Percentage of 2 different Sets of Birthday Out of 14 People
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 '18 at 23:17
Jenna Terral
212
add a comment |
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 '18 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 '18 at 23:51
The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 '18 at 3:51
add a comment |
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 '18 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 '18 at 23:17
Jenna Terral
212
asked Nov 16 '18 at 23:17
Jenna Terral
212
asked Nov 16 '18 at 23:17
Jenna Terral
212
asked Nov 16 '18 at 23:17
Jenna Terral
212
asked Nov 16 '18 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 '18 at 23:51
The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 '18 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 '18 at 23:51
The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 '18 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 '18 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 '18 at 23:51
The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 '18 at 3:51
The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 '18 at 3:51
add a comment |
1 Answer
1
active
oldest
votes
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 '18 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 '18 at 18:09
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001761%2fpercentage-of-2-different-sets-of-birthday-out-of-14-people%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 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 '18 at 18:09
add a comment |
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 '18 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 '18 at 18:09
add a comment |
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 '18 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 '18 at 4:58
Parcly Taxel
41.2k137199
answered Nov 17 '18 at 4:58
Parcly Taxel
41.2k137199
answered Nov 17 '18 at 4:58
Parcly Taxel
41.2k137199
answered Nov 17 '18 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 '18 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 '18 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 '18 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 '18 at 18:09
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001761%2fpercentage-of-2-different-sets-of-birthday-out-of-14-people%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 '18 at 23:51
The probability for precisely two people for each of both dates of birth
– Jenna Terral
Nov 17 '18 at 3:51