Please verify: a twist on the Birthday Problem

Hello,

I was assigned some optional practice on the birthday question. I got all the questions on the birthday problem itself correct, but although I got an answer for this last one (n >= 612.257), it seems VERY unintuitive to me. Can you please confirm whether it's right or what I did wrong? Here is the problem:

**Given n people randomly chosen, find the smallest value of n so that the probability at least one of them were born on the same day of the year that you are born is at least 50%.**

Now, here's what I thought:

P(X >= 1) >= 0.5

X~Binomial(n, 1/365)

P(X >= 1) = 1 - P(X=0) - P(X=1)

1 - P(X=0) - P(X=1) >= 0.5

0.5 >= P(X=0) + P(X=1)

0.5 >= (nCr(n,0) * (1/365)^0 * (364/365)^n) + (nCr(n,1) * (1/365) * (364/365)^(n-1))

Now, nCr(anything, 0) is always 1, and nCr(n, 1) is always n. So:

0.5 >= (364/365)^n + (n/365)*(364/365)^(n-1)

Plugging this into wolfram alpha, I get that n >= 612.257. This result seems very unintuitive to me. Is it correct?

Re: Please verify: a twist on the Birthday Problem

Hello, Gui!

Quote:

Given n people randomly chosen, find the smallest value of n so that the probability that

at least one of them were born on the same day of the year that you are born is at least 50%.

We will find the probability that *none* of them have your borthday.

The probability that *none* of the $\displaystyle n$ people have your birthday is:.$\displaystyle \left(\frac{364}{365}\right)^n$

. . We want this to be *less** *__than__ 50%.

We have: .$\displaystyle \left(\frac{364}{365}\right)^n \,<\:0.5$

Take logs: .$\displaystyle \ln\!\left(\frac{364}{365}\right)^n \,<\;\ln(0.5) \quad\Rightarrow\quad n\!\cdot\!\ln\!\left(\frac{364}{365}\right) \,<\:\ln(0.5) $

Therefore: .$\displaystyle n \:<\:\dfrac{\ln(0.5)}{\ln\left(\frac{364}{365} \right)} \:<\:252.6519836$

. . It will take 253 people.