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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!

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Ha! No, no too basic, and credit for letting us know what you've tried. –  Simon Hayward Dec 10 '12 at 22:14
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1 Answer

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.

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Thank you, this is very helpful! –  Andrew Dec 10 '12 at 23:33
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