What's the probability to choose the 3rd person moving up to the next round? Suppose I'm part of a group of $10$ people where $8$ of us are going to the 'next round' (it's a contest). The group's leader reads the names of the winners first and leaves the last $2$ losers to the end. What is the probability that I'm the $3$rd guy to move on to the next round?
My approach to this goes like this: the probability of the first person to be read is $\frac{7}{10}$ because $2$ are out and I don't want to be the first guy. 
The probability for the second person: $\frac{6}{9}$
The probability for the third person (me): $\frac{6}{8}$ because I want to be read this time!
so we have: $\frac{7}{10} \cdot \frac{6}{9} \cdot \frac{6}{8}$. Am I on the right track? Or should I use the binomial distribution (but it doesn't make sense to me to use constant success/failure if there's no replacement)?
 A: I'm not sure what do you mean by "the probability of the first person to be read is $\frac{7}{10}$". The name of each person will be read exactly once with probability $1$, won't it?
Without additional context, we assume that the situation is symmetric for all contestants, and the probability that your name will be the $k$-th is the same for all $k$ (i.e. you have equal chances to be the first person as being the third person or the seventh person). 
So under this assumption that all permutations of the list of the names are distributed uniformly, the answer is merely $\dfrac{1}{10}$.
A: Of the $\dfrac{8}{10}$ chances that you will be picked to go through to the next round,  $\dfrac{1}{8}$ of these will be with you being picked in the 3rd position
So the probability of you going through to the next round AND being picked in the third position is $$(\dfrac{8}{10})(\dfrac{1}{8}) = \dfrac{8}{80} = \dfrac{1}{10} $$
A: Your solution method is, on an intuitive level, flawed, because it is unclear what you are actually counting.  This is frequently a consequence of not properly defining the events and outcomes of interest.  Specifically, what does $7/10$ count in your scenario?  I presume that you intend that there are $7$ possible winners, excluding yourself, that could be chosen as the first winner announced.  But is this really how you wish to count?
I claim it is not for at least the following reason:  you have not explicitly stated that you are assumed to be a winner.  If you had, you might have phrased the question as, "what is the probability I am the third winner to be announced, given that I am among those selected to proceed to the next round?  Rather, you are simply asking for the probability you are announced third; in which case, your enumeration clearly is incorrect:  there are $9$ ways to choose someone that is not yourself to be the first name announced.  It can only be $7$ if you knew in advance who were not chosen to proceed.
To properly count the desired probability, it is important to clearly and unambiguously define the event of interest, and the set of possible elementary outcomes.  If we are to interpret your question as it is written, we can address it by simply considering the order of the names announced:  the last two names are automatically deemed the "losers" who will not proceed.  In such a case, it is entirely possible that your name could be among those not selected to proceed.
Clearly, there are $10!$ distinct ways of ordering the group, each of which is assumed to be equally likely to occur if we do not make any assumptions about whether you are one of the winners.  Among these $10!$ total elementary outcomes, there are exactly $9!$ such outcomes in which you are the third name announced.  This is because once you fix your position as the third name in the ordered list of $10$ people, there are $9$ remaining open spots, and $9!$ permutations.  The desired probability is therefore $$\frac{9!}{10!} = \frac{9!}{10(9!)} = \frac{1}{10},$$ which is as we should expect.
This of course raises the question of what the answer would be if you had qualified your question with the additional condition that you know in advance you are one of the winners.  Then the counting of elementary outcomes is not so simple.  But we already know that there are $9!$ ways in which you are third, so the numerator does not change.  It is the denominator that changes to exclude those cases in which you were not a winner; i.e. your name was among the last two announced.  You could be dead last in $9!$ ways (for reasons analogous to being third); or you could be second-to-last in, again, $9!$ ways.  So the number of ways you are a winner is $$10! - 9! - 9! = 10(9!) - 2(9!) = 8(9!)$$ ways, and the desired probability of being third given you are a winner, is simply $1/8$.  And I think you should be able to see that again, this is entirely intuitively sensible:  there are $8$ winners, and your name is equally likely to be in any of those positions; thus the chance you are third is $1/8$.
