# The Secretary Problem: How do you find the most suitable applicant?

general probability s To win, sum P=\sum_{k=r+1}^{n}\frac{1}{n}\cdot\frac{r}{k-1}. The $$result of this expression can be rewritten to simplify later. To do this, you offset the start and end points of the sum and adjust the terms to be added so that the result does not change:$$P=\frac{r}{n}\cdot\sum_{k=r}^{n-1}\frac{1} {k}. $$If there are few cards on the table (ie, n small), the amount can be quickly calculated. The result depends on truly Off – you just have to know which one truly The probability of winning is greater. ### What is the minimum number of cards you must hand? But when there are hundreds or even thousands of pieces of paper, it becomes confusing. Statisticians then resort to a common trick: they approximate the sum using an integral, which makes it easy to evaluate:$$P=\frac{r}{n}\cdot\sum_{k=r}^{n-1}\frac{1 } {k} \ almost \ frac {r} {n} \ cdot \ int_r ^ {n} \frac {1} {k} dk. $$for this one can define the antiderivative and insert the terms of the integral: } \ text {ln} \frac {r } {n}.$$

Now you have a good rough estimate of the probability of making the right decision if you are at the beginning truly He ignores the note and bets on the next higher value. The question is: How many scraps of paper? truly One must put aside the chance of winning s To make it as big as possible? Translated into a mathematical language: for her sake truly Become s Maximum? If you remember your school days, you know you have to have a curve discussion of this. This means that you are driving the job s distance truly Then select zeros. Then there is the maximum value. The first derivative of the profit function is: P'(r) \almost -\frac{1}{n}\left (\text{ln}\frac{r}{n}+1 \right). $this$ expression becomes empty when it disappears inside the parentheses, so: ln (r / n) = -1. This is the case when truly = n/e – this is about 37 percent of n.