chance on throwing a six with 6 dice

With $6$ dice there are $6^6$ possible outcomes. Of these, $5^6$ don't lead to six on any of the dice. So the number of outcomes that lead to at least one six are $6^6-5^6$, so the probability of at least one six is $\frac{6^6-5^6}{6^6} \approx 0.6651$

The expected number of dice showing $6$ is $1$ when throwing six dice. The probability to see at least one $6$ is $1 - (5/6)^6$, as explained in Timothy's answer.

The probability of the union of independent events (unlike disjoint events) is not the sum of the individual probabilities. If $E_i$ denotes the event "$6$ is obtained on the $i$th throw", then $E_i$ are independent events, and it does not hold ${\rm P}(E_1 \cup \cdots \cup E_6 ) = {\rm P}(E_1 ) + \cdots + {\rm P}(E_6 )$. The left-hand side probability can be found as follows. The complement of the event $E_1 \cup \cdots \cup E_6$ is $E_1^c \cap \cdots \cap E_6^c $, where $E_i^c$ is the event "$6$ is not obtained on the $i$th throw". Hence, $$ {\rm P}(E_1 \cup \cdots \cup E_6 ) = 1 - {\rm P}(E_1^c \cap \cdots \cap E_6^c ). $$ Now, the probability of the intersection of independent events is the product of their individual probabilities. So, ${\rm P}(E_1^c \cap \cdots \cap E_6^c ) = {\rm P}(E_1^c) \cdots {\rm P}(E_6^c) = (5/6)^6$. Hence, ${\rm P}(E_1 \cup \cdots \cup E_6 ) = 1 - (5/6)^6$.