probability

favourable sums
2 (1 case)
3 (2 cases)
5 (4 cases)
7 (6 cases)
11 (2 cases)

4 ( 3 cases)
9 ( 3 cases)
P(sum=9) = 3/36

P( 9 on first try) = 4/36
P( 9 on second try ) = 14/36 * 4/36
P( 9 on 3rd try ) = (14/36)^2 * 4/36
.
.
.
.
Pnet = 4/36 / (1-14/36) = 2/11