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What is the approximate probability (in percentage) that at least $2$ people in a group of $6$ randomly-selected have a birthday on the same day of the week?

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closed as off-topic by Jean-Claude Arbaut, AlexR, Najib Idrissi, Surb, N. F. Taussig Apr 15 '15 at 9:49

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    $\begingroup$ en.wikipedia.org/wiki/Birthday_problem $\endgroup$ – Surb Apr 15 '15 at 8:51
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    $\begingroup$ The general problem even has the same name. Plugging it into google will give you a lot of helpful references wich you can use to at least do something of your own and show your work. $\endgroup$ – AlexR Apr 15 '15 at 9:02
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$1-\frac{365*(365-7)*(365-14)*(362-21)*(361-28)*(360-35)}{365^6}$

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  • $\begingroup$ Why do you subtract multiples of $7$? $\endgroup$ – Jean-Claude Arbaut Apr 15 '15 at 9:23
  • $\begingroup$ Because in the second term I am calculating the probability of no two people sharing the same birthday week. Hence $-7$. $\endgroup$ – Aditya Agarwal Apr 15 '15 at 9:43
  • $\begingroup$ There are roughly $52$ weeks in a year: $52$ mondays, $52$ tuesdays... Here no two people must share the same day of week. $\endgroup$ – Jean-Claude Arbaut Apr 15 '15 at 9:46
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    $\begingroup$ Ok, I interpreted the problem improperly. I thought they shouldn't share the same birthday week. $\endgroup$ – Aditya Agarwal Apr 15 '15 at 11:05

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