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Find the probability of n different people having the same birthday month.

Is the following the right way to do:

Choose any one of the n people. Then the rest must have the same birthday month as the chosen one. So the probability is:

$n (\frac{1}{12^{n-1}})$.

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  • $\begingroup$ Just $(\frac{1}{12^{n-1}})$. You don't need to multiply by $n$. $\endgroup$ – RandomUser Apr 27 '14 at 6:35
  • $\begingroup$ Are there a fixed set of $n$ people? You shouldn't have the leading $n$ factor (e.g. imagine choosing the first of the $n$ people...) $\endgroup$ – ShreevatsaR Apr 27 '14 at 6:35
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Out of the $ n $ people, let us find the probability that a group of $ n - 1 $ people having the same birthday month as the one person we choose randomly.

In other words, the one person we choose randomly can have his/her birthday in any month. We find the probability of the rest of the people have the same birthday month. That is just $ \boxed{\dfrac{1}{12^{n-1}}} $

As an example, consider a group of 14 people. Let one person have his/her birthday in January. Now, what is the probability that the $ 13^{\text{th}} $ person has his birthday month as January? It is $ \dfrac{1}{12} $ Similarly, what is the probability that the $ 12^{\text{th}} $ person has his birthday month as January? $ \dfrac{1}{12} $ We can do the same for all the $ n - 1 $ people. And for all these events to happen together, we multiply them. Hence,

$ \dfrac{1}{12} \times \dfrac{1}{12} \times \dots \dfrac{1}{12}$ (13 times) is $ \dfrac{1}{12^{13}} $

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In this case, you just need to know the birthday month assigned to each person, not the order in which they are assigned. You choose a month number and assign it to n different people (simultaneously). So there is 12 ways for n people to have the same birthday month and the answer for your problem would be $\frac{1}{12^{n-1}}$.

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