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$>1$
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 $$ \mathbb{P}(n|\text{bigger than the first $k-1$})=\frac{k}{n} $$