Question

Suppose there are \(n\) types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of the \(n\) types. Suppose one continues collecting until a complete set of at least one of each type is obtained. (a) Find the probability that there is exactly one type \(i\) coupon in the final collection. Hint: Condition on \(T\), the number of types that are collected before the first type \(i\) appears. (b) Find the expected number of types that appear exactly once in the final collection.

Short answer

The short answer for this problem is: (a) The probability of having exactly one type i coupon in the final collection can be expressed as: \[P(\text{exactly one type i}) = \sum_{k = 0}^{n - 1} \frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\] (b) The expected number of types that appear exactly once in the final collection can be calculated using the linearity of expectation as: \[E(\text{number of types that appear exactly once}) = n \sum_{k = 0}^{n - 1} \frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\]

Step-by-step solution

(a) Probability of having exactly one type i coupon in the final collection

To solve this problem, we need to find the probability of having exactly one type i coupon in the final collection. Let's condition on T: - If T=0, it means that the collection starts with type i. Then we need to collect the remaining n-1 types without getting another type i. This is similar to distributing n-1 distinct items (types of coupons) in n-1 distinct boxes (remaining coupon types), which can be expressed using the Stirling numbers of the second kind. The probability of this happening can be expressed as \[\frac{S(n - 1, n - 1)}{n^{n-1}}.\] - If T=k (1 ≤ k ≤ n-1), it means that there are k types collected before the first type i appears. And we need to collect the remaining n-k-1 types without getting another type i coupon. This can be expressed by distributing n-k-1 distinct items in n-k distinct boxes. The probability of this happening can be written as \[\frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\] Now, we can compute the overall probability by summing these probabilities for T = 0 to T = n-1: \[P(\text{exactly one type i}) = \sum_{k = 0}^{n - 1} \frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\]

(b) Expected number of types that appear exactly once in the final collection

Now, we want to find the expected number of types that appear exactly once in the final collection. To do this, we'll use the linearity of expectation, which states that the expected value of the sum of random variables is the sum of their expected values. So, for each type of coupon, we can compute the probability of having only one of that type in the final collection, and then sum these probabilities: \[E(\text{number of types that appear exactly once}) = \sum_{i = 1}^{n} P(\text{exactly one type i}).\] Using the result from part (a), we have: \[E(\text{number of types that appear exactly once}) = n \sum_{k = 0}^{n - 1} \frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\]

Key concepts

Introduction to Probability Theory

Probability theory is a fundamental aspect of mathematics that deals with the likelihood of different outcomes. It helps us understand and quantify the chances of various events happening. Let's think about the scenario provided: collecting coupons. Imagine you have a collection of coupons, and you want to know the probability of certain outcomes, such as having exactly one of a particular type of coupon.

When dealing with probability, we operate under the assumption that each event is independent and has an equal chance of occurring. In the coupon collecting problem, this means that each time a new coupon is collected, it could be any of the available types with equal likelihood. This is where probability theory provides methods to compute the likelihood of complex events, such as completing a collection with specific constraints.

Stirling Numbers of the Second Kind Explained

Stirling numbers of the second kind, denoted as {\(S(n, k)\)}, play a pivotal role in combinatorics, which is a branch of mathematics. These numbers represent the count of ways to partition a set of \(n\) objects into \(k\) non-empty subsets. In simpler terms, think of having \(n\) distinct coupons and \(k\) distinct albums to fill with these coupons. Stirling numbers of the second kind will tell us how many ways we can distribute the coupons among the albums.

In the context of the coupon collecting problem, Stirling numbers help calculate scenarios where coupons need to be distributed into different 'types' without leaving any type empty. Understanding these numbers is crucial for solving part (a) of the problem, as it directly relates to finding the probability of having exactly one type \(i\) coupon in the final collection, given a condition on the number of types collected previously.

Linearity of Expectation

The linearity of expectation is an incredibly useful concept in probability theory since it simplifies the process of calculating expected values. It states that the expected value of the sum of random variables is equal to the sum of their individual expected values, regardless of whether the variables are independent or not.

This principle allows us to approach complex problems by breaking them down into simpler, more manageable parts. For example, in part (b) of the coupon collecting problem, we want to determine the expected number of types that appear exactly once in the final collection. Instead of tackling the entire collection at once, linearity of expectation lets us calculate the probability for each type independently and sum them up for the final result. Such a straightforward method proves invaluable in solving many statistical and probabilistic problems.