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Suppose a room contains $n$ people. What is the probability that at least two people share the same birthday?

Let $A$ be the probability that at least two people have the same birthday. I know that the way to solve this question is actually to find the complement of A and solve $1 - P(A^c)$. However, I'm confused on why $A^c$ is the probability that no one shares the same birthday (everyone has different birthdays), and not the probability that at most two people share the same birthday. Isn't the opposite or complement of "at least two people share the same birthday" equal to "at most two people share the same birthday?"

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  • $\begingroup$ If "there are at least two persons with common birth date" is false, then "no two persons have the same birth date" is true. $\endgroup$ – Alvin Lepik Oct 3 '19 at 7:06
  • $\begingroup$ But if "no two persons have the same birth date" is false, then wouldn't "two people will have the same birthday" be true? How does "two people will have the same birthday" imply "AT LEAST two people will have the same birthday?" I guess I'm just really confused with the wording. $\endgroup$ – John Oct 3 '19 at 7:18
  • $\begingroup$ The question is asking for the probability of a coincidence between people's birthdays. The opposite of a coincidence is that there are no coincidences - i.e., no two birthdays are the same, i.e., all birthdays are distinct. $\endgroup$ – pre-kidney Oct 3 '19 at 7:19
  • $\begingroup$ @John If two persons share the same birth date, then it is true of at least two persons in the room. On the other hand, if at least two share the same birth date, assuming $n>2$, there could be more than two with the same birth date. Neither is it excluded that all of them have the same birth date. $\endgroup$ – Alvin Lepik Oct 3 '19 at 7:29
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Let $D$ denote the number of days in a year, so $D=365$ or $D=366$ (or something else) depending on how you are counting (and which planet you are living on).

The probability that no one shares the same birthday is the product of the probabilities that the second person doesn't share their birthday with the first $(D-1)/D$ times the probability the third doesn't share with the first two $(D-2)/D$ and so on down the line, until we get $$ \mathbb P(\text{no common birthdays})=\frac{D-1}{D}\cdots \frac{D-n+1}{D}. $$ So the probability that at least two people share a birthday is $1$ minus this.

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