# Probability of Weighted Dice

A game uses a 12-sided die which is rolled once. Each of the faces are labelled 1-12. The dice is weighted so it is four times as likely to roll a 6, and twice as likely to roll an 8, 10 or 12.

The game costs \$20 to play.

If an even number is rolled, the player wins 30 dollars.

If a multiple of 3 is rolled, the player wins 40 dollars.

If any other number is rolled, they player loses the bet.

1. What is the probability of winning:

a. Exactly 30 dollars

b. Exactly 40 dollars

2. What is the expected return for each roll of the die?

I have calculated that the probability of rolling a 6 is 2/9.

The probability of rolling an 8, 10 or 12 is 1/9 each.

I do not know how to find the probability of winning EXACTLY 30 and 40 dollars, along with the expected return per roll.

Thanks!

• Welcome to Math SE! Please show context by including your work – have you calculated the probability that each side will come up? Jun 16 '19 at 6:20
• Yes, I have calculated the following probabilities: Rolling a 6 has a 4/18 probability Rolling an 8, 10 or 12 has a 2/18 probability, to a total of 6/18 probability. From this we can get the probability of earning a 30 or 40 dollar prize. I am having an issue with calculating the probability of winning EXACTLY 30 and 40 dollars respectively. Thanks! Jun 16 '19 at 6:26
• You should now put all the work you have in the question body to get a better answer. Jun 16 '19 at 6:27
• Done, thanks for your help! Jun 16 '19 at 6:30
• There is some ambiguity in your question: does 'four times as likely' mean four times more likely than chance (as you have calculated), or four times more likely than the other numbers? My guess is that the latter option is true, since the probabilities for the other numbers would be reduced under the first option. Jun 16 '19 at 6:35

Hint: You have calculated the probabilities correctly. Given this, to win exactly $$30$$ dollars, only even numbers that are not multiples of $$3$$ must be rolled. In the numbers $$1$$ to $$12$$, only $$2, 4, 8, 10$$ satisfy the condition. To win exactly $$40$$ dollars, only numbers that are a multiple of $$3$$ but not even should be rolled. From this, you should be able to calculate the probabilities.
$$P(\text{event } 1) \cdot P(\text{expected return from event } 1) + P(\text{event } 2) \cdot P(\text{expected return from event } 2) + \cdots + P(\text{event } n) \cdot P(\text{expected return from event } n)$$
Bear in mind many of the expected returns are $$0$$, so this should simplify your calculation a lot.