# The Natural Constant “e” and Finding The Best Girlfriend

Recently, a close friend of mine has been troubled by whether or not to accept a girl’s confession of love. He sought my advice. In matters of love, I am a blank slate and cannot provide any valuable guidance. However, I suggest transforming this issue into a mathematical problem for consideration.

Suppose a person can have only one lifelong partner—although many people may not agree with this premise, let’s temporarily set it for mathematical modeling. How can this person maximize the probability of obtaining an outstanding partner in reality?

Once we establish this model, it becomes evident that this problem is similar to Socrates’ wheat problem. Around 2700 years ago, Socrates instructed his students to walk from one end of a wheat field to the other, without turning back, and to pick only one ear of wheat. The challenge was to select the largest ear of wheat without the option to switch or replace it.

To ensure the highest probability of picking the largest ear of wheat, we can approach this problem as follows:

Let the length of an ear of wheat be denoted as “n.”

When n=1, only that particular ear of wheat can be selected, and the probability is 100%.

When n=2, the probability of selecting the largest ear of wheat is 50%.

For n=3, the probability of selecting the largest ear of wheat is 33.33% without any strategy. We need a strategy. Consider the following approach:

The first ear of wheat is not selected regardless of its size. If the second ear is larger than the first, choose the second ear; otherwise, choose the third ear. This strategy ensures a 50% probability of selection. (There are 3!=6 arrangements of the 3 ears of wheat.)

Arrangements:

1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, 3 2 1

After discarding the first ear of wheat, the largest is selected from the remaining 3 species, resulting in a probability of 3/6 × 100% = 50%.

For n>3, we continue using this strategy. However, the strategy involves skipping the previous k ears. If the next ear encountered is larger than the longest wheat among the previous k ears, choose it.

We create a sequence: 1, 2, 3, …, k, k+1, k+2, k+3, …, n, where each ear of wheat has a size represented by ai.

If the k+1 ear is the longest, the optimal probability of selection is 100%.

If the k+2 ear is the longest, considering the k+1 wheat, if we choose the k+1 ear, then obviously:

9. Select the optimal probability as k/(k+1).

10. Similarly, if the i-th ear is the longest, the probability of selecting the best is k/i.

11. Overall, the optimal probability of selection is:

12. The above equation can be approximated using the Newton-Leibniz formula.

13. Given that Z = k/n, the original formula is ln(Z)/Z. Since k < n, it’s evident that Z > 1. After deriving further, it was discovered that when Z = e, there exists a maximum value approximately equal to 1/e.

Consequently, we obtain k ≈ n/e.

Applying this insight, we can address the friend’s dilemma regarding choosing a life partner. Suppose my friend intends to compare 30 girls during the process of finding his ultimate girlfriend. The optimal strategy for him would be to discard the initial 11 girls (approximately 30 * 37%) and establish a benchmark based on these 11 individuals. Starting from the 12th girl, he should wholeheartedly pursue anyone who surpasses the previously set standard. This approach represents the best possible solution for him.

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