Ranking N candidates combinatorics problem. I have figured it out 1 way, seeking help in doing it a different way. I came across a question from grinstead and snell:
Let's say you have N candidates, they are to be ranked by a 3 person committee. A candidate will be selected if they are ranked #1 by at least two people. Assume the members on the committee ranks the candidates randomly. Find the probability that a candidate will be accepted.
I am having a very difficult time starting. I have the correct answer but no idea how to calculate it. I approach all probability problems as such: What the is the total probability space i.e. total # of ways, and what are the # of desirable outcomes. 
In this case my attempt is as follows:
Total probability space= # of different possible rankings. Which I reason to be $n!*3$ because let's say there are n candidates, candidate "1" can have a total of n rankings which leaves n-1 for the next, etc. There are 3 different lists; which is why I multiplied by 3. 
So $3*n!$ should be my denominator. Now I calculate the # of desirable outcomes. A candidate could be ranked 1st by 2 or by 3 people. He can be ranked by 1st by 2 people 3 different ways, and 1st by 3 people in only 1 way. That gives me 4 totaL ways a candidate can be selected
Thus my answer is $4/(3*n!)$
This is however, incredibly incorrect. The real answer is $(3n-2)/n^3$Could someone please guide me in the right direction? 
Also fwiw: so far in the book it hasn't covered N choose K formulae, so I feel like I should be able to reason this out purely through counting. 
EDIT: I have figured out the correct answer using simple probability math, HOWEVER, I would love it of somoene could help convert the answer into: desirable outcomes / total # of ways. 
Answer:
Probability of being chosen= 3 ways of being chosen by 2 members + 1 way of being chosen by all 3
$3*((1/n)*(1/n)*(1-(1/n)) + (1/n)^3 $ = $(3n-2)/n^3$
But I hate the way I did it. I really want to start by figuring out the total sample space and divide that by the # of desirable outcomes. Any way to do that?
 A: First of all, I think you interpreted the question incorrectly (it may have been worded poorly). You were answering: "what is the probability that anyone at all is chosen?" I believe the question meant to ask "what is the probability that a given person is chosen?"
That said, you also have to be careful not to over-count. When you count the ways that a person can get two votes, you allow for the third vote to be anything. This means that you are counting cases where an applicant gets three votes once as an instance of getting two votes, and again as an instance of getting three votes.
The chance a person gets exactly two votes is $$\frac 1N\cdot\frac 1N-\frac1{N^3}=\frac{N-1}{N^3}.$$
Remember there are three ways to get exactly two votes, so this term becomes
$$\frac{3N-3}{N^3}$$
The chance for a person to get all three votes is simply $\frac 1{N^3}$. All together, we have
$$\frac{3N-3}{N^3}+\frac{1}{N^3}=\frac{3N-2}{N^3}.$$
A: Each committee member chooses a candidate randomly, so therefore each of the $N$ candidates has a $1/N$ chance of being ranked number $1$ (or indeed any rank from $1$ to $N$ as we are considering the choice is taken at random) by a member. There are $N$ candidates in total and we need at least $2$ committee members out of the $3$ on the panel to place a candidate at number $1$ for them to win. For each member there are $N$ people to choose from, so $3N$ in total when all members are taken into account. Out of this amount we remove the $2$ that are needed to be picked as number $1$ to give $3N-2$. Then, since each of the $N$ candidates has a $1/N$ chance of being picked by a member, we multiply $3N-2$ by this probability $3$ times giving the required probability as 
$$(3N-2)\cdot\frac{1}{N}\cdot\frac{1}{N}\cdot\frac{1}{N}$$
Note since the candidates ranking is chosen randomly this probability is the same for a candidate to be picked by at least $2$ members at any ranking from $1$st to $N$th.
