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.

  • $\begingroup$ Where do you get $\binom{464}{100}$ from? $\endgroup$ – Lord Shark the Unknown May 26 '18 at 9:32
  • $\begingroup$ 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. $\endgroup$ – shiva May 26 '18 at 9:45

The problem is that the persons are distinguishable objects.

To understand the flaw in your reasoning consider this extreme example. According to your solution there's only one configuration in which all people are born on 1 January, which is right. However again according to your solution there is a single configuration in which one person is born on 2 January and the rest are on 1 January. But in reality there are 100 such configurations, as the the first person can be born on 2 January, then the second and so on.

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.

  • $\begingroup$ I got your point that my solution will be wrong as persons are distinguishable objects and I am not considering which person is born on which day. However I fail to understand how that is relevant when all we care is that all birthdays in the set should be different. $\endgroup$ – shiva May 26 '18 at 12:43
  • $\begingroup$ @shiva As mentioned above you are counting too many configurations as one, when in fact they are different. What you are doing can be put as this: You tell everybody to write down their birthday and then you collect the papers and you only check them. Now imagine the following let's the first person be born on 2 January and the rest on 1 January. Also let the second one be born on 2 January and the rest on 1 January. According to your method these two are same, but in fact they are obviously different. $\endgroup$ – Stefan4024 May 26 '18 at 12:47
  • $\begingroup$ So if I am counting them as one configuration in the numerator, then I am doing the same in the denominator as well, so doesn't it cancel out? Sorry but I am still a little confused. $\endgroup$ – shiva May 26 '18 at 16:47
  • $\begingroup$ @shiva Is $\frac{1}{2} = \frac{2}{3}$? $\endgroup$ – Stefan4024 May 26 '18 at 16:54
  • $\begingroup$ will that solution be correct if people were indistinguishable? $\endgroup$ – shiva May 26 '18 at 17:02

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