problem: there are n persons in a room, what is the probability that no two of them celebrate the same birthday in a year? Here is my thought process, The sample space is $|\{(b_1,b_2,\dots,b_n): b_1,b_2\in\{1,2,\dots 365\}\}|=(365)^n$ , and I got stuck at counting the Event, $|\{(b_1,b_2,\dots,b_n)：b_i\neq b_j \forall i\neq j \}|=365*364*\dots*1$ but what if n>365? how do I count that?
$\begingroup$
$\endgroup$
6
-
7$\begingroup$ Let's ignore leap day for a moment. So working with $365$ days. If $n > 365$ then by the pigeon-hole principal there has to be at least two people that share a birthday. $\endgroup$– oliverjonesApr 11, 2022 at 22:43
-
2$\begingroup$ If n>365, two people will have the same birthday with probability one. $\endgroup$– econbernardoApr 11, 2022 at 22:43
-
1$\begingroup$ If $n \gt 365$, what does the pigeonhole principle tell you about the probability that no two people have the same birthday? $\endgroup$– Robert ShoreApr 11, 2022 at 22:44
-
$\begingroup$ Hi, thanks for the comments! Totally forgot about the pigeon hole principal, that makes sense now! $\endgroup$– RemuApr 12, 2022 at 0:16
-
$\begingroup$ For what it's worth, the probability of no birthday matches, for $n \in \{1,2,\cdots, 365\}$ can be more tersely expressed as $$\frac{(365)!}{[365 - n]!} \times \frac{1}{(365)^n}.$$ As you indicated, the denominator of the 2nd term above represents the sample space, while the first factor represents the number of ways of sampling $n$ birthdays, where the selection is done without replacement. Here, it is easiest to consistently regard the order of selection as pertinent in both the numerator and denominator, since that facilitates the $(365)^n$ expression. $\endgroup$– user2661923Apr 12, 2022 at 1:46
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
Your first calculation is $365$,
your second calculation $365\times 364$,
...,
your $365$th calculation $365\times 364\times \cdots\times 1$,
and your $366$th calculation $365\times 364\times \cdots\times 1\times 0$ which is $0$.
You can carry on further, but you will always have the $\times 0$ term with more people. So whenever you have more people than possible birthdays, you never get them each having a different birthday.
