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What is the probability that at least one person in a room of $n$ people (excluding me) has the same birthday as I and that nobody else in the room shares a birthday on a different date? For practical purposes, pretend that leap years don't exist.


As a follow-up, what is the expected number of people that have the same birthday as I in a room of $n$ people (excluding me) if nobody else in the room shares a birthday on a different day?

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The probability is:

[Explanantion:Spoiler]

Choose n birthdays from 364 days(except the one which you have), all different dates then select one person whose birthday you want to change to yours and arrange the rest.Total ways of selecting birthdays will be $365^n$. In case of 366 persons select one person to have same birthday as you and arrange the rest 364 persons with the remaining 364 dates.

$$\frac{\binom{364}{n}\binom{n}{1}(n-1)!}{365^n}:\quad n\le 364\\\frac{\binom{365}1364!}{365^n}:n=365\\0:n\ge366$$ Sorry I know combinatorics and haven't yet covered probability or expected value.

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  • $\begingroup$ What about "and that nobody else in the room shares a birthday on a different date"? $\endgroup$ – user171358 Nov 4 '14 at 10:57
  • $\begingroup$ @DigitalBrain that should be easy provided the solution to first part $\endgroup$ – RE60K Nov 4 '14 at 10:58
  • $\begingroup$ I guess you've only calculated probability of "at least two people sharing birthdates" and not "exactly two" $\endgroup$ – user171358 Nov 4 '14 at 11:00
  • $\begingroup$ @DigitalBrain don't guess, revisit it $\endgroup$ – RE60K Nov 4 '14 at 11:01
  • $\begingroup$ I'm afraid you're wrong, You're making sure that there's only one person sharing birthday with you, but what about birthdays shared by people except you two? $\endgroup$ – user171358 Nov 4 '14 at 11:07

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