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A pair of fair dice is rolled once. Suppose that you lose ​$8 if the dice sum to 10 and win ​$11 if the dice sum to 11 or 12. How much should you win or lose if any other number turns up in order for the game to be​ fair?

Respuesta :

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First find the possible dices rolls that give you 10, 11, and 12

Gives 10:
5,5
4,6
6,4

Gives 11 or 12:
5,6
6,5
6,6

The dice rolls have 6 * 6 = 36 possible combinations

The probability of getting a 10 is 3/32
The probability of getting an 11 or 12 is 3/32

multiplying those probabilities together with their reward/loss we have:

3/32 * -8 = -3/4
3/32 * 11 = 33/32

to make the game "fair" we need the probable amount of money you can win to be $0, so we can use the equation

-3/4 + 33/32 + (32/32 - 6/32)*x = 0

then you can just solve for x

Answer:

Loss of $ 0.3.

Step-by-step explanation:

Since, when two dices are rolled,

Then the all possible outcomes = 36,

Also, the possible way of getting the sum of 10 are,

(4, 5), (5, 4), (5, 5),

So, the possibility of getting the sum of 10 = [tex]\frac{3}{36}=\frac{1}{12}[/tex]

Now, the possible way of getting the sum of 11 or 12 are,

(5, 6), (6, 5), (6, 6),

So, the possibility of getting the sum of 11 or 12 = [tex]\frac{3}{36}=\frac{1}{12}[/tex]

Now, the possible number of ways of getting other number= 36 - 3 - 3 = 30,

Thus, the possibility of getting other number = [tex]\frac{30}{36}=\frac{5}{6}[/tex]

Given, for getting the sum of 10 profit is -$ 8, for the sum of 11 or 12 the profit is $11,  ( '-' sign shows the loss )

Let x be the profit of getting other number,

So, the expected value of the game = [tex]\frac{1}{12}\times -8+\frac{1}{12}\times 11 + \frac{5}{6}\times x[/tex]

[tex]=-\frac{8}{12}+\frac{11}{12}+\frac{5x}{6}[/tex]

If the game is fair,

Expected value of game = 0

[tex]\implies -\frac{8}{12}+\frac{11}{12}+\frac{5x}{6}=0[/tex]

[tex]\frac{3+10x}{12}=0[/tex]

[tex]3+10x=0[/tex]

[tex]x=-0.3[/tex]

Hence, there should be a loss of $ 0.3 if any other number turns up in order for the game to be​ fair.

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