In a game of dice, the probability of rolling a 12 is 1/36. The probability of rolling a 9, 10, or 11 is 9/36. The probability of rolling any other number is 26/36. If the player rolls a 12, the player wins $5. If the player rolls a 9, 10, or 11, the player wins $1. Otherwise, the player loses $1. What is the expected value of this game? If the game is played 100 times what are the expected winnings (or losses) of the player?

Question

In a game of dice, the probability of rolling a 12 is 1/36.  The probability of rolling a 9, 10, or 11 is 9/36.  The probability of rolling any other number is 26/36.  If the player rolls a 12, the player wins $5.  If the player rolls a 9, 10, or 11, the player wins $1.  Otherwise, the player loses $1.  What is the expected value of this game?  If the game is played 100 times what are the expected winnings (or losses) of the player?

Please show more than one approach.

Tamar · Answer

do you divide 1/36 and times it by 5. It would make it five over thirty six. For the amounts of 9, 10, 11, you do the same thing, which makes it 18/36. You add them together 23/36? That is where I get lose.