In a box, we have $5$ dice. Three of them are proper, one has two sixes, and a second one has six sixes.

We’re given the events 

* $A$: after the first throw, we have a six.
* $B$: after the next two throws, we have six and something else.

We want to find the probability $$P(A \cap B \mid \text{proper dice}).$$

My attempt is:
$$ (1/6)\times(1/6)\times(5/6)=\left(5/6^3\right).$$
I know that this is the wrong answer. The question is, why?