Hi Kennedy,

I am going to indicate the 3 possibilities as hw, home win, aw, away win and d, draw. Thus for the first game there are 3 possible outcomes, hw, aw or d.

For each of these possibilities there are 3 possibilities for game 2, and thus there are $3 \times 3 = 3^2$ possibilities for the first two games. You can list them

hw, hw
hw, aw
hw, d
aw ,hw
aw, aw
aw, d
d, hw
d, aw
d, d.

For each of the nine possibilities for the first two games there are 3 possibilities for game 3. Hence there are $3 \times 3 \times 3 = 3^3$ possibilities for the first 3 games. Again you can list them if you want

hw, hw, hw
hw, hw, aw
hw, hw, d
hw, aw, hw
and so on.

Do you see now the number of possibilities for 13 games?

Penny