Sorry for that i read question wrongly ( sandra also throws 2 dices )... But for my satisfaction i keep it as it is
DICE = {1,2,3,4,5,6}
Sum possible with rolling Two dices = {2,3,4,5,6,7,8,9,10,11,12}
favorable cases for
SUM = 2 ------ { (1,1) } =====> $\frac{1}{36}$
SUM = 3 ------ { (1,2),(2,1) } =====> $\frac{2}{36}$
SUM = 4 ------ { (1,3),(2,2),(3,1) } =====> $\frac{3}{36}$
SUM = 5 ------ { (1,4),(2,3),(3,2),(4,1) } =====> $\frac{4}{36}$
SUM = 6 ------ { (1,5),(2,4),(3,3),(4,2),(5,1) } =====> $\frac{5}{36}$
SUM = 7 ------ { (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) } =====> $\frac{6}{36}$
SUM = 8 ------ { (2,6),(3,5),(4,4),(5,3),(6,2) } =====> $\frac{5}{36}$
SUM = 9 ------ { (3,6),(4,5),(5,4),(6,3) } =====> $\frac{4}{36}$
SUM = 10 ------ { (4,6),(5,5),(6,4) } =====> $\frac{3}{36}$
SUM = 11 ------ { (5,6),(6,5) } =====> $\frac{2}{36}$
SUM = 12 ------ { (6,6) } =====> $\frac{1}{36}$
Eric Sum would be {2,3,4,5,6,7,8,9,10,11,12}
when Eric Sum=2, Sandra's Sum should be grater than 2
when Eric Sum=3, Sandra's Sum should be grater than 3 and etc..
Required Probability
= { ( Eric=2 and Sandra >2 ) + ( Eric=3 and Sandra >3 ) + ( Eric=4 and Sandra >4 ) + ...... + ( Eric=12 and Sandra >12 ) }
= $ ( \frac{1}{36} * \frac{35}{36} )+ ( \frac{2}{36} * \frac{33}{36} )+ ( \frac{3}{36} * \frac{30}{36} )+ ( \frac{4}{36} * \frac{26}{36} )+ ( \frac{5}{36} * \frac{21}{36} )+ ( \frac{6}{36} * \frac{15}{36} )+ ( \frac{5}{36} * \frac{10}{36} )+ ( \frac{4}{36} * \frac{6}{36} )+ ( \frac{3}{36} * \frac{3}{36} )+ ( \frac{2}{36} * \frac{1}{36} )+ ( \frac{1}{36} * \frac{0}{36} ) $
= $ \frac{1}{36^2} $ . {(35*1)+(33*2)+(30*3)+(26*4)+(21*5)+(15*6)+(10*5)+(6*4)+(3*3)+(2*1)+(1*0)}
= $ \frac{1}{1296} $ . {(35)+(66)+(90)+(104)+(105)+(90)+(50)+(24)+(9)+(2)+(0)}
= $ \frac{1}{1296} $ . {(575)}
= $ \frac{575}{1296} $