Probability for first person in room having same name as someone else in the room? There are 7 people in the group whom names are picked from 12 randomly picked names. What is a probability that first one in the room will have the same name as someone else in the room?
 A: Hint:
This is just a rewording of the question:

What is the probability that you roll at least one $1$ within six throws of a fair twelve-sided die?  (The twelve-sided die having each of the faces $1,2,3,\dots,12$)

Are you able to see why these are the same question?
To continue, try to find the probability of the complementary event, that none of the times you roll the die it comes up $1$.  I.e. the probability that no-one shares the same name as the first person.
A: I am going to assume this is what you meant:

A group consists of seven people. Each person has one of twelve names. The name of one person does not affect the name of another person. They then enter the room and line up, one by one. What is the probability that person one shares his/her name with at least one of the remaining six?

Here are the different combinations of names that satisfy our condition, where $×$ indicates a name that matches that of person one and $·$ indicates that it doesn’t. Beside each of these combinations is the probability of that combination occurring.
$$\begin{array}{cccccc|c}
× & · & · & · & · & · & \left(\frac{1}{12}\right)^1\left(\frac{11}{12}\right)^5\\
× & × & · & · & · & · & \left(\frac{1}{12}\right)^2\left(\frac{11}{12}\right)^4 \\
×  & ×  & × & · & · & · & \left(\frac{1}{12}\right)^3\left(\frac{11}{12}\right)^3\\
×  & ×  & × & ×  & · & · & \left(\frac{1}{12}\right)^4\left(\frac{11}{12}\right)^2\\
×  & ×  & × & ×  & ×  & · & \left(\frac{1}{12}\right)^5\left(\frac{11}{12}\right)^1\\
×  & ×  & × & ×  & ×  & ×  & \left(\frac{1}{12}\right)^6\left(\frac{11}{12}\right)^0\\
\end{array}$$
Next, we want to compute the number of ways to rearrange each of the above combinations since we have decided that order matters. After all, “$\,×\,·\,·\,·\,·\,·\,$” and “$\,·\,×\,·\,·\,·\,·\,$” are both lineups that would fall into our solution set.
We can’t use the standard permutation formula, because we do not want to count the lineup “$\,×\,×\,·\,·\,·\,·\,$” twice by switching the two $\times$s and pretending doing so is significant.
To compute the number of ways to arrange a lineup with $n$ “$×$”s and $k$ “$\cdot$”s, we can use a formula:

A set of $n$ objects with $a$ identical objects of one kind and $b$ identical objects of another kind can be permuted in $\dfrac{n!}{a!\cdot b!}$ recognizable ways.

Thus, the number of ways to arrange a lineup with $n$ “$×$”s and $k$ “$\cdot$”s is $\dfrac{6!}{n!k!}=\dfrac{720}{n!(6-n)!}$.
Finally, to compute the overall probability that at least one of people two through seven has a name that matches that of person one, we will multiply the number of ways to make a valid lineup times the probability of that lineup and then add this product for all valid lineups:
$$\begin{align}
\text{probability} &= \sum_{n=1}^{6} \frac{720}{n!(6-n)!}\left( \frac{1}{12}\right)^n \left( \frac{11}{12} \right) ^{6-n} \\
&= 0.40678055 \\
\end{align}$$
