How many people do we need so that the probability of someone having the same birthday as Person A is 100%?

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Suppose you have some designated Person A. How many people do we need so that the probability that one of those people has the same birthday as person A is 100%?




The first thing that came to my mind was $365$, but that doesn't have to be true. Some of those $365$ people could have the same birthday, so the Person A's birthday doesn't have to be covered. I thought of calculating probability that $n$ people don't share the birthday with Person A. Then $1$ minus that probability, I don't know how to calculate that. It seems to me that the probability I am asking for will never be 100%.




probability combinatorics birthday




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  • 1




    Your suspicion is correct.
    – N. F. Taussig
    Aug 26 at 10:28










  • To me that seems weird and kind of beautiful
    – Milan Stojanovic
    Aug 26 at 10:29






  • 1




    If you have $367$ people, you can be certain that they do not all have different birthdays ($366$ if you ignore leap years) so some must share. But you cannot be absolutely certain a particular person shares a birthday, even with more people, unless you have more information (such as them being in a room with their twin)
    – Henry
    Aug 26 at 10:36














up vote
1
down vote

favorite













Suppose you have some designated Person A. How many people do we need so that the probability that one of those people has the same birthday as person A is 100%?




The first thing that came to my mind was $365$, but that doesn't have to be true. Some of those $365$ people could have the same birthday, so the Person A's birthday doesn't have to be covered. I thought of calculating probability that $n$ people don't share the birthday with Person A. Then $1$ minus that probability, I don't know how to calculate that. It seems to me that the probability I am asking for will never be 100%.




probability combinatorics birthday




share|cite|improve this question


















  • 1




    Your suspicion is correct.
    – N. F. Taussig
    Aug 26 at 10:28










  • To me that seems weird and kind of beautiful
    – Milan Stojanovic
    Aug 26 at 10:29






  • 1




    If you have $367$ people, you can be certain that they do not all have different birthdays ($366$ if you ignore leap years) so some must share. But you cannot be absolutely certain a particular person shares a birthday, even with more people, unless you have more information (such as them being in a room with their twin)
    – Henry
    Aug 26 at 10:36












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Suppose you have some designated Person A. How many people do we need so that the probability that one of those people has the same birthday as person A is 100%?




The first thing that came to my mind was $365$, but that doesn't have to be true. Some of those $365$ people could have the same birthday, so the Person A's birthday doesn't have to be covered. I thought of calculating probability that $n$ people don't share the birthday with Person A. Then $1$ minus that probability, I don't know how to calculate that. It seems to me that the probability I am asking for will never be 100%.




probability combinatorics birthday




share|cite|improve this question















Suppose you have some designated Person A. How many people do we need so that the probability that one of those people has the same birthday as person A is 100%?




The first thing that came to my mind was $365$, but that doesn't have to be true. Some of those $365$ people could have the same birthday, so the Person A's birthday doesn't have to be covered. I thought of calculating probability that $n$ people don't share the birthday with Person A. Then $1$ minus that probability, I don't know how to calculate that. It seems to me that the probability I am asking for will never be 100%.





probability combinatorics birthday





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 28 at 22:48









Mike Pierce

11k93574




11k93574










asked Aug 26 at 10:24









Milan Stojanovic

529




529







  • 1




    Your suspicion is correct.
    – N. F. Taussig
    Aug 26 at 10:28










  • To me that seems weird and kind of beautiful
    – Milan Stojanovic
    Aug 26 at 10:29






  • 1




    If you have $367$ people, you can be certain that they do not all have different birthdays ($366$ if you ignore leap years) so some must share. But you cannot be absolutely certain a particular person shares a birthday, even with more people, unless you have more information (such as them being in a room with their twin)
    – Henry
    Aug 26 at 10:36












  • 1




    Your suspicion is correct.
    – N. F. Taussig
    Aug 26 at 10:28










  • To me that seems weird and kind of beautiful
    – Milan Stojanovic
    Aug 26 at 10:29






  • 1




    If you have $367$ people, you can be certain that they do not all have different birthdays ($366$ if you ignore leap years) so some must share. But you cannot be absolutely certain a particular person shares a birthday, even with more people, unless you have more information (such as them being in a room with their twin)
    – Henry
    Aug 26 at 10:36







1




1




Your suspicion is correct.
– N. F. Taussig
Aug 26 at 10:28




Your suspicion is correct.
– N. F. Taussig
Aug 26 at 10:28












To me that seems weird and kind of beautiful
– Milan Stojanovic
Aug 26 at 10:29




To me that seems weird and kind of beautiful
– Milan Stojanovic
Aug 26 at 10:29




1




1




If you have $367$ people, you can be certain that they do not all have different birthdays ($366$ if you ignore leap years) so some must share. But you cannot be absolutely certain a particular person shares a birthday, even with more people, unless you have more information (such as them being in a room with their twin)
– Henry
Aug 26 at 10:36




If you have $367$ people, you can be certain that they do not all have different birthdays ($366$ if you ignore leap years) so some must share. But you cannot be absolutely certain a particular person shares a birthday, even with more people, unless you have more information (such as them being in a room with their twin)
– Henry
Aug 26 at 10:36










3 Answers
3






active

oldest

votes

















up vote
5
down vote



accepted










You are correct.



Given person $A$'s birthday, for each other person there is a probability of $364/365$ (ignoring leap years) that they do not share a birthday with $A$.



Therefore if there are $n$ others, then there is a probability of $(364/365)^n$ that none share a birthday with $A$, hence a probability of $1-(364/365)^n<1$ that at least one does share a birthday with $A$.






share|cite|improve this answer
















  • 2




    Thanks. The calculation wasnt hard but the result seemed weird to me so i thouht i was doing something wrong.
    – Milan Stojanovic
    Aug 26 at 10:33

















up vote
2
down vote













There is no way you can achieve a probability of 100% for this problem. Imagine that A is born on January 1st, but, being very unlucky, all the other persons are born on January 2nd.






share|cite|improve this answer



























    up vote
    1
    down vote













    Yes, you are correct in saying that we can never get $100 $ percent probability of having someone with the same birthday as person A.






    share|cite|improve this answer




















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      5
      down vote



      accepted










      You are correct.



      Given person $A$'s birthday, for each other person there is a probability of $364/365$ (ignoring leap years) that they do not share a birthday with $A$.



      Therefore if there are $n$ others, then there is a probability of $(364/365)^n$ that none share a birthday with $A$, hence a probability of $1-(364/365)^n<1$ that at least one does share a birthday with $A$.






      share|cite|improve this answer
















      • 2




        Thanks. The calculation wasnt hard but the result seemed weird to me so i thouht i was doing something wrong.
        – Milan Stojanovic
        Aug 26 at 10:33














      up vote
      5
      down vote



      accepted










      You are correct.



      Given person $A$'s birthday, for each other person there is a probability of $364/365$ (ignoring leap years) that they do not share a birthday with $A$.



      Therefore if there are $n$ others, then there is a probability of $(364/365)^n$ that none share a birthday with $A$, hence a probability of $1-(364/365)^n<1$ that at least one does share a birthday with $A$.






      share|cite|improve this answer
















      • 2




        Thanks. The calculation wasnt hard but the result seemed weird to me so i thouht i was doing something wrong.
        – Milan Stojanovic
        Aug 26 at 10:33












      up vote
      5
      down vote



      accepted







      up vote
      5
      down vote



      accepted






      You are correct.



      Given person $A$'s birthday, for each other person there is a probability of $364/365$ (ignoring leap years) that they do not share a birthday with $A$.



      Therefore if there are $n$ others, then there is a probability of $(364/365)^n$ that none share a birthday with $A$, hence a probability of $1-(364/365)^n<1$ that at least one does share a birthday with $A$.






      share|cite|improve this answer












      You are correct.



      Given person $A$'s birthday, for each other person there is a probability of $364/365$ (ignoring leap years) that they do not share a birthday with $A$.



      Therefore if there are $n$ others, then there is a probability of $(364/365)^n$ that none share a birthday with $A$, hence a probability of $1-(364/365)^n<1$ that at least one does share a birthday with $A$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Aug 26 at 10:29









      Robert Chamberlain

      3,8521421




      3,8521421







      • 2




        Thanks. The calculation wasnt hard but the result seemed weird to me so i thouht i was doing something wrong.
        – Milan Stojanovic
        Aug 26 at 10:33












      • 2




        Thanks. The calculation wasnt hard but the result seemed weird to me so i thouht i was doing something wrong.
        – Milan Stojanovic
        Aug 26 at 10:33







      2




      2




      Thanks. The calculation wasnt hard but the result seemed weird to me so i thouht i was doing something wrong.
      – Milan Stojanovic
      Aug 26 at 10:33




      Thanks. The calculation wasnt hard but the result seemed weird to me so i thouht i was doing something wrong.
      – Milan Stojanovic
      Aug 26 at 10:33










      up vote
      2
      down vote













      There is no way you can achieve a probability of 100% for this problem. Imagine that A is born on January 1st, but, being very unlucky, all the other persons are born on January 2nd.






      share|cite|improve this answer
























        up vote
        2
        down vote













        There is no way you can achieve a probability of 100% for this problem. Imagine that A is born on January 1st, but, being very unlucky, all the other persons are born on January 2nd.






        share|cite|improve this answer






















          up vote
          2
          down vote










          up vote
          2
          down vote









          There is no way you can achieve a probability of 100% for this problem. Imagine that A is born on January 1st, but, being very unlucky, all the other persons are born on January 2nd.






          share|cite|improve this answer












          There is no way you can achieve a probability of 100% for this problem. Imagine that A is born on January 1st, but, being very unlucky, all the other persons are born on January 2nd.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 26 at 10:32









          J.-E. Pin

          17.3k21753




          17.3k21753




















              up vote
              1
              down vote













              Yes, you are correct in saying that we can never get $100 $ percent probability of having someone with the same birthday as person A.






              share|cite|improve this answer
























                up vote
                1
                down vote













                Yes, you are correct in saying that we can never get $100 $ percent probability of having someone with the same birthday as person A.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Yes, you are correct in saying that we can never get $100 $ percent probability of having someone with the same birthday as person A.






                  share|cite|improve this answer












                  Yes, you are correct in saying that we can never get $100 $ percent probability of having someone with the same birthday as person A.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 26 at 10:31









                  Mohammad Riazi-Kermani

                  30.5k41852




                  30.5k41852



























                       

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