If there are 100 people in a room then what is the probability that at least 2 of them share birthdays? I read that answer should be $1-\left(\frac{365!}{265!\times365^{100}} \right)$, and I understood why it is so.

However if I solve it another way I am getting the wrong answer. If we think of the problem as assigning birthdays to people, then there are ${365 \choose 100}$ ways to assign each person with a different birthday and ${464 \choose 100}$ ways so that one birthday could be assigned to more than one person. So why the following equation gives me the wrong answer: $$1-\frac{{365 \choose 100}}{{464 \choose 100}}$$

I don't understand why the order of people matters in this question.

• Where do you get $\binom{464}{100}$ from? May 26, 2018 at 9:32
• If we consider a binary sequence of 100 0s and 364 1s, then number of 0s to the left of leftmost 1 will give the number of people assigned the birthday 1st January and so on. May 26, 2018 at 9:45

To illustrate it even better according to your solution the probability that all people are born on 1 January is same as the probability that they are born on 1 January, 2 January, 3 January ... Now imagine the first person coming then there's a $\frac{1}{365}$ chance the first configuration is satisfied, but there's $\frac{100}{365}$ the second one is satisfied. Similarly for the second person you have $\frac{1}{365}$ chance the first confifuration is true, while $\frac{99}{365}$ the second one is.
• @shiva Is $\frac{1}{2} = \frac{2}{3}$? May 26, 2018 at 16:54