What's the smallest $n$ so that it's more likely than not that someone among the $n$ people has the same birthday as you? Suppose that the birthdays of different people in a group of $n$ people are independent, each equally likely to be on the $365$ possible days. (Pretend there's no such thing as a leap day.)
What's the smallest $n$ so that it's more likely than not that someone among the $n$ people has the same birthday as you? (You're not part of this group.)
At first, I get $n=23$ by using the complement rule to get the probability that someone has the same birthday as me $(P\{\text{same birthday}\} = 1 - P\{\text{different birthday}\})$ and use $e^{-x}=1-x$.

But I don't think the answer is $23$ since I am not part of the group.
 A: HINT
The probability that none of the $n$ people have the same birthday as you is:
$$\big( \frac{364}{365} \big)^n$$
So, set this equal to $\frac{1}{2}$ and let WolframAlpha do its magic.
And as you alrady suspected, this problem is not like the infamous 'Birthday Problem'
A: $n=23$ famously appears in the Birthday Paradox, where we find the probability that there's at least one pair of people who share the same birthday — but there's no restrictions on when that common birthday is, i.e. it can be any day of the year.
Here your question is different because you're looking for a specific birthday date in the group. Your idea of using the complement rule
$$(P\{\text{same birthday}\} = 1 - P\{\text{different birthday}\})$$
is the right one! You just need to implement it better. Note that here $P\{\text{different birthday}\})$ refers to the probability that all people in the group have their birthdays on any day of the year other than your own birthday. In other words, each member of the group has $364$ possibilities for their birthday. Since we consider their birthdays to be independent, this probability is
$$P\{\text{different birthday}\})=\left(\frac{364}{365}\right)^n.$$
