probability! help!!!

Total 6 x 6 = 36 cases :
1+1 = 2 , (1 case)
1+2 or 2+1 = 3 (2 cases)
1+3 or 3+1 or 2+2 = 4 (3 cases)
1+4 or 4+1 or 2+3 or 3+2 = 5 (4 cases)
1+5 or 5+1 or 2+4 or 4+2 or 3+3 = 6 (5 cases)
1+6 or 6+1 or 2+5 or 5+2 or 3+4 or 4+3 = 7 (6 cases)
2+6 or 6+2 or 3+5 or 5+3 or 4+4 = 8 (5 cases)
3+6 or 6+3 or 4+5 or 5+4 = 9 (4 cases)
deduce that :
Sum = 10 (3 cases) , 11(2 cases) and 12(1 case)
But the sum of John throws only have 6 cases :
2 , 4 , 6 , 8 , 10 , 12
a)
P( John will win )
= P(John 4 win Mary(2 to 3)) + P(6 win (2 to 5)) + P(8 win (2 to 7))
+ P(10 win (2 to 9)) + P(12 win (2 to 11))
= (1/6) (1+2)/36 + (1/6) (1+2+3+4)/36 + (1/6) (1+2+3+4+5+6)/36
+ (1/6) (1+2+3+4+5+6+5+4)/36 + (1/6) (1+2+3+4+5+6+5+4+3+2)/36
= (1/6)(1/36)(3 + 10 + 21 + 30 + 35)
= 11/24
c)
P(There is a tie game)
= P(John 2 draw Mary 2) + P(4 draw 4) + P(6 draw 6) + P(8 draw 8)
+ P(10 draw 10) + P(12 draw 12)
= (1/6) [(1/36) + (3/36) + (5/36) + (7/36) + (9/36) + (11/36)]
= 1/6
b)
P(Mary will win)
= 1 - P( John will win ) - P(There is a tie game)
= 1 - 11/24 - 1/6
= 3/8

2010-05-19 23:07:50 補充：
Corrections :

c)P(There is a tie game)

= P(John 2 draw Mary 2) + P(4 draw 4) + P(6 draw 6) + P(8 draw 8)
+ P(10 draw 10) + P(12 draw 12)

= (1/6) [(1/36) + (3/36) + (5/36) + (5/36) + (3/36) + (1/36)]

= 1/12

b) P(Mary will win)

= 1 - P( John will win ) - P(There is a tie game)

= 1 - 11/24 - 1/12

= 11/24

Sorry!!