(1) only touches it once, and if it is odd, it will win the prize, so the probability of Xiao Ming winning the prize is 23; ? 1 2 3 1 ( 1, 1) (2, 1) (3, 1) 2 ( 1,2) (2,2) (3,2)

Then p (Xiao Mingsheng) = 49;

(2) To sum up, let Xiao Ming touch n times, all of which are odd numbers, so the probability of Xiao Ming winning the prize is (23) n;

(3) Draw the following tree diagram:

There are 27 equal possibilities, 8 of which are odd.

Then p = 827.

So the answer is: (1) 23; 49; (2)(23)n； (3)827.