Conditional probability question with cards I was given the following homework question:

There are $n$ cards labeled from $1$ to $n$ and ordered randomly in a
  pack. The cards are dealt one by one. Let $k\leq n$. Given that the
  label on the $k$-th card dealt is the largest among the first $k$
  cards dealt, what is the probability that the label on this $k$-th
  card dealt is in fact the largest in the entire pack?

My thoughts:
I will denote $P(n)$ as the probability of being labeled as $n$.
We want $$P(n|\text{bigger then the first k-1)}=\frac{P(n\cap\text{bigger then the first k-1)}}{\text{P(bigger then the first }k-1)}=\frac{P(n)}{P(\text{bigger then the first k-1)}}=\frac{1}{n}\cdot\frac{1}{P(\text{bigger then the first k-1})}$$
And so I need to calculate the probability that the $k-th$ is bigger
then all the cards before it.
In total there can be $\binom{n}{k-1}$ cards in the first $k-1$
places, so I need to find how many combination from that satisfy what
I want.
This is where my problem is, I thought about dividing for cases if
the $k-th$ card is $k,k+1,...$ and that gives me $\binom{k-1}{k-1}+\binom{k}{k-1}+...+\binom{n}{k-1}$
which I can't calculate.
Am I correct until this point ? How can I complete the calculation ?
I think that I have a mistake, because I say the the total number of options is $\binom{n}{k-1}$ and I sum
it up to be in the denominator so it will give some number$>1$ 
 A: Your conditional probability calculation goes through fine. Whatever the first $k$ cards are, the probability the $k$-th is the biggest among these is $\dfrac{1}{k}$. So using your calculation we have that our conditional probability is $\dfrac{\frac{1}{n}}{\frac{1}{k}}$. 
A: What you have done is right. Now let's compute the probability that the $k$ th card is the maximum of these $k$ cards.
Denote $C_1,C_2,\ldots C_k$ the first $k$ cards. Since you know nothing of the draws, each permutation is as equally likely to have occured. Denote $M=\max_{1\leq i\leq k}\{C_1,C_2,\cdots, C_k\}$. On the $k!$ possible permutation, there are exactly $(k-1)!$ that have $M$ has the $k$ th card. (Fix it as last card and permute the $k-1$ remaining so the probability that $M$ is at position $k$ is 
$$
\frac{(k-1)!}{k!}=\frac{1}{k}
$$
Add this to your $P(n)=1/n$ to get 
$$
\mathbb{P}(n|\text{bigger than the first $k-1$})=\frac{k}{n}
$$
A: The probability that the maximum occurs within the first $k$ cards is $\frac kn$.
This is also the answer to the problem question because whether the maximum of all cards appears among the first $k$ cards is independent from the relative rank of the $k$th card among the first $k$ cards.
