Simple (hopefully) probability question

[DCJ311]

Here are the prerequisites:

Hand A stands to win 1/3 of showdowns.
Hand B stands to win 2/3 of showdowns.
Ties are impossible.

Hands A and B do battle precisely 4 times.

What is the probability of:

A) Hand A defeating Hand B exactly 2 times out of 4.

B) Hand A defeating Hand B 2 or more times out of 4.

Any comments on how to do this problem are appreciated, as well as answers!

[Tom1975]

probability a wins 4 =(1/3)^4 =0.01234567

probability a wins 3 =(1/3)^3*(2/3)^1*COMBIN(4,3) =0.098765432

probability a wins 2 =(1/3)^2*(2/3)^2*COMBIN(4,2)=0.296296296

probability a wins 1 =(1/3)^1*(2/3)^3*COMBIN(4,1)=0.39506172

probability a wins 0 =(2/3)^4 = 0.197530864


So the probability of A winning exactly 2 is ~.296 and 2 or more is ~.407