Agame is played with the following rules. two ordinary number cubes, each with six sides labeled 1 to 6, are rolled. after each roll a player is given a certain number of points depending on the outcome. if the sum is 2 or 12, the player receives +6 points if the sum is 3 or 11, the player receives +2 points. if the sum is any other number, the player receives –1 point. what is the expected value of the number of points a player will receive after a roll?

- 0.28

Explanation:

1) Expected value: is the weighted average of the values, being the probabilities the weight.

That is: ∑ of prbability of event i × value of event i.

In this case: (probability of getting 2 or 12) × (+6) + (probability of gettin 3 or 11) × (+2) + (probability of any other sum) × (-1).

2) Sample space:

Sum              Points awarded
    
1+ 1 = 2              +6

1 + 2 = 3             +2

1 + 3 = 4             -1

1 + 4 = 5             -1

1 + 5 = 6             -1

1 + 6 = 7             -1

2 + 1 = 3             +2

2 + 2 = 4             -1

2 + 3 = 5             -1

2 + 4 = 6             -1

2 + 5 = 7             -1

2 + 6 = 8             -1

3 + 1 = 4             -1

3 + 2 = 5             -1

3 + 3 = 6             -1

3 + 4 = 7             -1

3 + 5 = 8             -1

3 + 6 = 9             -1

4 + 1 = 5             -1

4 + 2 = 6             -1

4 + 3 = 7             -1

4 + 4 = 8            -1

4 + 5 = 9            -1

4 + 6 = 10          -1

5 + 1 = 6            -1

5 + 2 = 7            -1

5 + 3 = 8            -1

5 + 4 = 9            -1

5 + 5 = 10          -1

5 + 6 = 11          +2

6 + 1 = 7            -1

6 + 2 = 8            -1

6 + 3 = 9            -1

6 + 4 = 10          -1

6 + 5 = 11          +2

6 + 6 = 12          +6

2) Probabilities

From that, there is:
- 2/36 probabilities to earn + 6 points.
- 4/36 probabilites to earn + 2 points
- the rest, 30/36 probabilities to earn - 1 points

3) Expected value = (2/36)(+6) + (4/36) (+2) + (30/36) (-1) = - 0.28

For a 0.25 pfor b o.3pp (0.25+0.3)=0.28