For many probability problems, you can look at them two ways: The probability of getting your result, or the probability of not getting your result. If you know one, you will know the other, because they are mutually exclusive (you either get your result, or you don't).

In this case, the most straightforward way is to simply say, "What are the odds of never rolling a six in six attempts?" Then subtract that from 100%. As you've seen, this is a straightforward calculation, since the odds of NOT rolling a six are 5/6 and you need to do that 6 times, so it's: (5/6)(5/6)(5/6)(5/6)(5/6)*(5/6) = (5/6)⁶ = ~33.5%. Subtract it from 100% to get the answer.

Now, you could do it the other way. To do so in this case would be calculating the chance of the desired result on a given roll added together: (1/6)+(1/6)(5/6)+(1/6)(5/6)^{2+(1/6)(5/6)3+(1/6)(5/6)4+(1/6)*(5/6)5}

And even this is a simplified form. Because, for just two rolls, the logic is really: The chances of getting a 6 in JUST the first roll, plus the chance of getting a 6 in JUST the second roll, plus the chance of getting a 6 in both, as all those satisfy the condition. That is: (1/6)(5/6)+(5/6)(1/6)+(1/6)(1/6). This simplifies, of course, to: (1/6)+(5/6)(1/6) Now, that logic expands out to even grander scales as you add rolls, but it keeps simplifying, so for three rolls it is (simplified): (1/6)+(1/6)(5/6)+(1/6)(5/6)² And you keep on going until you hit six die rolls (the series I posted above).

And that is why people are taught to simply do 1-(5/6)⁶

More importantly, though, it helps demonstrate the important logical concept of using negations to help structure an investigative technique.