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
asked Nov 16 at 23:17
Jenna Terral
212
add a comment |
up vote
2
down vote
favorite
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
asked Nov 16 at 23:17
Jenna Terral
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
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
asked Nov 16 at 23:17
Jenna Terral
212
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
asked Nov 16 at 23:17
Jenna Terral
212
asked Nov 16 at 23:17
Jenna Terral
212
asked Nov 16 at 23:17
Jenna Terral
212
asked Nov 16 at 23:17
Jenna Terral
212
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
add a comment |
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
add a comment |
1 Answer
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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%.
answered Nov 17 at 4:58
Parcly Taxel
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
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
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%.
answered Nov 17 at 4:58
Parcly Taxel
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
add a comment |
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%.
answered Nov 17 at 4:58
Parcly Taxel
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
add a comment |
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%.
answered Nov 17 at 4:58
Parcly Taxel
41.2k137199
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%.
answered Nov 17 at 4:58
Parcly Taxel
41.2k137199
answered Nov 17 at 4:58
Parcly Taxel
41.2k137199
answered Nov 17 at 4:58
Parcly Taxel
41.2k137199
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
add a comment |
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
add a comment |
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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