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So I saw this interesting problem: https://en.wikipedia.org/wiki/Birthday_problem

And I am not the best at probability, so my question is why I cant calculate the probability with

P (2 in n same birthday) = 1/365 * 2/365 * ... * n-1/365

and have to use this instead?

P (2 in n same birthday) = 1 − P (2 in n not same birthday)

I understand how it works, my problem is that this would not be my first approach on this problem.

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  • $\begingroup$ How does your formula go when n is 364? 365? 366? $\endgroup$ – theonetruepath Mar 25 at 17:26
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    $\begingroup$ because it is easier to count. otherwise you need to count probability of two person sharing birthday, three person sharing birthday, and so on, also two person sharing birthday while another two sharing another birthday and all those.. $\endgroup$ – Rezha Adrian Tanuharja Mar 25 at 17:26
  • $\begingroup$ @RezhaAdrianTanuharja, but I am not doing it by saying 2 people have a 2/365, 3 people have a 3/365 chance to have birthday on the same day? $\endgroup$ – Neeliy Mar 25 at 17:28
  • $\begingroup$ If we have one birthday, no duplicate is possible. The second birthday has probability $\frac{364}{365}$ not to cause a duplicate, the next $\frac{363}{365}$ and so son since every new birthday gives one free day less. $\endgroup$ – Peter Mar 25 at 17:29
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    $\begingroup$ that is not what i said, what i mean is that having duplicate birthday can be: 2 person sharing birthday, 3 person, 4 person, etc., maybe 2 person share one day and the other 2 share another, etc.. Too many that i cannot even think of listing the possibilities, let alone counting the probability :) $\endgroup$ – Rezha Adrian Tanuharja Mar 25 at 17:39
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Let's see how your formula works with small, easily calculated numbers. What is the probability that two people out of $n$ share the same birth season? You claim it would be $$\frac 14 \frac 24 \frac 34 \cdots \frac {n-1}4$$ So

  • a group of two would have $\frac 14$ chance of sharing a birthseason (correct!)
  • a group of 3 would have a $\frac 14\frac 24 = \frac 18$ chance of sharing a birthseason (adding another person made the probability go down??)
  • a group of 4 would have a $\frac 3{32}$ chance of sharing a birthseason (???)
  • a group of 5 would also have only a $\frac 3{32}$ chance of sharing a birthseason!

Which is ridiculous because for $n > 4$ the real probability is obviously always $100\%$.

I'm not sure why you thought this was a valid scheme for calculating the probability, but it is not even close.

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