chance on throwing a six with 6 dice The chance to throw a 6 with one die is 1/6
And 6 times 1/6 = 1
So, if I throw with 6 dice, the chance to throw at least 1 six should be 1.
But when I throw 6 dice, I sometimes don't throw any 6 at all..
How come?
 A: 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.
A: 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, ${\rm P}(E_1  \cup  \cdots  \cup E_6 ) = 1 - (5/6)^6$.
A: If you flip a coin, there's a 50-50 chance you get heads. If you flip two coins there's a $3/4$ chance you get at least one head, since there's a $1/2 \times 1/2 = 1/4 $ chance they're both tails. So similarly, there's a $5/6$ chance each die is not $6$. The chance they're all not $6$ is $(5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) = (5/6)^6$. So the chance that at least one is a six is $1 - (5/6)^6$. 
A: 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 $\frac{6^6-5^6}{6^6} \approx 0.6651$
A: You expect to get one $6$ on average.  But sometimes you get more than one $6$. 
Since the expected number of $6$s is one, this means that sometimes you have to get fewer than one $6$, i.e. zero $6$s. 
A: The classic method to calculate this is considering the probability of not throwing a six which equals 5/6 , six times and then subtracting that probability from the probability of all possible outcomes, 1, which gives the probability of having thrown at least one six
= 1 - (5/6)6
= 0.665
But consider another method, the probability of the first die being a 6, which is 1/6, then consider that there is an additional probability contribution of a 6 from the next die from the remaining possible events which are now 5/6 of the total, which is 5/6*1/6, and so on until one gets the following sum
= 1/6 + (5/6) / 6 + (5/6)2 / 6 + (5/6)3 / 6 + (5/6)4 / 6 + (5/6)5 / 6
= 0.665 or
= 67%
