# 高级算法 (Fall 2019)/Balls into bins

# Balls into Bins

Consider throwing [math]m[/math] balls into [math]n[/math] bins uniformly and independently at random. This is equivalent to a random mapping [math]f:[m]\to[n][/math]. Needless to say, random mapping is an important random model and may have many applications in Computer Science, e.g. hashing.

We are concerned with the following three questions regarding the balls into bins model:

- birthday problem: the probability that every bin contains at most one ball (the mapping is 1-1);
- coupon collector problem: the probability that every bin contains at least one ball (the mapping is on-to);
- occupancy problem: the maximum load of bins.

## Birthday Problem

There are [math]m[/math] students in the class. Assume that for each student, his/her birthday is uniformly and independently distributed over the 365 days in a years. We wonder what the probability that no two students share a birthday.

Due to the pigeonhole principle, it is obvious that for [math]m\gt 365[/math], there must be two students with the same birthday. Surprisingly, for any [math]m\gt 57[/math] this event occurs with more than 99% probability. This is called the **birthday paradox**. Despite the name, the birthday paradox is not a real paradox.

We can model this problem as a balls-into-bins problem. [math]m[/math] different balls (students) are uniformly and independently thrown into 365 bins (days). More generally, let [math]n[/math] be the number of bins. We ask for the probability of the following event [math]\mathcal{E}[/math]

- [math]\mathcal{E}[/math]: there is no bin with more than one balls (i.e. no two students share birthday).

We first analyze this by counting. There are totally [math]n^m[/math] ways of assigning [math]m[/math] balls to [math]n[/math] bins. The number of assignments that no two balls share a bin is [math]{n\choose m}m![/math].

Thus the probability is given by:

- [math]\begin{align} \Pr[\mathcal{E}] = \frac{{n\choose m}m!}{n^m}. \end{align} [/math]

Recall that [math]{n\choose m}=\frac{n!}{(n-m)!m!}[/math]. Then

- [math]\begin{align} \Pr[\mathcal{E}] = \frac{{n\choose m}m!}{n^m} = \frac{n!}{n^m(n-m)!} = \frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-(m-1)}{n} = \prod_{k=1}^{m-1}\left(1-\frac{k}{n}\right). \end{align} [/math]

There is also a more "probabilistic" argument for the above equation. Consider again that [math]m[/math] students are mapped to [math]n[/math] possible birthdays uniformly at random.

The first student has a birthday for sure. The probability that the second student has a different birthday from the first student is [math]\left(1-\frac{1}{n}\right)[/math]. Given that the first two students have different birthdays, the probability that the third student has a different birthday from the first two students is [math]\left(1-\frac{2}{n}\right)[/math]. Continuing this on, assuming that the first [math]k-1[/math] students all have different birthdays, the probability that the [math]k[/math]th student has a different birthday than the first [math]k-1[/math], is given by [math]\left(1-\frac{k-1}{n}\right)[/math]. By the chain rule, the probability that all [math]m[/math] students have different birthdays is:

- [math]\begin{align} \Pr[\mathcal{E}]=\left(1-\frac{1}{n}\right)\cdot \left(1-\frac{2}{n}\right)\cdots \left(1-\frac{m-1}{n}\right) &= \prod_{k=1}^{m-1}\left(1-\frac{k}{n}\right), \end{align} [/math]

which is the same as what we got by the counting argument.

There are several ways of analyzing this formular. Here is a convenient one: Due to Taylor's expansion, [math]e^{-k/n}\approx 1-k/n[/math]. Then

- [math]\begin{align} \prod_{k=1}^{m-1}\left(1-\frac{k}{n}\right) &\approx \prod_{k=1}^{m-1}e^{-\frac{k}{n}}\\ &= \exp\left(-\sum_{k=1}^{m-1}\frac{k}{n}\right)\\ &= e^{-m(m-1)/2n}\\ &\approx e^{-m^2/2n}. \end{align}[/math]

The quality of this approximation is shown in the Figure.

Therefore, for [math]m=\sqrt{2n\ln \frac{1}{\epsilon}}[/math], the probability that [math]\Pr[\mathcal{E}]\approx\epsilon[/math].

## Coupon Collector

Suppose that a chocolate company releases [math]n[/math] different types of coupons. Each box of chocolates contains one coupon with a uniformly random type. Once you have collected all [math]n[/math] types of coupons, you will get a prize. So how many boxes of chocolates you are expected to buy to win the prize?

The coupon collector problem can be described in the balls-into-bins model as follows. We keep throwing balls one-by-one into [math]n[/math] bins (coupons), such that each ball is thrown into a bin uniformly and independently at random. Each ball corresponds to a box of chocolate, and each bin corresponds to a type of coupon. Thus, the number of boxes bought to collect [math]n[/math] coupons is just the number of balls thrown until none of the [math]n[/math] bins is empty.

**Theorem**- Let [math]X[/math] be the number of balls thrown uniformly and independently to [math]n[/math] bins until no bin is empty. Then [math]\mathbf{E}[X]=nH(n)[/math], where [math]H(n)[/math] is the [math]n[/math]th harmonic number.

**Proof.**Let [math]X_i[/math] be the number of balls thrown while there are *exactly*[math]i-1[/math] nonempty bins, then clearly [math]X=\sum_{i=1}^n X_i[/math].When there are exactly [math]i-1[/math] nonempty bins, throwing a ball, the probability that the number of nonempty bins increases (i.e. the ball is thrown to an empty bin) is

- [math]p_i=1-\frac{i-1}{n}. [/math]

[math]X_i[/math] is the number of balls thrown to make the number of nonempty bins increases from [math]i-1[/math] to [math]i[/math], i.e. the number of balls thrown until a ball is thrown to a current empty bin. Thus, [math]X_i[/math] follows the geometric distribution, such that

- [math]\Pr[X_i=k]=(1-p_i)^{k-1}p_i[/math]

For a geometric random variable, [math]\mathbf{E}[X_i]=\frac{1}{p_i}=\frac{n}{n-i+1}[/math].

Applying the linearity of expectations,

- [math] \begin{align} \mathbf{E}[X] &= \mathbf{E}\left[\sum_{i=1}^nX_i\right]\\ &= \sum_{i=1}^n\mathbf{E}\left[X_i\right]\\ &= \sum_{i=1}^n\frac{n}{n-i+1}\\ &= n\sum_{i=1}^n\frac{1}{i}\\ &= nH(n), \end{align} [/math]

where [math]H(n)[/math] is the [math]n[/math]th Harmonic number, and [math]H(n)=\ln n+O(1)[/math]. Thus, for the coupon collectors problem, the expected number of coupons required to obtain all [math]n[/math] types of coupons is [math]n\ln n+O(n)[/math].

- [math]\square[/math]

Only knowing the expectation is not good enough. We would like to know how fast the probability decrease as a random variable deviates from its mean value.

**Theorem**- Let [math]X[/math] be the number of balls thrown uniformly and independently to [math]n[/math] bins until no bin is empty. Then [math]\Pr[X\ge n\ln n+cn]\lt e^{-c}[/math] for any [math]c\gt 0[/math].

**Proof.**For any particular bin [math]i[/math], the probability that bin [math]i[/math] is empty after throwing [math]n\ln n+cn[/math] balls is - [math]\left(1-\frac{1}{n}\right)^{n\ln n+cn} \lt e^{-(\ln n+c)} =\frac{1}{ne^c}. [/math]

By the union bound, the probability that there exists an empty bin after throwing [math]n\ln n+cn[/math] balls is

- [math] \Pr[X\ge n\ln n+cn] \lt n\cdot \frac{1}{ne^c} =e^{-c}. [/math]

- [math]\square[/math]

## Occupancy Problem

Now we ask about the loads of bins. Assuming that [math]m[/math] balls are uniformly and independently assigned to [math]n[/math] bins, for [math]1\le i\le n[/math], let [math]X_i[/math] be the **load** of the [math]i[/math]th bin, i.e. the number of balls in the [math]i[/math]th bin.

An easy analysis shows that for every bin [math]i[/math], the expected load [math]\mathbf{E}[X_i][/math] is equal to the average load [math]m/n[/math].

Because there are totally [math]m[/math] balls, it is always true that [math]\sum_{i=1}^n X_i=m[/math].

Therefore, due to the linearity of expectations,

- [math]\begin{align} \sum_{i=1}^n\mathbf{E}[X_i] &= \mathbf{E}\left[\sum_{i=1}^n X_i\right] = \mathbf{E}\left[m\right] =m. \end{align}[/math]

Because for each ball, the bin to which the ball is assigned is uniformly and independently chosen, the distributions of the loads of bins are identical. Thus [math]\mathbf{E}[X_i][/math] is the same for each [math]i[/math]. Combining with the above equation, it holds that for every [math]1\le i\le m[/math], [math]\mathbf{E}[X_i]=\frac{m}{n}[/math]. So the average is indeed the average!

Next we analyze the distribution of the maximum load. We show that when [math]m=n[/math], i.e. [math]n[/math] balls are uniformly and independently thrown into [math]n[/math] bins, the maximum load is [math]O\left(\frac{\log n}{\log\log n}\right)[/math] with high probability.

**Theorem**- Suppose that [math]n[/math] balls are thrown independently and uniformly at random into [math]n[/math] bins. For [math]1\le i\le n[/math], let [math]X_i[/math] be the random variable denoting the number of balls in the [math]i[/math]th bin. Then
- [math]\Pr\left[\max_{1\le i\le n}X_i \ge\frac{3\ln n}{\ln\ln n}\right] \lt \frac{1}{n}.[/math]

- Suppose that [math]n[/math] balls are thrown independently and uniformly at random into [math]n[/math] bins. For [math]1\le i\le n[/math], let [math]X_i[/math] be the random variable denoting the number of balls in the [math]i[/math]th bin. Then

**Proof.**Let [math]M[/math] be an integer. Take bin 1. For any particular [math]M[/math] balls, these [math]M[/math] balls are all thrown to bin 1 with probability [math](1/n)^M[/math], and there are totally [math]{n\choose M}[/math] distinct sets of [math]M[/math] balls. Therefore, applying the union bound, - [math]\begin{align}\Pr\left[X_1\ge M\right] &\le {n\choose M}\left(\frac{1}{n}\right)^M\\ &= \frac{n!}{M!(n-M)!n^M}\\ &= \frac{1}{M!}\cdot\frac{n(n-1)(n-2)\cdots(n-M+1)}{n^M}\\ &= \frac{1}{M!}\cdot \prod_{i=0}^{M-1}\left(1-\frac{i}{n}\right)\\ &\le \frac{1}{M!}. \end{align}[/math]

According to Stirling's approximation, [math]M!\approx \sqrt{2\pi M}\left(\frac{M}{e}\right)^M[/math], thus

- [math]\frac{1}{M!}\le\left(\frac{e}{M}\right)^M.[/math]

Due to the symmetry. All [math]X_i[/math] have the same distribution. Apply the union bound again,

- [math]\begin{align} \Pr\left[\max_{1\le i\le n}X_i\ge M\right] &= \Pr\left[(X_1\ge M) \vee (X_2\ge M) \vee\cdots\vee (X_n\ge M)\right]\\ &\le n\Pr[X_1\ge M]\\ &\le n\left(\frac{e}{M}\right)^M. \end{align} [/math]

When [math]M=3\ln n/\ln\ln n[/math],

- [math]\begin{align} \left(\frac{e}{M}\right)^M &= \left(\frac{e\ln\ln n}{3\ln n}\right)^{3\ln n/\ln\ln n}\\ &\lt \left(\frac{\ln\ln n}{\ln n}\right)^{3\ln n/\ln\ln n}\\ &= e^{3(\ln\ln\ln n-\ln\ln n)\ln n/\ln\ln n}\\ &= e^{-3\ln n+3\ln\ln\ln n\ln n/\ln\ln n}\\ &\le e^{-2\ln n}\\ &= \frac{1}{n^2}. \end{align} [/math]

Therefore,

- [math]\begin{align} \Pr\left[\max_{1\le i\le n}X_i\ge \frac{3\ln n}{\ln\ln n}\right] &\le n\left(\frac{e}{M}\right)^M &\lt \frac{1}{n}. \end{align} [/math]

- [math]\square[/math]

When [math]m\gt n[/math], Figure 1 illustrates the results of several random experiments, which show that the distribution of the loads of bins becomes more even as the number of balls grows larger than the number of bins.

Formally, it can be proved that for [math]m=\Omega(n\log n)[/math], with high probability, the maximum load is within [math]O\left(\frac{m}{n}\right)[/math], which is asymptotically equal to the average load.