Jimmy vs. Richard

Suppose Dice 1 (D1) rolled 1, so I got 6 possibilities for Dice 2 (D2), this way:

D1 + D2 = Sum
1 + 1 = 2 Jimmy wins
1 + 2 = 3 Jimmy wins
1 + 3 = 4 Jimmy wins
1 + 4 = 5 Jimmy wins
1 + 5 = 6 Jimmy wins
1 + 6 = 7 Richard wins
Now, for those six possible outcomes, five of them benefits Jimmy and only one is in favour to Richard. But as long Dice 1 can show higher results, Jimmy's chances will get smaller, while Richard's ones will get increased. At the point where D1 rolled 6, Jimmy has no chance anymore. So:

Jimmy's = 5+4+3+2+1+0 = 15
Richard's = 1+2+3+4+5+6 = 21

X= sum of two dice in 1 round.
Jimmy's probability to win = P(X<7)=15/36=5/12 Jimmy's probability to lose =7/12
Richard's probability to win = P(X>=7)=21/36=7/12 Richard's probability to lose = 5/12
Expected of Jimmy's money =(5)(P(X<7))+(-4)P(X>=7)= (5)(5/12)+(-4)(7/12) = -3/12 =-1/4
Expected of Richard's money =(-5)P(X<7)+(4)P(X>=7)= (-5)(5/12)+(4)(7/12) = +3/12 =1/4
So. ......

It may look like that that probability partition is equal for those two but it's not. 2 rolls do not have 12 possible outcomes, they have 11. Since you can't have 1.

So Richard's probability to win : 6/11 (More chances to win)

Jimmy's probability to win : 5/11