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
Best Answer
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.
$$\binom{365}{363,1,1}=\frac{365!}{363!\cdot 1!\cdot 1!}=365\cdot 364$$
This is right. Therefore
$$\text{P}(\text{matching birthday})=\frac{365\ }{365\cdot 364 +365 } =\frac{1}{365}$$