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

Odds of winning the secretary problem | You want to find the most suitable out of twelve candidates. If you are the first truly Rejecting applicants and then hiring the person who was better than all applicants is a possibility s the best choice. Depending on the number of people directly rejected truly The graph shows the possibilities s. bigger falls on them truly= 4 off.

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.

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if you put truly = n/e In the original equation for the probability of winning, you get the result: s = 1/e? 0.37 … This means that the probability of picking the largest number immediately using the mentioned strategy is about 37 nbs; percent. But as many of Myself studies I found that most people are impatient: People who encounter similar problems tend to make a decision very quickly — rather than dismissing 37 percent of opportunities. In this way, one usually achieves a weaker result. So it is advisable to be patient.

What is your favorite math theory? Feel free to write it in the comments – and maybe it will be the topic of this column soon!

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