Question 1190569: suppose it cost $8 to roll a pair of dice. we get paid the sum of the numbers in dollars that appear on the dice, is it a fair game
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
X = net earnings
The possible sums on the dice are
{2,3,4,5,6,7,8,9,10,11,12}
Subtract 8 from each of those items
2-8 = -6
3-8 = -5
etc, until
12-8 = 4
The possible values of X are {-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4}
For instance, if you roll a 2 you earn $2 but pay $8. So in reality you lose $6 (ie 2-8 = -6)
Another example: You roll a 12, so you earn $12 and walk away with 12-8 = 4 dollars.
Use a sum of dice chart to see all the possible ways to roll two dice.
There's only one way to roll a 2, out of 36 ways total.
This means 1/36 is the probability of rolling a 2.
It's the probability of X being -6
As another example, there are 2 ways to roll a 3 (1+2 = 3 and 2+1 = 3)
2/36 is the probability of rolling a 3 and earning $3
This leads to a net earnings of X = -5 as calculated earlier.
Continue this process until you reach rolling a 12.
This is what the probability distribution looks like
| Rolling a... | X | P(X) | | 2 | -6 | 1/36 | | 3 | -5 | 2/36 | | 4 | -4 | 3/36 | | 5 | -3 | 4/36 | | 6 | -2 | 5/36 | | 7 | -1 | 6/36 | | 8 | 0 | 5/36 | | 9 | 1 | 4/36 | | 10 | 2 | 3/36 | | 11 | 3 | 2/36 | | 12 | 4 | 1/36 |
What we do from here is multiply each X and P(X) value.
Example: -6*(1/36) = -6/36
I won't reduce any of the fractions.
This way I can keep the same denominator to add the fractions fairly easy in the next step
| Rolling a... | X | P(X) | X*P(X) | | 2 | -6 | 1/36 | -6/36 | | 3 | -5 | 2/36 | -10/36 | | 4 | -4 | 3/36 | -12/36 | | 5 | -3 | 4/36 | -12/36 | | 6 | -2 | 5/36 | -10/36 | | 7 | -1 | 6/36 | -6/36 | | 8 | 0 | 5/36 | 0/36 | | 9 | 1 | 4/36 | 4/36 | | 10 | 2 | 3/36 | 6/36 | | 11 | 3 | 2/36 | 6/36 | | 12 | 4 | 1/36 | 4/36 |
Every item in the X*P(X) column is something over 36
Focus on the numerators only. Add them up to get
-6+(-10)+(-12)+(-12)+(-10)+(-6)+0+4+6+6+4 = -36
Then divide that sum over the common denominator
-36/36 = -1
The expected value is -1.
The expected net earnings is -1, which means the player expects (on average) to lose $1 each time they play the game.
This expected value is not 0, so the game is not mathematically fair.
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Another approach:
Y = winnings only (ignore the cost)
Y is an integer from the set {2,3,4,5,6,7,8,9,10,11,12} which are the possible outcomes of the dice
This is the probability distribution for the random variable Y
| Y | P(Y) | | 2 | 1/36 | | 3 | 2/36 | | 4 | 3/36 | | 5 | 4/36 | | 6 | 5/36 | | 7 | 6/36 | | 8 | 5/36 | | 9 | 4/36 | | 10 | 3/36 | | 11 | 2/36 | | 12 | 1/36 |
Then compute the Y*P(Y) column
| Y | P(Y) | Y*P(Y) | | 2 | 1/36 | 2/36 | | 3 | 2/36 | 6/36 | | 4 | 3/36 | 12/36 | | 5 | 4/36 | 20/36 | | 6 | 5/36 | 30/36 | | 7 | 6/36 | 42/36 | | 8 | 5/36 | 40/36 | | 9 | 4/36 | 36/36 | | 10 | 3/36 | 30/36 | | 11 | 2/36 | 22/36 | | 12 | 1/36 | 12/36 |
Once again, everything is over 36 so we add the numerators to get
2+6+12+20+30+42+40+36+30+22+12 = 252
Divide that sum over 36
252/36 = 7
We get an expected value of 7, which makes sense because rolling a 7 is the most frequent item. It's at the center of the distribution, and it's the mean value.
On average, the player expects to win $7 on any given roll.
However, we have to factor in the $8 cost to play the game (reality catches up to us unfortunately). So the player expects to net 7-8 = -1 dollars per roll, on average.
This matches the -1 found at the conclusion of the previous section.
Whichever method you use, the game is not mathematically fair. We would need the expected value to be 0 when accounting for the cost to play the game.
It's probably fairly obvious at this point that if the cost to play the game was $7, then the final expected value would be 0.
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Answer: No, it is not a fair game. The expected value is not 0.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The numbers on each die are 1 through 6; the average is 3.5.
The average sum of the numbers on two dice is 2(3.5)=7.
So the average payout on rolling two dice is $7, while the cost is $8 -- so it is not a fair game.
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