Probability help please?

Question

In the New Jersey and Nevada gaming casinos, craps is a popular gambling game. One of the many bets available is the "Hard-way 8." A $1 bet in this fashion will win the player $4 if in the game the pair of dice come up 4 and 4 prior to one of the other ways of totaling 8. For a $1 bet, what is the expected result?
If you can get the answer please explain how you got it. Thanks

Answer 1

The ways to get 8 are:
2 6
3 5
4 4
5 3
6 2

So 20% of the time you get it the hard way, 80% of the time you don't.

20% of the time you will win $4.
80% of the time you will win $0.

So the expected winning are:
0.20 x 4 + 0.80 x 0 = 0.80

You're expected to win 80 cents on your $1 bet.

Answer 2

There are 5 ways to get a total of 8 with a pair of 6 sided dice:
2 & 6
3 & 5
4 & 4
5 & 3
6 & 2

so the chances of getting 8 rolling 4 & 4 = 1 in 5 chance

Answer 3

Did you leave something out.. looks like you answered your own question.

you bet $1.. the expected result

a) You win... you get $4
b) You lose.. you get experience

you keep making that bet, you will get pour because the odds of doing it are only 1 in 36 (only one way to roll that combo) but the odds of rolling others are 1 in 9 (4 ways).

So the house odds are 4 to 1 against you winning.

If you like that I have a bridge for sale....

Answer 4

Ways to make 8: {(2,6), (3,5), (4,4), (5,3), (6,2)}

A hard 8 is one of five way to make an 8, so the set of returns is:
{0, 0, $4, 0, 0}

The expected return is (4+0+0+0+0)/5 or $0.80 on a one dollar bet.