關於期望值的數學題(急~)

let P(X) is the probability of the sum of the numbers on top is X
P(10)=3/36
P(11)=2/36
P(12)=1/36
the expected value of this game
=10*3/36+11*2/36+12* 1/36-3
=(30+22+12)/36-3
=64/36-3
=(64-108)/36
=-44/36
a “fair” cost to play the game, the expected value of this game would become 0
let the cost is x
then
64/36-x=0
x=64/36
a “fair” cost
=64/36

2006-10-25 21:13:56 補充：
parislei_1987我個腦點思考這條問題?我會分成可賺和必蝕兩部份點數1 - 9時, 會蝕左3蚊一舖~呢個係必需計入的~ 是呀﹐而且點數10 - 12時也會蝕左3蚊即是以概率1蝕3元點數1 - 9時﹐我攞到相應ge錢是0﹐點數10-12則拿到10-12元期望值=p(1)*0 p(2)*0 ... p(9)*0 p(10)*10 p(11)*11 p(12)*12 1*(-3)=10*3/36 11*2/36 12* 1/36-3=-44/36=-11/9

2006-10-25 21:14:30 補充：
加號再次不能顯示

The probabilities for the sum of 2 dice are listed below:

P (Sum = 2) = 1/36
P (Sum = 3) = 2/36
P (Sum = 4) = 3/36
P (Sum = 5) = 4/36
P (Sum = 6) = 5/36
P (Sum = 7) = 6/36
P (Sum = 8) = 5/36
P (Sum = 9) = 4/36
P (Sum = 10) = 3/36
P (Sum = 11) = 2/36
P (Sum = 12) = 1/36

Hence, we can see
P (Sum of 2 dice less than 10) = 1 - 3/36 - 2/36 - 1/36 = 30/36 = 5/6

Expected value of the game
= 5/6 (-3) + 3/36 (10 - 3) + 2/36 (11 - 3) + 1/36 (12 - 3)
=-5/2 + 1/12 (7) + 1/18 (8) + 1/36 (9)
=-5/2 + 7/12 + 8/18 + 1/4
=-5/2 + 7/12 + 4/9 + 1/4
=-90/36 + 21/36 + 16/36 + 9/36
=(-90 + 21 + 16 + 9) /36
=-44/36
= -11/9

Let P(x) be the probability of the overall money invloved for each throw
x be the overall money invloved for each throw
the probability of the overall money if throwing 10 =P(10-3)=3/36
the probability of the overall money if throwing 11 =P(11-3)=2/36
the probability of the overall money if throwing 12 =P(12-3)=1/36
So, the expected value required:
(10-3) 3/36+(11-3) 2/36 + (12-3) 1/36 + (-3) 30/36
= -11/9