Question

If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?

Answer 1

Here is the complete question.

1). What are the odds for rolling a sum of 5 in a single roll of two fair dice?

2). If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?

Answer:

1).  $\frac{1}{8}$

2).  $8.

Step-by-step explanation:

1).

The sample space (S) for rolling two fair dice is given as the following parameters illustrated below:

$\left[\begin{array}{cccccc}(1,2)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\end{array}\right] \left[\begin{array}{cccccc}(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}\right]$

Let F represent the event that the sum of both dice turns up is 5.

F = [(1,4),(2,3),(3,2),(4,1)}

∴ The probability for an event F  is illustrated below as:

$\frac{P(F)}{P(F')}$  $=\frac{\frac{4}{36} }{1-\frac{4}{36} }$

$=\frac{\frac{4}{36} }{\frac{32}{36} }$

$={\frac{4}{36}}*\frac{36}{32}$

= $\frac{1}{8}$

2). From above, we can see that the probability for rolling a sum of 5 are 1 to 8. Therefore, if you roll a sum of 5, in order for the game to be fair, the house is required to pay $8.