Question

(Brainliest Question) You are setting up to run a game of dice. The players will roll 2 dice. If the players roll a sum of 7 on the days they earn $12, If they roll a some of 2 or 12 they earn $36, If they roll anything else they lose. How much should you charge to have a fair game?

Answer 1

Using the expected value, it is found that you should charge $4 to have a fair game.

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• A probability is the number of desired outcomes divided by the number of total outcomes.
• The expected value is the sum of each outcome multiplied by it's respective probability.

• When you roll two dice, there are $6^2 = 36$ possible outcomes.
• We suppose a cost of x.
• In 6 of them, the sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1).
• Thus, $\frac{6}{36}$ probability of earning a profit of 12 - x.

• 2 of them have sums of 2 or 12, which are (1,1) and (6,6).
• Then, $\frac{2}{36}$ probability of earning a profit of 36 - x.

• If the outcome is any of the remaining 28, the player loses, thus $\frac{28}{36}$ probability of losing x.

• The game is fair if the expected value is 0, then:

$\frac{6}{36}(12 - x) + \frac{2}{36}(36 - x) - \frac{28}{36}x = 0$

$\frac{6(12 - x) + 2(36 - x) - 28x}{36} = 0$

$72 + 72 - 36x = 0$

$36x = 144$

$x = \frac{144}{36}$

$x = 4$

A value of $4 should be charged.

A similar problem is given at brainly.com/question/24855677