Probability and statistics
posted by Rosmery .
A dice game involves rolling 2 dice. If you roll a 2, 3, 4, 10, 11, or a 12 you win $5. If you roll a 5, 6, 7, 8, or 9 you lose $5. Find the expected value you win (or lose) per game.

Prob(2) = 1/36
prob(3) = 2/36
prob(4) = 3/36
prob(12) = 1/36
prob(11) = 2/36
prob(10) = 3/36
total of above , your cases of winning = 12/36
so the prob of remaining cases = 24/36
expected value of game
= (12/36)(5) + (24/36)(5)
= (1/3)(5)  (2/3)(5
= 5/3
You would be expected to lose $1.67 
Prob(2) = 1/36
prob(3) = 2/36
prob(4) = 3/36
prob(12) = 1/36
prob(11) = 2/36
prob(10) = 3/36
total of above , your cases of winning = 12/36
so the prob of remaining cases = 24/36
expected value of game
= (12/36)(5) + (24/36)(5)
= (1/3)(5)  (2/3)(5
= 5/3
You would be expected to lose $1.67