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A game 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?

User Nik Kober
by
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2 Answers

0 votes

Answer:

-5/18

Step-by-step explanation:


User Occhiso
by
5.9k points
1 vote
Answer: - 0.28


Step-by-step 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
User JakeHova
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5.8k points