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My first intuition was to think $P(\{\text{No two bdays coincide}\}) = \displaystyle{\frac{\binom{365}{n}}{365}}$ given $365 \ge n$. Turns out $P(\{\text{No two bdays coincide}\}) = \displaystyle{\frac{365 \cdot 364 \cdots (365 - n + 1)}{365^n}}$. Why is it important we line up the people instead of simply dealing with subsets? What do we undercount when we solve this problem according to my initial intuition? Thanks.

Edit: Came here to understand the combinatorial reasoning behind the question, but I got a much better (obviously, personal opinion) answer based on probabilistic reasoning. That answer might be beneficial to future students struggling to grok the classical solution to this problem. Here it is below:

Let $A_i$ be an event where the $i$th person shares bday with none of the $n−1$ people. Then we have $P(\bigcap_{i-1}^n A_i) = P(A_1)P(A_2\mid A_1)P(A_3\mid A_1, A_2)\cdots P(A_n\mid A_1,\ldots A_{n-1}) = \frac{365}{365} \cdot \frac{364}{365} \cdots \frac{365 - n + 1}{365}$

user825005
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    Let's take an example: Suppose there are two people. Your intuition would give an 18200% probability that no two birthdays coincide. – SlipEternal Sep 14 '20 at 14:06
  • Now... it sounds as though you were wanting to just look at the set of days that appeared and not take order of people into account in which case $\binom{365}{n}$ is the correct count of such cases where all $n$ are different days. Meanwhile via stars and bars $\binom{n+365-1}{365-1}$ is the number of different distributions of days with repeats allowed and order not mattering, far more than just $365$ alone. Even the answer of $\dfrac{\binom{365}{n}}{\binom{n+365-1}{365-1}}$ is wrong however since those outcomes counted by stars and bars are not equally likely to have occurred. – JMoravitz Sep 14 '20 at 14:14
  • @JMoravitz, I am not sure this problem is all that amenable to combinatorial thinking. Think the following works better? Let $A_i$ be an event where the $i$th person shares bday with none of the $n - 1$ people. Then we want $P(\bigcap_{i=1}^n A_i) = \frac{365}{365} \cdots \frac{365 - n + 1}{365}$. But that supposes these events are independent. Are they? Also, is this method sound? Thanks. – user825005 Sep 14 '20 at 16:13
  • No, $A_1,A_2,A_3,\dots$ are not independent, however the product there you see isn't $P(A_1)\times P(A_2)\times P(A_3)\cdots$ so we never assumed that nor used that. The product you see is $P(A_1)P(A_2\mid A_1)P(A_3\mid A_1,A_2)P(A_4\mid A_1,A_2,A_3)\cdots$ and expanding the product like this is valid regardless of independence. – JMoravitz Sep 14 '20 at 16:22
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    As for "not that amenable to combinatorial thinking"... sure it is. Just consider the people distinct and thus the "order of the birthdays" (or more accurately, who it was who received which birthday) as relevant... it gives a probability of $\dfrac{365\frac{n}{~}}{365^n}$ using basic multiplication principle of counting (here $a\frac{b}{~}$ is notation for falling factorial). – JMoravitz Sep 14 '20 at 16:25
  • I see. Product of events conditioned on previous events looks even better because it's, basically, word for word description of the given answer. No more questions. Thanks. – user825005 Sep 14 '20 at 16:31

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