Happy Probabilistic Birthday to Me

Today is my birthday. And I wonder if I ask people their birthdays at work, one at a time, then on average how many do I have ask until I get someone having the same birthday as me. 

The answer is 253. We can ask this question in a similar but probabilistic way. The trick is to keep in mind about "average". So "mid-way", right?

Basically, this is the same as asking the question, "How many people should I ask so that the probability that at least someone shares the same birthday as me is greater than 0.5?" Or let me even rephrase it again. "How many people should I ask so that the probability that none shares my birthday is less than 0.5?" That's complimentary theory.

So, let's forget about leap year and let's assume that there are 365 days in a year. And I will be having a party, letting people enter into the party room one at time and asking their birthday one at a time as they enter. I don't care if they share the same birthday between themselves. They only need to have a different birthday from me to be let into my party.

The first person enters and the probability that he or she has a different birthday from me is 364/365. The same goes for the second person, the third person and so on. By k-th person, we have 
          (364/365)^k   <  1/2 
And if we solve the inequality, we get the smallest k to be 253.

QED.