probability - Throw a fair die repeatedly until I have thrown three sixes

I play a game in which I have to throw a fair die repeatedly until I have thrown three sixes, after which I stop and note the total number of throws. What is the probability that I take six throws?

I did $(1/6)^3 \times (5/6)^3$ and this gave me $0.00268$ to 3.s.f.

This doesn't seem right as couldn't this be achieved in many ways by throwing the die to be 6 in different orders? E.g. $6,6,6$, non six,non six, non six OR 6,non six, non six, 6, 6, non six

Could anyone explain this?

Answer

In the first $5$ throws exactly $2$ sixes must be thrown.

The probability of this event is $\binom52\left(\frac16\right)^2\left(\frac56\right)^3$.

After that a six must be thrown so we end up with a probability of: $$\binom52\left(\frac16\right)^3\left(\frac56\right)^3$$