Q:

What is wrong with this reasoning in a probability problem?

Consider the problem: what is the probability that when throwing six dice, we get exactly one six, at least one six and exactly two six?
I tried to solve the first part as follows, but it clearly went wrong. The outcomes of the experiment are tuples $(a_1,\dots,a_6)$ with $a_i\in \{1,\dots, 6\}$ where $a_i$ is what comes out of the $i$-th dice.
If $p = \frac{1}{6}$ is the probability of getting any number throwing the dice, then the probability for any such tuple is
$$q = p^6.$$
On the other hand: how many tuples are there with exactly one six? Well, there are 6 ways of choosing where the 6 comes, and then $5^5$ ways of choosing how to place the other numbers, since they can be equal. Thus it seems there are $\Omega = 6\times 5^5$ such tuples.
In that case, it seems the probability of exactly one six is
$$P(\text{one six})=\Omega q=\dfrac{1}{6^6}6\times 5^5=\left(\frac{5}{6}\right)^5=0,4$$
but this seems wrong, the probability is too high for me, I think it should be quite smaller.
For the other parts I would follow the same reasoning, changing the number of possible tuples. So for instance, to get at least one six, I want the number of tuples with six somewhere regardless of the rest. 
So we have $6$ ways of placing the one six, and $6^5$ ways of determining the other numbers, giving $\Omega = 6^6$ tuples. This is obviously wrong, this is the number of all the tuples, not of some subset of them.
So I believe that the issue is that I'm not computing correctly the number of tuples $\Omega$. So is that my mistake here? If so, what am I missing? What is the right way to do it?

A:

Your reasoning is right :
Let $X$ be the number of six among your six dice.
$$P(X=k)=\binom{6}{k}\frac{1}{6^k}\frac{5^{6-k}}{6^{6-k}}=\binom{6}{k}\frac{5^{6-k}}{6^{6}}$$
Your can then check that the probalities sum up to 1 :
$$\sum_{k=0}^{6}{P(X=k)}=\sum_{k=0}^{6}{\binom{6}{k}\frac{5^{6-k}}{6^{6}}}$$
$$=\frac{1}{6^6}\sum_{k=0}^{6}{\binom{6}{k}5^{6-k}}=\frac{1}{6^6}(1+5)^6=1$$
And for the second event , i.e. $X \ge 1$ you have
$$P(X \le 1) = \sum_{k=1}^{6}{P(X=k)}=1-P(X=0)=1-\frac{5^6}{6^6}$$

