Problem (skip to part d): Ignoring leap days, the days of the year can be numbered $1$ to $365$. Assume that birthdays
are equally likely to fall on any day of the year. Consider a group of $n$ people, of which
you are not a member. An element of the sample space $Ω$ will be a sequence of n birthdays
(one for each person).
(a) Define the probability function $P$ for $Ω$.
(b) Consider the following events:
$A$: “someone in the group shares your birthday”
$B$: “some two people in the group share a birthday”
$C$: “some three people in the group share a birthday”
Carefully describe the subset of $Ω$ that corresponds to each event.
(c) Find an exact formula for $P(A)$. What is the smallest $n$ such that $P(A) > .5$?
(d) Justify why $n$ is greater than $\frac{365}{2}$ without doing any computation. (We are looking for
a short answer giving a heuristic sense of why this is so.)
Solution provided for part d: While $\frac{365}{2}$ different birthdays would have a $50$ percent chance of matching your
birthday, $\frac{365}{2}$ people probably don’t all have different birthdays, so they have a less than
$50$ percent chance of matching.
My confusion: I get that a person's birthday has a 50% chance of landing in one half of all possible birthdays. How does other people having the same birthday as you affect this fact? And what is the justification for why $n$ is greater than $\frac{365}{2}$? I understood all other previous parts of the question but on this part I am now lost.
Best Answer
You said that you get how a person's birthday has a $50\%$ chance of landing in one half of all possible birthdays. Imagine if $\frac{365}{2}$ distinct days were chosen, and then your birthday was randomly chosen (out of all possible days). There's a $50\%$ chance your birthday is one of those days. Other people having the same birthday as you doesn't affect this. However,
There are almost definitely overlaps amongst themselves for the birthdays of the $\frac{365}{2}$ people, so they likely don't all have distinct birthdays. Instead, it's less than $\frac{365}{2}$ distinct birthdays. Thus, $n > \frac{365}{2}$ to make up for this discrepancy.