To win k dollars requires n1 wins and n2 losses such that n1+n2=n, and

n1-n2=k.

This last imples that 2 n1 - (n1+n2) = k, so:

2 n1 = k+n

Hence k+n is even so the probabilty of winning k dollars in n trials is 0 if k+n is odd.

Also if k+n is even then the number of wins is (k+n)/2, and the probability of this is:

RonL