3
$\begingroup$

How many people must be in a room so that at least 10 have a birthday on a Friday?

edit: Assume that no two people share the same birthday

I'm somewhat confused and see two different ways to solve the problem. First, since all birthdays fall on one of seven days, if there are 64 people in the room one would at least one day of the week to hold 10 people. This is a similar problem: picking certain number of days.

However, there are 366 possible birthdays and only 52 or 53 of those are Fridays, giving $\displaystyle366 - 52= 314$ days of the year that are not Friday. So to be absolutely certain there are 10 people with Friday birthdays, would one need $314+10=324$ people?

$\endgroup$
  • 3
    $\begingroup$ Perhaps I misunderstand the question, but: What if everyone has a birthday on Thursday? There's no number large enough to guarantee an answer. $\endgroup$ – user61527 Dec 5 '13 at 3:42
  • $\begingroup$ Ha! Great minds think alike... $\endgroup$ – vadim123 Dec 5 '13 at 3:43
  • 1
    $\begingroup$ Everybody has a birthday on Friday some year, but if that's not what the question means, then you can have as many people as will fit in the room, and still not have even one with a Friday birthday. $\endgroup$ – Thomas Andrews Dec 5 '13 at 3:43
  • $\begingroup$ Do you mean how many people with different birthdays (pairwise dates) need to be there so at least $10$ have their birthday on friday this year? This is going to be very hard to compute $\endgroup$ – clark Dec 5 '13 at 3:56
  • $\begingroup$ This question was given in an exam without any other specifications, but I am assuming it means people with different birthdays. Otherwise, like others have pointed out, there will never be enough people to be certain. $\endgroup$ – display_name Dec 5 '13 at 4:13
2
$\begingroup$

The latter reasoning of the OP is correct. Assuming that we are in a normal year with 365 days in it and that this year contains 52 Fridays and that no two people share the same birthdate, then there are $365-52=313$ ways to assign people so that no one has a birthday on a Friday. After that, you are forced to fill up the empty Fridays, thus you need exactly 10 more people to do that.

Thus any collection of $313+10=323$ people will have at least 10 Friday birthdays.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Even on a normal year, someone may have been born on Feb. 29th and not have a birthday on any day. Like my sister. $\endgroup$ – ex0du5 Dec 5 '13 at 16:13
  • $\begingroup$ @ex0du5: Your point gives an interesting view on the question: can two people without a birthday (this year, because it would have been the non-existent Feb.$~29$) be said to share the same birthday? Strictly speaking I would say no, and you could fill the room with as many people born some Feb.$~29$ as you can find; there is no limit. $\endgroup$ – Marc van Leeuwen Dec 7 '13 at 12:44
4
$\begingroup$

You can never be certain that at least 10 people have birthdays on Friday; even with a billion people they might all have birthdays on Thursday.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.