I've forgotten my Bike combo...any Math Savvy people out there?

Well, the total lock has 4 digits, and I have the choices of #'s 1-6
Can anyone calculate the number of combos I have to try in order to find the right one?

Assuming you can use the same number more than once, you have six choices for the first digit, six choices for the 2nd, six choices for the 3rd and six choices for the 4th. So the total number of choices is 6*6*6*6 = 6^4 = 1296 combinations.

If by combo you mean permutation ...

that's how a combination lock really works since

ABCDE
and ACBED are two different combinations

then
6*6*6*6 = ANSWER

Now, the probability that you'll get it right on the first try is
1/6^4
First or second trial
= 1/6^4 *2
Probability that you'll get it right by the third trial is
3/6^4 ...

so for your second question
you'll have to try it at least
6^4*0.5 /6^4 to have a 50% shot of getting it right

For the first number you have 6 choices. Now, for each choice you have for the first digit, you have 6 separate choices for the second digit. And for each choice of the first two digits, you have 6 choices for the third digit. This reasoning should convince you that there total number of possibilities is

(6 choices for the first) * (6 choices for the second) * (6 choices for the third) * (6 choices for the fourth)

which is merely 6^4 = 1296.

Hope this helps.