# Odds probability

Problem : Jason gets two chances to roll a sum of 8 with the two dice. To further clarify, he can win by either rolling an 8 on his first roll, and if he fails to roll an 8 on his first roll, he can win by rolling an 8 on his second roll. The wager for each game is $2.

The rules are as follows: *If an 8 is rolled on either the first roll or the second roll, Jason will win a profit of $5. If neither roll is an 8, Jason will lose his $2 wager. *

- Should Jason expect to win money or lose money playing this game?

- What would Jason’s profit for winning need to be for him to have the same likelihood of winning and losing this game?

## 1 Expert Answer

Joseph F. answered • 05/15/20

Joe's Math, Science and Chess

Hi Jason!

This is an expected value calculation. To complete it, you need to figure out the probabilities that Jason will win on either roll, and multiply that probability by +3, which is what he would win if he paid 2 and won 5. The remaining probability, that he doesn't win on either roll, must be multiplied by -2, for he paid 2 and didn't win.

The probabilities are:

P(8 on first roll) = 10 rolls out of 36 possibilities = 5/18. Multiply this by 3.

P(8 on second roll after NOT 8 on first roll) = [1-5/18]*5/18 = (5*13)/18^2=65/324. Also multiply this by 3.

P(no 8 on either roll) = 1- 5/18 - 65/324 = (324 - 5*18 - 65)/324 = 169/324. Multiply this by -2.

Add all these products up. If the result is positive, Jason will win over the long term. If it is negative, Jason will lose.

Now, a "fair game" is one in which Jason will neither win nor lose, that is, the expected value is zero. To design a fair game, solve for what the amount won per game, less the amount paid, must be for the expected value to be zero.

Cheers,

Joe

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Jason W.

Dices are shape of cube and the numbers are 1-605/14/20