Hey guys, anybody got any ideas on how to do this ?

We all know that the probability of finding two people in a room with the same birthday becomes better that a half when there are more than 23 people in the room. How many have to be in the room before the probability of two pairs of people with the same birthday is better than a half?

Thanks

2. Nice question...

I have no idea.

3. Originally Posted by Bruce
Nice question...

I have no idea.
Originally Posted by Rule #17
Only respond to questions where you have a sufficient knowledge of the topic to give helpful input. It is not helpful to make a hopeful guess or stab in the dark if you have no idea what's going on. This might in fact cause a delay in the question getting a competent response.

4. Originally Posted by jessismith
Hey guys, anybody got any ideas on how to do this ?

We all know that the probability of finding two people in a room with the same birthday becomes better that a half when there are more than 23 people in the room. How many have to be in the room before the probability of two pairs of people with the same birthday is better than a half?

Thanks

Eg.

Birthday problem - Wikipedia, the free encyclopedia

The Birthday Paradox :: curiousmath :: math is an attitude

HowStuffWorks "Someone told me that if there are 20 people in a room, there's a 50/50 chance that two of them will have the same birthday. How can that be?"

etc. etc. etc.

5. Originally Posted by jessismith
Hey guys, anybody got any ideas on how to do this ?

We all know that the probability of finding two people in a room with the same birthday becomes better that a half when there are more than 23 people in the room. How many have to be in the room before the probability of two pairs of people with the same birthday is better than a half?

Thanks
Suppose there are $N$ people. Let $X$ be the number of pairs of people with the same birthday.

What "we all know" is how to compute $P(X\geq 1)$, and this is done by writing $P(X\geq 1)=1-P(X=0)=1-\frac{365\cdots (365-N+1)}{365^N}$.

What you want now is $P(X\geq 2)=1-P(X=0)-P(X=1)$, so it remains to find $P(X=1)$.

We have $P(X=1)=\frac{\mbox{\# sequences of N birthdays with exactly one matching pair}}{\mbox{\# sequences of N birthdays}}$. The denominator is of course $365^N$. As for the numerator, it decomposes into the number of ways to choose the matching pair and the number of ways to choose a sequence of $N-1$ different birthdays. So we get: $P(X=1)=\frac{\frac{N(N-1)}{2}365\cdots(365-(N-1)+1)}{365^N}$.

Finally, $P(X\geq 2)=1-\frac{365\cdots (365-N+1)}{365^N}-\frac{\frac{N(N-1)}{2}365\cdots(365-(N-1)+1)}{365^N}$. It would be possible to simplify the writing a bit. Anyway, you can have the values computed by a software or a calculator for, say $N=20,\ldots, 40$. And you find (well, at least I find, you should check it) that 36 people are enough to make the probability of double matching greater than a half.