SOLUTION: Suppose someone was playing the following dice game. The game costs $5 to play. You roll two dice and take the sum. If the sum is 2 or 12, you get $10. If the sum is 3 or 11, you g

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose someone was playing the following dice game. The game costs $5 to play. You roll two dice and take the sum. If the sum is 2 or 12, you get $10. If the sum is 3 or 11, you g      Log On


   

Question 1200826: Suppose someone was playing the following dice game. The game costs $5 to play. You roll two dice and take the sum. If the sum is 2 or 12, you get $10. If the sum is 3 or 11, you get $9. If the sum is 4 or 10, you get $8. If the sum is 5 or 9, you get $5 (you get your money back). If the sum is 6, 7, or 8, you get nothing.
a. What is the expected value of this game, taking into consideration the $5 cost of playing the game? What does this mean?
b. Does the game favor you, or the people who play it?
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Part (a)
Refer to this sum of dice chart.
+123456
1234567
2345678
3456789
45678910
567891011
6789101112

Eg: adding a blue 1 with a red 6 gets 1+6 = 7 (top right corner).

That dice chart shown above leads to this probability distribution table.
X = net winnings = prize - cost
XP(X)
52/36
44/36
36/36
08/36
-516/36

Example: The sum 2 and 12 show up once each, so a total of 1+1 = 2 times. That leads to 2/36 for X = 5 (the amount the player walks away with after winning $10).
I'll avoid reducing to keep each denominator the same.
Spreadsheet software is recommended to quickly compute values, and to organize the data in a tidy fashion.

Form a third column multiplying each X and corresponding P(X) value.
XP(X)X*P(X)
52/3610/36
44/3616/36
36/3618/36
08/360/36
-516/36-80/36

Then add up the items in that third column.
E[X] = expected value = expected net winnings
E[X] = sum of the X*P(X) values
E[X] = 10/36 + 16/36 + 18/36 + 0/36 + (-80/36)
E[X] = (10+16+18+0+(-80))/36
E[X] = -36/36
E[X] = -1

Expected value = -1
Interpretation: The player expects to lose $1, on average, each time they play the game.

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Part (b)

The negative expected value is in reference to the player. Meaning the game favors the person hosting the game (and not the person playing the game).

This is typical in many settings. Casinos make sure that they have the odds tilted in their favor to earn the most money possible.