Math
posted by Hannah .
Suppose you play a game in which two fair standard dice are rolled. If the numbers showing on the dice are different, you lose $2. If the numbers showing are the same, you win $2 plus the dollar value of the sum of dice. Complete the next table, in which the values in the first column are the outcomes for the rolls of the two dice. The second column has the probabilities for those outcomes, and the third column has the payoff values for each outcome. What is the expected value of the game?

Suppose you play the game 36 times. (all possible outcomes)
Your critical outcomes are
1,1 expected return = (1/36)(2) + 2 = 2.0555
2,2 expected ret;urn = (1/36)(2) + 4 = 4.0555
.
.
6,6 expected return = (1/36)(2) + 12 = 12.0555
the expected winnings are 2.0555+4.0555+6.0555+8.0555+10.0555+12.0555
= 42.3333
the other 30 outcomes would all result in a loss of 2
expected losses = (30/36)(2) = 1.6666666
The expected value of the game is $40.67 for 36 game.
So the expected value of one game is 40.67/36 or $1.13