<s>[INST] Probability Two People are Born on the Same Date (alternative way) 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 
 [/INST] 
$\{ \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
$$\text{P}(\text{matching birthday})=\frac{365\ }{365\cdot 364 +365 } =\frac{1}{365}$$
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