There is a set of $n$ cards, numbered 1 to $n$. The cards are distributed at random to $n$ people. Assume that $n$ is large enough that we are sure that every person in the group has a card with his/her age on it.
(a) If there are $n$ cards and $n$ people, what is the expected number of cards that are given to people whose age matches the number on the card?
(b)Now, suppose there are $n$ cards and $\frac{n}{k}$ people, assuming that $\frac{n}{k}$ is an integer. Each of the $\frac{n}{k}$ people is given $k$ cards. What is the expected number of cards given to people such that the card matches the person's age?
My Answer
(a) Call event $A_i$ the event that the $i^{th}$ person gets distributed a card with his/her age on it. Let $X_i$ be an indicator random variable that is equal to 1 if $A_i$ occurs and 0 if $A_i$ does not occur. Then, the number of cards that are distributed to people that match the age of the person, X, is given by:
$$ X = \sum_{i = 1}^{n}{X_i} $$
So, the expected value of $X$ is the following:
$$ E(X) = E\left(\sum_{i = 1}^{n}{X_i}\right) = \sum_{i = 1}^{n}{E(X_i)} = nP(A_i) = n \frac{1}{n} = 1$$
I say that the $P(A_i)$ is $\frac{1}{n}$ because if the cards are numbered $1$ to $n$, then for the $i^{th}$ person who only has $1$ age, there is only $1$ card out of the $n$ that has his/her age on it.
(b) So in the first case, I think my reasoning is valid because the number of people who have a card that matches their age is the same as the number of cards distributed to a person where the number on the card matches that person's age. I believe this is still valid in the second case because the cards are numbered $1$ to $n$ and each person has only one age, which is a number in the set $1, …, n$.
Taking $A_i$, $X_i$, and $X$ to mean the same thing as part (a), the key question to answer is what is the $P(A_i)$, the probability that the $i^{th}$ person has a card with his/her age on it. The only difference being that the $i^{th}$ person has $k$ cards instead of $1$. I am kind of confused as to how to compute this probability.
Best Answer
There is still one card with the person’s age on it, and the person gets $k$ out of $n$ cards. At most one of them has their age on it, and the probability for one of them to be the one with that age is $\frac kn$; so the sum over the $\frac nk$ people is again $1$.