5 + 0 - 5 \;=\;0\)


The expected value is $0.
The player can expect to "break even".
. . Therefore, the game is fair.