Asked 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.
Answers
Answered by
Reiny
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(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
Answered by
Barack Obama
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(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
Answered by
felicita
$1.67
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