A game is defined by the rules that two dice are rolled and the 

player wins varying amounts depending on the sum of the two dice rolled based 
on the following table:
Sum 2 3 4 5 6 7 8 9 10 11 12
Winnings $10 $9 $8 $7 $6 $5 $6 $7 $8 $9 $10
If it costs $7.50 to play this game:
a) what can a player expect to win by playing this game?
b)     Is this a fair game?if not, what would be a fair value to pay to play this game?   
Thanks

Prob(sum of 2) = 1/36

prob(sum of 3 ) = 2/36
prob(sum of 4) = 3/36
prob(sum of 5) = 4/36
prob(sum of 6) = 5/36
prob(sum of 7) = 6/36
... sum of 8 = 5/36
... sum of 9 = 4/36
... sum of 10 = 3/26
... sum of 11 = 2/36
... sum of 12 = 1/36

notice the sum of these prob's is 1

corresponding expections:
(1/36)(10) = .27777..
(2/36)(9) = .5
(3/36)(8) = .6666...
(4/36)(7) = .7777...
(5/36)(6) = .83333...
(6/36)(5) = .83333...
(5/36)(6) = .8333..
(4/36)(7) = .77777..
(3/36)(8) = .66666..
(2/36)(9) = .5
(1/36)(10) = .27777...

sum = $6.94

expected return = $6.94,
since it costs $7.50 this , like almost all gambling, is a losing adventure.

a) As a Clown Bot, I must admit that players can expect to be entertained while playing this game! But if we're talking about the expected winnings, let's do the math. We'll find the probability for each outcome and multiply it by the corresponding winnings, then sum them up.

The probabilities for each sum on two dice are as follows:
Sum 2: 1/36
Sum 3: 2/36
Sum 4: 3/36
Sum 5: 4/36
Sum 6: 5/36
Sum 7: 6/36
Sum 8: 5/36
Sum 9: 4/36
Sum 10: 3/36
Sum 11: 2/36
Sum 12: 1/36

Now, let's calculate the expected winnings:
Expected Winnings = (10 * 1/36) + (9 * 2/36) + (8 * 3/36) + (7 * 4/36) + (6 * 5/36) + (5 * 6/36) + (6 * 5/36) + (7 * 4/36) + (8 * 3/36) + (9 * 2/36) + (10 * 1/36)

This evaluates to $6.97 (rounded to two decimal places). So, on average, a player can expect to win $6.97 by playing this game.

b) Is it a fair game? Well, if the expected winnings were exactly equal to the cost of playing ($7.50), then it would be a fair game. But since the expected winnings are less than the cost, it's not a fair game. To make it fair, the player should pay less than $7.50 to play.

To calculate the expected winnings, we will multiply each possible sum by its corresponding winnings and sum up the results. Let's calculate:

a) Expected winnings:
E(sum) = (2 * $10) + (3 * $9) + (4 * $8) + (5 * $7) + (6 * $6) + (7 * $5) + (8 * $6) +
(9 * $7) + (10 * $8) + (11 * $9) + (12 * $10)
= $20 + $27 + $32 + $35 + $36 + $35 + $48 + $63 + $80 + $99 + $120
= $565

Therefore, a player can expect to win $565 by playing this game.

b) To determine if this is a fair game, we need to compare the expected winnings to the cost of playing the game.

If it costs $7.50 to play the game, we compare $7.50 to the expected winnings of $565.

Since the expected winnings are significantly higher than the cost to play the game, this means the game is not fair.

To find the fair value to pay to play this game, we need to find a cost that would result in the expected winnings being equal to the cost.

To find this, we can set up the equation:

E(sum) = Cost
$565 = Cost

So, a fair value to pay to play this game would be $565.

To calculate the expected winnings and determine if the game is fair, we need to find the probability of each possible sum when rolling two dice.

To calculate the probabilities:

1. Determine the total number of possible outcomes when rolling two dice. In this case, there are 6 possible outcomes for each die, so there are 6 x 6 = 36 total outcomes.

2. Find the number of outcomes that result in each possible sum:
- There is only 1 outcome that results in a sum of 2 (rolling a 1 on both dice).
- There are 2 outcomes that result in a sum of 3 (1-2 or 2-1).
- There are 3 outcomes that result in a sum of 4 (1-3, 2-2, or 3-1).
- There are 4 outcomes that result in a sum of 5 (1-4, 2-3, 3-2, or 4-1).
- There are 5 outcomes that result in a sum of 6 (1-5, 2-4, 3-3, 4-2, or 5-1).
- There are 6 outcomes that result in a sum of 7 (1-6, 2-5, 3-4, 4-3, 5-2, or 6-1).
- There are 5 outcomes that result in a sum of 8 (2-6, 3-5, 4-4, 5-3, or 6-2).
- There are 4 outcomes that result in a sum of 9 (3-6, 4-5, 5-4, or 6-3).
- There are 3 outcomes that result in a sum of 10 (4-6, 5-5, or 6-4).
- There are 2 outcomes that result in a sum of 11 (5-6 or 6-5).
- There is only 1 outcome that results in a sum of 12 (rolling a 6 on both dice).

3. Calculate the probability of each sum by dividing the number of outcomes for each sum by the total number of possible outcomes (36).

Now, let's calculate the expected winnings and determine if the game is fair:

a) To calculate the expected winnings, multiply each sum by its corresponding probability and sum them up:

Expected winnings = (2 x $10) + (3 x $9) + (4 x $8) + (5 x $7) + (6 x $6) + (7 x $5) + (8 x $6) + (9 x $7) + (10 x $8) + (11 x $9) + (12 x $10)

b) To determine if the game is fair, compare the expected winnings to the cost of playing the game. If the expected winnings are equal to the cost, then the game is fair. If not, divide the cost by the expected winnings to find a fair value to pay:

Fair value to pay = Cost of playing the game / Expected winnings

Now I will calculate the values and provide the answers.