Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

First, if this question is too basic for math.stackexchange, I apologize. I wasn't sure where else to ask, but if you have a suggestion I'll happily take the question elsewhere.

I'm totally mathematically unsophisticated, and I was asked a theoretical question I don't know how to answer. The question is:

I want to invite as many people as possible to my birthday party.
I do not want people who have the same birthday as me to attend.
Everyone I invite who does not share my birthday will attend.
People who do share my birthday will be jealous and only have
a 1 in 3 chance of attending the party if they are invited.

How many invitations should I send if I want there to be no more than
a 50% chance of someone with the same birthday as me showing up?

Now, I tried to approach this as follows:

The chance that a random other person would have the same birthday as you is 1/365. The chance that someone who has the same birthday as you would accept the invitation is 1/3. Therefore the combined probability of someone having the same birthday and choosing to attend is 1/1095.

Every time another person attends we are adding one more chance of the above happening, so we can think of this as:

0.5 = n/1095


n = 0.5 / (1/1095) = 547.5

Well, I was told that the above is not correct, but the reason why I was wrong and the correct approach to understanding this problem was not explained to me.

Could anyone explain my mistake and how to correctly calculate this probability problem? Thanks!

share|cite|improve this question
Ha! No, no too basic, and credit for letting us know what you've tried. – Simon Hayward Dec 10 '12 at 22:14
up vote 4 down vote accepted

The probability that someone is invited, shares your birthday, and attends, as you pointed out, is $\dfrac{1}{1095}$. So the probability that with one invitation, this doesn't happen, is $p=\dfrac{1094}{1095}$. It follows that the probability this doesn't happen with $n$ invitations is $p^n$.

We want the largest $n$ such that $p^n \ge \dfrac{1}{2}$.

Solve the equation $$\left(\frac{1094}{1095}\right)^x=\frac{1}{2}.$$ Taking logarithms, we get $$x\log(1094/1095)=\log(1/2).$$ The calculator gives $x\approx 758.65$.

So $n=758$ will have the probability of a clash just under $1/2$, and inviting one more would make the probability of a clash a bit over $1/2$.

Remark: It is often difficult to explain why a certain procedure is wrong, apart from the fact that it gives an incorrect answer. Perhaps here one can say a little more.

Imagine we invite people one after the other, they reply, and we stop inviting after the first clash. Then it turns out that the median number invited is $\dfrac{1095}{2}$. This was your suggested answer, and is based on reasonable intuition.

share|cite|improve this answer
Thank you, this is very helpful! – Andrew Dec 10 '12 at 23:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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