Secretary Problem

“Imagine there are eight roses in your garden; you inspect them one-by-one randomly. Take the first three roses you encounter as sample and go on. Once you encounter a rose more beautiful than any of the first three roses, you pick it up and stop. How likely is it that the one you picked up is the most beautiful one?”

At the first sight of the problem, I thought that I could solve it using the methods learned from the second chapter of Concrete Mathematics, and I did managed to. In this post, I will generalize the problem above into the case of $N$ roses and $K$ worth of sample size, then show you how to derive the probability in a closed form, using combination and summation strategies.

Let’s say that these roses have “beauty index” from $1$ through $N$, respectively, without duplication, and that $a_n$ is such index of the $n$-th rose you encounter. In this setup, ‘$N$’ is the most beautiful rose; if we will successfully pick up the most beautiful rose, the roses must follow

$$\overbrace{a_1\;a_2\;\ldots\;a_K}^{\text{$K$ terms}}\;\overbrace{a_{K+1}\;\ldots\;a_j}^{\text{$(j-K)$ terms}}\;N\;\overbrace{a_{j+2}\;\ldots\;a_N}^{\text{$(N-j-1)$ terms}},\quad\text{for $K\leq j\lt N$,}$$

where you pick up the $(j+1)$-th rose you inspect, which is also the most beautiful rose. In order for this to happen, all terms from $a_{K+1}$ through $a_j$ must be less than $k=\max(a_1,a_2,\ldots,a_K)$; there are $k$ such terms, but $K$ of them are already in $a_1\,a_2\,\ldots\,a_K$. So, at most $(k-K)$ terms are available to settle between ‘$a_K$’ and ‘$N$’, hence $j$ is at most $k$.

That means, if $k=\max(a_1,a_2,\ldots,a_K)$, the probability that you pick up the right rose is $\sum_{j=K}^k\frac{(N-j-1)!(k-K)^{\underline{j-K}}}{(N-K)!}$. (Note that $n^{\underline k}$ can denote the number of permutations of $k$ elements out of $n$ elements.)

(This refers to the notation of falling factorial powers, defined by the rule $n^{\underline k}=n(n-1)\cdots(n-k+1)$. More of its properties are described in Concrete Mathematics Chapter 2, partially available in my notes.)

Alright, we will readily be able to evaluate the probability if we know how often $\max(a_1,a_2,\ldots,a_K)$ equals to a specific $k$; let the probability be $Q(k)$. What we do is randomly choosing $K$ roses out of $N$ roses, and the largest of them is $k$. In other words, one of the choices must be the rose $k$, and the rest $(K-1)$ choices are made within $(k-1)$ roses that are less beautiful than the rose $k$. It’s now easy to see that

$$\begin{aligned}Q(k)&=\frac{\binom{k-1}{K-1}}{\binom NK}\\&=\frac K{N^{\underline K}}(k-1)^{\underline{K-1}},\quad\text{for $k\geq K$.}\end{aligned}$$

To sum up all what was said above, let $P_{N,K}$ be the ultimate probability we’ve been pursuing. We have

$$\begin{aligned}P_{N,K}&=\sum_{k=K}^{N-1}Q(k)\sum_{j=K}^k\frac{(N-j-1)!(k-K)^{\underline{j-K}}}{(N-K)!}\\&=\sum_{k=K}^{N-1}\frac K{N^{\underline K}}(k-1)^{\underline{K-1}}\sum_{j=K}^k\frac{(N-j-1)!(k-K)^{\underline{j-K}}}{(N-K)!}\\&=\frac K{N^{\underline K}(N-K)!}\sum_{k=K}^{N-1}\sum_{j=K}^k(N-j-1)!(k-1)^{\underline{K-1}}(k-K)^{\underline{j-K}}\\&=\frac K{N!}\sum_{k=K}^{N-1}\sum_{j=K}^k(N-j-1)!(k-1)^{\underline{K-1}}(k-K)^{\underline{j-K}}\\&=\frac K{N!}\sum_{k=K}^{N-1}\sum_{j=K}^k(N-j-1)!(k-1)^{\underline{j-1}}\\&=\frac K{N!}\sum_{K\leq k\lt N}\sum_{K\leq j\leq k}(N-j-1)!(k-1)^{\underline{j-1}}\\&=\frac K{N!}\sum_{K\leq j\lt N}\sum_{j\leq k\lt N}(N-j-1)!(k-1)^{\underline{j-1}}\\&=\frac K{N!}\sum_{K\leq j\lt N}(N-j-1)!\sum_{j\leq k\lt N}(k-1)^{\underline{j-1}}\\&=\frac K{N!}\sum_{K\leq j\lt N}(N-j-1)!\frac{(N-1)^{\underline j}-(j-1)^{\underline j}}j\\&=\frac K{N!}\sum_{K\leq j\lt N}\frac{(N-j-1)!(N-1)^{\underline j}}j\\&=\frac K{N!}\sum_{K\leq j\lt N}\frac{(N-1)!}j\\&=\frac{K(N-1)!}{N!}(H_{N-1}-H_{K-1})\\&=\frac KN(H_{N-1}-H_{K-1}),\end{aligned}$$

where $H_k$ is the $k$-th harmonic number, $H_k=1+\frac12+\cdots+\frac1k$.

Here are some critical observations about the steps above:

From line 6 to line 7, I interchanged the order of summation, such that $(N-j-1)!$ could be factored out.

From line 8 to line 9, I used the rule

$$\sum_{a\leq n\lt b}x^{\underline m}=\frac{b^{\underline{m+1}}-a^{\underline{m+1}}}{m+1},\quad\text{for $m\neq-1$.}$$

In line 9, $(j-1)^{\underline j}=0$, by the definition of falling factorial powers.

Back to our initial problem, we have $N=8$ and $K=3$; thus



So far, I have derived means of evaluating the probability given $N$ and $K$. Further derivations such as methods of optimizing the probability are still under investigation.