Probability Two People are Born on the Same Date (alternative way)

Question

What is the probability two people (individuals) will have the same exact birthday?

There are 365 days in a year and I assume that any person can be born on any random day, so uniformly.

I like to use a slots method when I look at combinations/permutations

Slots of possibilities person 1 was born on (e.g. born on Jan 2)

$\{ \text{Person 1} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 1 \quad }{2} \frac{ \quad 0 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $

Slots of possibilities person 2 was born on (e.g. born on Jan 3)

$\{ \text{Person 2} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 0 \quad }{2} \frac{ \quad 1 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $

Add these two slots and you get these two slot possibilities

The two people are not born the same date

$\{ \text{Slot 1 + Slot 2} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 1 \quad }{2} \frac{ \quad 1 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $

There $365 \choose 2$ ways of arranging two $1$'s and three hundred sixty three $0$'s

OR

Both people are born on the same day (e.g. Jan 3)

$\{ \text{Slot 1 + Slot 2} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 0 \quad }{2} \frac{ \quad 2 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $

There are 365 ways of arranging one 2 and three hundred sixty four $0$'s

So the probability of two people have matching birthdays

$$\text{P}(\text{matching birthday})=\frac{365 }{{365 \choose 2} +365 } \approx 0.005$$

But this answer is incorrect. I know what the correct answer is and I know how to do it another way. My question is why is the method I laid out not correct? I am less concerned with the answer, where am I wrong in my thinking in looking at all the possibilities? Thank you

Answer 1

$\{ \text{Slot 1 + Slot 2} \} = \frac{ \quad 0 \quad }{1} \frac{ \quad 1 \quad }{2} \frac{ \quad 1 \quad }{3} \quad \cdot \cdot \cdot \cdot \cdot \cdot \quad \frac{ \quad 0 \quad }{364} \frac{ \quad 0 \quad }{365} $

There $365 \choose 2$ ways of arranging two $1$'s and three hundred sixty three $0$'s

Here is the flaw. You have two different days of birthdays (a,b). So basically you arrange 365 elements. 363 have the same label $x$. Then you have additionally two different labels. So you are looking for the number of ways to arrange the following elements

$$\underbrace{xx...xx}_{=363}ab$$

So you have three different types of elements. Here you use the multinomial coefficient.

$$\binom{365}{363,1,1}=\frac{365!}{363!\cdot 1!\cdot 1!}=365\cdot 364$$

There are 365 ways of arranging one 2 and three hundred sixty four 0's

This is right. Therefore

$$\text{P}(\text{matching birthday})=\frac{365\ }{365\cdot 364 +365 } =\frac{1}{365}$$