Probability - Birthday Permutation

This is an example I read it from the book, “Introduction to Probability”. The solution is:

Assume that there are 365 possible birthdays for each person(ignore leap years). We must find the probability that have no duplication of birthdays.

For two people, it is…

$$\frac{365 \times 364}{365^2}$$

For three people, it is…

$$\frac{365 \times 364 \times 363}{365^3}$$

For more people, it is…

$$\frac{365 \times 364 \times \dotsm \times (365 - i + 1)}{365^i} \text{, where $i$ = 2, 3, ..., 365}$$

So the probability that two people in the room will have the same birthday is…

$$1 - \frac{365 \times 364 \times \dotsm \times (365 - i + 1)}{365^i} \text{, where $i$ = 2, 3, ..., 365}$$

It turns out that

Number of people, Probability that two ppl have same birthday
20                0.411438
21                0.443688
22                0.475695
23                0.507297 <-- the answer
24                0.538344
25                0.568700

The Python code is:

def calc_prob(ppl_num):
    """
    Calc the probability of that ppl have no duplication of birthdays.
    """
    prob_no_dup = 1
    if ppl_num > 1:
        for i in range(1, ppl_num):
            prob_no_dup *= (365. - i if i <= 365 else 0) / 365.
    return 1. - prob_no_dup

And the plot is like:

It is tempting to guess that we would need about $\frac{1}{2}$ this number, or $183$, but the answer is $23$. From the chart we can see the probability that two people have same birthday is almost $1$ when there’re more than $100$ people in a room.

[code]