Percentage of 2 different Sets of Birthday Out of 14 People











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I am trying to calculate the odds of in a group of 14 people, the percentage of there being 2 different birthday matches (2 people having one Date of Birth, then another 2 people having a different Date of Birth).



I know the odds are about 20% of 2 people out of 14 having the same birthday.



So what would the odds be of there being 2 different birthday matches out of a group of 14?






probability birthday







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    Do you want the probability of precisely two people for each of both dates of birth, or of at least two people for each of both dates of birth?
    – Servaes
    Nov 16 at 23:51












  • The probability for precisely two people for each of both dates of birth
    – Jenna Terral
    Nov 17 at 3:51















up vote
2
down vote

favorite
1












I am trying to calculate the odds of in a group of 14 people, the percentage of there being 2 different birthday matches (2 people having one Date of Birth, then another 2 people having a different Date of Birth).



I know the odds are about 20% of 2 people out of 14 having the same birthday.



So what would the odds be of there being 2 different birthday matches out of a group of 14?






probability birthday







share|cite|improve this question


















  • 1




    Do you want the probability of precisely two people for each of both dates of birth, or of at least two people for each of both dates of birth?
    – Servaes
    Nov 16 at 23:51












  • The probability for precisely two people for each of both dates of birth
    – Jenna Terral
    Nov 17 at 3:51













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I am trying to calculate the odds of in a group of 14 people, the percentage of there being 2 different birthday matches (2 people having one Date of Birth, then another 2 people having a different Date of Birth).



I know the odds are about 20% of 2 people out of 14 having the same birthday.



So what would the odds be of there being 2 different birthday matches out of a group of 14?






probability birthday







share|cite|improve this question













I am trying to calculate the odds of in a group of 14 people, the percentage of there being 2 different birthday matches (2 people having one Date of Birth, then another 2 people having a different Date of Birth).



I know the odds are about 20% of 2 people out of 14 having the same birthday.



So what would the odds be of there being 2 different birthday matches out of a group of 14?






probability birthday




probability birthday






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 16 at 23:17









Jenna Terral

212




212








  • 1




    Do you want the probability of precisely two people for each of both dates of birth, or of at least two people for each of both dates of birth?
    – Servaes
    Nov 16 at 23:51












  • The probability for precisely two people for each of both dates of birth
    – Jenna Terral
    Nov 17 at 3:51














  • 1




    Do you want the probability of precisely two people for each of both dates of birth, or of at least two people for each of both dates of birth?
    – Servaes
    Nov 16 at 23:51












  • The probability for precisely two people for each of both dates of birth
    – Jenna Terral
    Nov 17 at 3:51








1




1




Do you want the probability of precisely two people for each of both dates of birth, or of at least two people for each of both dates of birth?
– Servaes
Nov 16 at 23:51






Do you want the probability of precisely two people for each of both dates of birth, or of at least two people for each of both dates of birth?
– Servaes
Nov 16 at 23:51














The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 at 3:51




The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 at 3:51










1 Answer
1






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1
down vote













We make the usual assumption that birthdays are uniformly distributed in a 365-day year.



The number of ways to assign birthdays to people such that there are exactly two pairs is the product of




  • the number of ways to choose the two duplicated birthdays: $binom{365}2$

  • the number of ways to assign the earlier birthday to two people, and the later birthday to two other people: $binom{14}2binom{12}2$

  • the number of ways to assign 10 out of the remaining 363 days to the rest of the people: $frac{363!}{353!}$


Dividing this product by the number of ways to distribute birthdays without restrictions $365^{14}$ yields the final probability of around $0.018776$, or 1.9%.






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  • The answer is 4.97% per my boss. But I am not sure how to get to that conclusion
    – Jenna Terral
    Nov 28 at 18:09











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up vote
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We make the usual assumption that birthdays are uniformly distributed in a 365-day year.



The number of ways to assign birthdays to people such that there are exactly two pairs is the product of




  • the number of ways to choose the two duplicated birthdays: $binom{365}2$

  • the number of ways to assign the earlier birthday to two people, and the later birthday to two other people: $binom{14}2binom{12}2$

  • the number of ways to assign 10 out of the remaining 363 days to the rest of the people: $frac{363!}{353!}$


Dividing this product by the number of ways to distribute birthdays without restrictions $365^{14}$ yields the final probability of around $0.018776$, or 1.9%.






share|cite|improve this answer





















  • The answer is 4.97% per my boss. But I am not sure how to get to that conclusion
    – Jenna Terral
    Nov 28 at 18:09















up vote
1
down vote













We make the usual assumption that birthdays are uniformly distributed in a 365-day year.



The number of ways to assign birthdays to people such that there are exactly two pairs is the product of




  • the number of ways to choose the two duplicated birthdays: $binom{365}2$

  • the number of ways to assign the earlier birthday to two people, and the later birthday to two other people: $binom{14}2binom{12}2$

  • the number of ways to assign 10 out of the remaining 363 days to the rest of the people: $frac{363!}{353!}$


Dividing this product by the number of ways to distribute birthdays without restrictions $365^{14}$ yields the final probability of around $0.018776$, or 1.9%.






share|cite|improve this answer





















  • The answer is 4.97% per my boss. But I am not sure how to get to that conclusion
    – Jenna Terral
    Nov 28 at 18:09













up vote
1
down vote










up vote
1
down vote









We make the usual assumption that birthdays are uniformly distributed in a 365-day year.



The number of ways to assign birthdays to people such that there are exactly two pairs is the product of




  • the number of ways to choose the two duplicated birthdays: $binom{365}2$

  • the number of ways to assign the earlier birthday to two people, and the later birthday to two other people: $binom{14}2binom{12}2$

  • the number of ways to assign 10 out of the remaining 363 days to the rest of the people: $frac{363!}{353!}$


Dividing this product by the number of ways to distribute birthdays without restrictions $365^{14}$ yields the final probability of around $0.018776$, or 1.9%.






share|cite|improve this answer












We make the usual assumption that birthdays are uniformly distributed in a 365-day year.



The number of ways to assign birthdays to people such that there are exactly two pairs is the product of




  • the number of ways to choose the two duplicated birthdays: $binom{365}2$

  • the number of ways to assign the earlier birthday to two people, and the later birthday to two other people: $binom{14}2binom{12}2$

  • the number of ways to assign 10 out of the remaining 363 days to the rest of the people: $frac{363!}{353!}$


Dividing this product by the number of ways to distribute birthdays without restrictions $365^{14}$ yields the final probability of around $0.018776$, or 1.9%.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 17 at 4:58









Parcly Taxel

41.2k137199




41.2k137199












  • The answer is 4.97% per my boss. But I am not sure how to get to that conclusion
    – Jenna Terral
    Nov 28 at 18:09


















  • The answer is 4.97% per my boss. But I am not sure how to get to that conclusion
    – Jenna Terral
    Nov 28 at 18:09
















The answer is 4.97% per my boss. But I am not sure how to get to that conclusion
– Jenna Terral
Nov 28 at 18:09




The answer is 4.97% per my boss. But I am not sure how to get to that conclusion
– Jenna Terral
Nov 28 at 18:09


















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