why calculate one minus the complement for probability of at least one six on four dice? If I roll a dice four times, it seems that the probability of at least one six on one of the four rolls is calculated as:
  $\text{P(At least 1 six)} = 1 - (P(1\mbox{ not six}) * P(2\mbox{ not six}) * P(3\mbox{ not six}) * P(4\mbox{ not six})) = 1 - (5/6)^4$
I was expecting the solution to be calculated like so:
  $\text{P(At least 1 six)} = P(1\mbox{ six}) + P(2\mbox{ six}) +P(3\mbox{ six}) +P(4\mbox{ six}) = 1/6 + 1/6 + 1/6 + 1/6$
Why is the first equation correct, but not the second one?
 A: Because in the second method you have to use inclusion/exclusion principle:


*

*Include the number of combinations with 1st die showing six

*Include the number of combinations with 2nd die showing six

*Include the number of combinations with 3rd die showing six

*Include the number of combinations with 4th die showing six

*Exclude the number of combinations with 1st and 2nd dice showing six

*Exclude the number of combinations with 1st and 3rd dice showing six

*Exclude the number of combinations with 1st and 4th dice showing six

*Exclude the number of combinations with 2nd and 3rd dice showing six

*Exclude the number of combinations with 2nd and 4th dice showing six

*Exclude the number of combinations with 3rd and 4th dice showing six

*Include the number of combinations with 1st, 2nd and 3rd dice showing six

*Include the number of combinations with 1st, 2nd and 4th dice showing six

*Include the number of combinations with 1st, 3rd and 4th dice showing six

*Include the number of combinations with 2nd, 3rd and 4th dice showing six

*Exclude the number of combinations with 1st, 2nd, 3rd and 4th dice showing six


That's a hell of a lot more work than the first method.
