Question 430365: Could someone help me with this problem?
We are playing a dice game with 2 fair dice. You roll the dice,I will pay you $1 if the total of the dice is 6 or greater,you will pay me $3 dollars if the total of the dice is $5 or less.What is the expected value of the game for me. I know the answer is .111 but I cannot figure our how to do this problem.
Thanks
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! We are playing a dice game with 2 fair dice. You roll the dice,I will pay you $1 if the total of the dice is 6 or greater,you will pay me $3 dollars if the total of the dice is 5 or less.
What is the expected value of the game for me. I know the answer is .111 but I cannot figure our how to do this problem.
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Sketch the 6 by 6 rectangle of 2-die sums.
234567
345678
456789
5678910
6789101112
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Random gain values for you::::: 1....-3
Corresponding probabilities::::
P(6 or greater) = 26/36
P(5 or less) = 10/36
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Expected gain for you: (26/36)1 -3(10/36)
= [26-30]/36
= -4/36
= -1/9
= -0.11
---
You can expect to lose 11 cents each time you play.
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Cheers,
Stan H.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! Could someone help me with this problem?
We are playing a dice game with 2 fair dice. You roll the dice,I will pay you $1 if the total of the dice is 6 or greater,you will pay me $3 dollars if the total of the dice is $5 or less.What is the expected value of the game for me. I know the answer is .111 but I cannot figure our how to do this problem.
Thanks
The other tutor has the payout and the income switched, and that's why he came
up with a negative answer.
We have to find the probabilities of rolling 6 or greater and 5 or less:
Here are all the possible dice rolls with two dice.
(1,1) (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)
(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)
I will color the ones red which are 6 or greater,
and the one 5 or less blue:
(1,1) (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)
(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)
Since there are 26 red rolls above and 10 blue rolls, the probability
of rolling 6 or greater is 26/36 or 13/18 and the probability of
rolling 5 or less is 10/36 or 5/18
So we make a probability distribution function:
dice roll x p(x)
6 or more -1 13/18
5 or less +3 5/18
E(x) = ∑[x·p(x)] = (-1)(13/18) + 3(5/18) = -13/18 + 15/18 = 2/18 = 1/9
and 1/9 as a decimal is .111···
Edwin
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