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If there are $12$ people in the room what the chances are that no two people will have the same birthday month?

I tried to think of it in this way:

The total number of options is $12^{12}$

The option that no one will have the same birthday month is:

$12 * 11 * ... 1$ which means $12!$

So I think the answer is:

$$\frac{12!}{12^{12}}=0.0000537$$

What do you think?

Thank you!

BAM
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    Looks good. Of course, you are assuming that all months are equally likely, which isn't technically true (even if you assume all birthdays are equally likely). Good approximation though. – lulu Mar 25 '16 at 17:47
  • I think this is correct. – shabo Mar 25 '16 at 17:48
  • Yaaa! Thanks! – BAM Mar 25 '16 at 18:01
  • Related to http://math.stackexchange.com/questions/770997/. – BruceET Mar 25 '16 at 18:01
  • It's not the same, the post you linked has the same birthday month – BAM Mar 25 '16 at 18:03
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    I solved pretty much the same question earlier today (http://math.stackexchange.com/questions/1712328/birthday-enumeration-question/1712990#1712990) – JKnecht Mar 25 '16 at 18:14
  • related: https://math.stackexchange.com/questions/1475378/what-is-the-probability-that-nobody-is-born-in-the-same-month – Henry Aug 09 '19 at 10:41

2 Answers2

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Just to close the question:

Assuming each person's birth month is independent and each month is equally likely, then your calculation of $\frac{12!}{12^{12}}\approx 0.0000537$ is correct

Incidentally, $\frac{n!}{n^{n}}$ is close to $\sqrt{2 \pi n}\, e^{-n}$ for large $n$, and for $n=12$ would give about $0.0000534$

Henry
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You can think about this in terms of conditional probability. Let $E_1$ be "first person have a birthday", which is obviously $P(E_1) = 1$. Then, let $E_2$ be "second person have a birthday on a different month than first person". $P(E_2|E_1) = \frac{11}{12}$, since second person have 11 different months to "choose" from. Then, you have $P(E_3|E_1 \cap E_2) = \frac{10}{12}$, $P(E_4|E_1 \cap E_2 \cap E_3) = \frac{9}{12}$, and so on. Then $$P(E_1 \cap \dots \cap E_{12}) = P(E_1) \cdot P(E_2|E_1) \cdot \dots \cdot P(E_{12}|E_1 \cap \dots \cap E_{11}) = 1 \cdot \frac{11}{12} \cdot \dots \cdot \frac{1}{12} = \frac{12!}{12^{12}}$$

Dark Archon
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