Counting card question

• Jun 17th 2009, 08:52 PM
pop
Counting card question
Let there be k unique cards (unique) in the set. Choose n of these cards (with replacement).
What is the probability that you choose one of each card?
What is the probability that you choose one of each card except one?
What is the probability that you choose one of each card except two?
More generally, find a probability function to calculate the probability of getting all except t cards of the set.
• Jun 18th 2009, 06:41 AM
HallsofIvy
Quote:

Originally Posted by pop
Let there be k unique cards (unique) in the set. Choose n of these cards (with replacement).
What is the probability that you choose one of each card?

Here we have to assume that $\displaystyle n\le k$
Choose a card- it can, of course, be any thing. Replace it and choose another card. It can be anything except the first card- the probability of that is (k-1)/k. Replace it and choose a third card. It can be anything except the first 2- the probability of that is (k-2)/k. Doing this n times, we have $\displaystyle \frac{k-1}{k}\frac{k-2}{k}\cdot\cdot\cdot\frac{k-n+1}{k}= \frac{(k-1)!}{(k-n)!k^n}$.

Quote:

What is the probability that you choose one of each card except one?
What is the probability that you choose one of each card except two?
More generally, find a probability function to calculate the probability of getting all except t cards of the set.
• Jun 18th 2009, 04:53 PM
pop
I don't think I was clear enough on the question. An example will make this clearer:

Suppose there are 10 different types of cards (say 1, 2, 3 ... 10). Now I buy 20 random cards. Each of these is either 1 or 2 or ... or 10. What is the probability that I will have bought at least one of each card?

So the answer I'm looking for will have probability = 0 if n < k. And P -> 1 as n -> infinite. I have found a answer myself but I question it's accuracy.

A hint is to remember that there are $\displaystyle binomial(n + k -1, n)$ ways to distribute n objects into k boxes.