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

Loss of $ 0.3.

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 =
`(3)/(36)=(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 =
`(3)/(36)=(1)/(12)`

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

Thus, the possibility of getting other number =
`(30)/(36)=(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 =
`(1)/(12)* -8+(1)/(12)* 11 + (5)/(6)* x`

`=-(8)/(12)+(11)/(12)+(5x)/(6)`

If the game is fair,

Expected value of game = 0

`\implies -(8)/(12)+(11)/(12)+(5x)/(6)=0`

`(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.