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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

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  • $\begingroup$ You are assuming that all of these events occur with equal probabilities. This is not true. $\endgroup$ Mar 21, 2020 at 20:09
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    $\begingroup$ @PeterForeman But this is not the problem of the OP. $\endgroup$ Mar 21, 2020 at 20:11
  • $\begingroup$ @callculus I don't understand what you mean. You cannot apply $\mathbb{P}(A)=|A|/|\Omega|$ without each event having equal probabilities. $\endgroup$ Mar 21, 2020 at 20:12
  • $\begingroup$ I look at the possibilities (weight) of each of the only two possible outscomes. A Birthdays Match and Birthdays No Match $\endgroup$
    – jessica
    Mar 21, 2020 at 20:13
  • $\begingroup$ @jessica Yes and each of these events are not equally likely to occur. So one cannot use the formula stated. Imagine I have a biased coin. There are two events - heads or tails. So therefore the probability of heads is $1/2$ (division of event required by total events). This obviously does not always give the required probability as shown in this case. $\endgroup$ Mar 21, 2020 at 20:15

1 Answer 1

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$\{ \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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    $\begingroup$ Thank you!! You are my hero! $\endgroup$
    – jessica
    Mar 21, 2020 at 21:59
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    $\begingroup$ You´re welcome. Stay healthy. $\endgroup$ Mar 21, 2020 at 22:27

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