Question

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?

Answer 1

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 2

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 = $\frac{3}{36}=\frac{1}{12}$

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 = $\frac{3}{36}=\frac{1}{12}$

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

Thus, the possibility of getting other number = $\frac{30}{36}=\frac{5}{6}$

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 = $\frac{1}{12}\times -8+\frac{1}{12}\times 11 + \frac{5}{6}\times x$

$=-\frac{8}{12}+\frac{11}{12}+\frac{5x}{6}$

If the game is fair,

Expected value of game = 0

$\implies -\frac{8}{12}+\frac{11}{12}+\frac{5x}{6}=0$

$\frac{3+10x}{12}=0$

$3+10x=0$

$x=-0.3$

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