Question 1190569
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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<table border = "1" cellpadding = "5"><tr><td>Rolling a...</td><td>X</td><td>P(X)</td></tr><tr><td>2</td><td>-6</td><td>1/36</td></tr><tr><td>3</td><td>-5</td><td>2/36</td></tr><tr><td>4</td><td>-4</td><td>3/36</td></tr><tr><td>5</td><td>-3</td><td>4/36</td></tr><tr><td>6</td><td>-2</td><td>5/36</td></tr><tr><td>7</td><td>-1</td><td>6/36</td></tr><tr><td>8</td><td>0</td><td>5/36</td></tr><tr><td>9</td><td>1</td><td>4/36</td></tr><tr><td>10</td><td>2</td><td>3/36</td></tr><tr><td>11</td><td>3</td><td>2/36</td></tr><tr><td>12</td><td>4</td><td>1/36</td></tr></table>


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<table border = "1" cellpadding = "5"><tr><td>Rolling a...</td><td>X</td><td>P(X)</td><td>X*P(X)</td></tr><tr><td>2</td><td>-6</td><td>1/36</td><td>-6/36</td></tr><tr><td>3</td><td>-5</td><td>2/36</td><td>-10/36</td></tr><tr><td>4</td><td>-4</td><td>3/36</td><td>-12/36</td></tr><tr><td>5</td><td>-3</td><td>4/36</td><td>-12/36</td></tr><tr><td>6</td><td>-2</td><td>5/36</td><td>-10/36</td></tr><tr><td>7</td><td>-1</td><td>6/36</td><td>-6/36</td></tr><tr><td>8</td><td>0</td><td>5/36</td><td>0/36</td></tr><tr><td>9</td><td>1</td><td>4/36</td><td>4/36</td></tr><tr><td>10</td><td>2</td><td>3/36</td><td>6/36</td></tr><tr><td>11</td><td>3</td><td>2/36</td><td>6/36</td></tr><tr><td>12</td><td>4</td><td>1/36</td><td>4/36</td></tr></table>


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<table border = "1" cellpadding = "5"><tr><td>Y</td><td>P(Y)</td></tr><tr><td>2</td><td>1/36</td></tr><tr><td>3</td><td>2/36</td></tr><tr><td>4</td><td>3/36</td></tr><tr><td>5</td><td>4/36</td></tr><tr><td>6</td><td>5/36</td></tr><tr><td>7</td><td>6/36</td></tr><tr><td>8</td><td>5/36</td></tr><tr><td>9</td><td>4/36</td></tr><tr><td>10</td><td>3/36</td></tr><tr><td>11</td><td>2/36</td></tr><tr><td>12</td><td>1/36</td></tr></table>

Then compute the Y*P(Y) column<table border = "1" cellpadding = "5"><tr><td>Y</td><td>P(Y)</td><td>Y*P(Y)</td></tr><tr><td>2</td><td>1/36</td><td>2/36</td></tr><tr><td>3</td><td>2/36</td><td>6/36</td></tr><tr><td>4</td><td>3/36</td><td>12/36</td></tr><tr><td>5</td><td>4/36</td><td>20/36</td></tr><tr><td>6</td><td>5/36</td><td>30/36</td></tr><tr><td>7</td><td>6/36</td><td>42/36</td></tr><tr><td>8</td><td>5/36</td><td>40/36</td></tr><tr><td>9</td><td>4/36</td><td>36/36</td></tr><tr><td>10</td><td>3/36</td><td>30/36</td></tr><tr><td>11</td><td>2/36</td><td>22/36</td></tr><tr><td>12</td><td>1/36</td><td>12/36</td></tr></table>
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.
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