Probability that two randomly chosen people out of 10 have birthday in the same month











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There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?




My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility






probability combinatorics







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    Your intuition is correct, well at least if you make the usual assumptions (every month equally likely to be a birth month, say).
    – lulu
    Nov 13 at 17:23















up vote
0
down vote

favorite













There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?




My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility






probability combinatorics







share|cite|improve this question


















  • 1




    Your intuition is correct, well at least if you make the usual assumptions (every month equally likely to be a birth month, say).
    – lulu
    Nov 13 at 17:23













up vote
0
down vote

favorite









up vote
0
down vote

favorite












There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?




My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility






probability combinatorics







share|cite|improve this question














There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?




My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility






probability combinatorics




probability combinatorics






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share|cite|improve this question











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asked Nov 13 at 17:21









Aemilius

1,664314




1,664314








  • 1




    Your intuition is correct, well at least if you make the usual assumptions (every month equally likely to be a birth month, say).
    – lulu
    Nov 13 at 17:23














  • 1




    Your intuition is correct, well at least if you make the usual assumptions (every month equally likely to be a birth month, say).
    – lulu
    Nov 13 at 17:23








1




1




Your intuition is correct, well at least if you make the usual assumptions (every month equally likely to be a birth month, say).
– lulu
Nov 13 at 17:23




Your intuition is correct, well at least if you make the usual assumptions (every month equally likely to be a birth month, say).
– lulu
Nov 13 at 17:23










1 Answer
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Since there are only 12 possible birth months, we have a 100% chance of some two birth months being the same for a group of 13 people or more by The Pigeonhole Principle.



For groups of less than 13, the analysis can go as follows. What are the chances that 2 people don't have the same birth month? 11/12 yielding a probability of 1/12 that they do.



There are 12 ways of picking a birthday for 1 person. There remain only 11 ways to pick a a second person so that you don't have matching birth months. There's a total of 12^2 =144 ways to pick two different birth months.



In general the formula is: $P(k)=frac{12!}{(12-k)!}frac{1}{12^k}$ where $k$



is the number of people in the group. Where $P(k)$ is the probability of not having a matching pair. So the probability of there being a matching pair is $1-P(k).$



So for 10 people the probability of two having the same birth month is 99.6132%.



This assumes all months have the same number of days and all months have the same chances of people being born in them. Both assumptions are slightly off.



The chances hit over 50% for any group over 4 members large.



If $k=2$, $P(2)=12*11/144$, and $1-P(2)=1/12$.



So prob of 2 people no matter what sized group they are chosen from is 1/12 by the same analysis.






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  • 1




    I think the question being asked is a little bit different. The question is what is the probability any two randomly chosen people have the same birth month. So given that person 1 has birth month A, there is a $frac{1}{12}$ chance that person B has the same birth month (assuming for simplicity that having a birthday in any month is equally likely).
    – Jack Moody
    Nov 13 at 19:35











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Since there are only 12 possible birth months, we have a 100% chance of some two birth months being the same for a group of 13 people or more by The Pigeonhole Principle.



For groups of less than 13, the analysis can go as follows. What are the chances that 2 people don't have the same birth month? 11/12 yielding a probability of 1/12 that they do.



There are 12 ways of picking a birthday for 1 person. There remain only 11 ways to pick a a second person so that you don't have matching birth months. There's a total of 12^2 =144 ways to pick two different birth months.



In general the formula is: $P(k)=frac{12!}{(12-k)!}frac{1}{12^k}$ where $k$



is the number of people in the group. Where $P(k)$ is the probability of not having a matching pair. So the probability of there being a matching pair is $1-P(k).$



So for 10 people the probability of two having the same birth month is 99.6132%.



This assumes all months have the same number of days and all months have the same chances of people being born in them. Both assumptions are slightly off.



The chances hit over 50% for any group over 4 members large.



If $k=2$, $P(2)=12*11/144$, and $1-P(2)=1/12$.



So prob of 2 people no matter what sized group they are chosen from is 1/12 by the same analysis.






share|cite|improve this answer



















  • 1




    I think the question being asked is a little bit different. The question is what is the probability any two randomly chosen people have the same birth month. So given that person 1 has birth month A, there is a $frac{1}{12}$ chance that person B has the same birth month (assuming for simplicity that having a birthday in any month is equally likely).
    – Jack Moody
    Nov 13 at 19:35















up vote
0
down vote













Since there are only 12 possible birth months, we have a 100% chance of some two birth months being the same for a group of 13 people or more by The Pigeonhole Principle.



For groups of less than 13, the analysis can go as follows. What are the chances that 2 people don't have the same birth month? 11/12 yielding a probability of 1/12 that they do.



There are 12 ways of picking a birthday for 1 person. There remain only 11 ways to pick a a second person so that you don't have matching birth months. There's a total of 12^2 =144 ways to pick two different birth months.



In general the formula is: $P(k)=frac{12!}{(12-k)!}frac{1}{12^k}$ where $k$



is the number of people in the group. Where $P(k)$ is the probability of not having a matching pair. So the probability of there being a matching pair is $1-P(k).$



So for 10 people the probability of two having the same birth month is 99.6132%.



This assumes all months have the same number of days and all months have the same chances of people being born in them. Both assumptions are slightly off.



The chances hit over 50% for any group over 4 members large.



If $k=2$, $P(2)=12*11/144$, and $1-P(2)=1/12$.



So prob of 2 people no matter what sized group they are chosen from is 1/12 by the same analysis.






share|cite|improve this answer



















  • 1




    I think the question being asked is a little bit different. The question is what is the probability any two randomly chosen people have the same birth month. So given that person 1 has birth month A, there is a $frac{1}{12}$ chance that person B has the same birth month (assuming for simplicity that having a birthday in any month is equally likely).
    – Jack Moody
    Nov 13 at 19:35













up vote
0
down vote










up vote
0
down vote









Since there are only 12 possible birth months, we have a 100% chance of some two birth months being the same for a group of 13 people or more by The Pigeonhole Principle.



For groups of less than 13, the analysis can go as follows. What are the chances that 2 people don't have the same birth month? 11/12 yielding a probability of 1/12 that they do.



There are 12 ways of picking a birthday for 1 person. There remain only 11 ways to pick a a second person so that you don't have matching birth months. There's a total of 12^2 =144 ways to pick two different birth months.



In general the formula is: $P(k)=frac{12!}{(12-k)!}frac{1}{12^k}$ where $k$



is the number of people in the group. Where $P(k)$ is the probability of not having a matching pair. So the probability of there being a matching pair is $1-P(k).$



So for 10 people the probability of two having the same birth month is 99.6132%.



This assumes all months have the same number of days and all months have the same chances of people being born in them. Both assumptions are slightly off.



The chances hit over 50% for any group over 4 members large.



If $k=2$, $P(2)=12*11/144$, and $1-P(2)=1/12$.



So prob of 2 people no matter what sized group they are chosen from is 1/12 by the same analysis.






share|cite|improve this answer














Since there are only 12 possible birth months, we have a 100% chance of some two birth months being the same for a group of 13 people or more by The Pigeonhole Principle.



For groups of less than 13, the analysis can go as follows. What are the chances that 2 people don't have the same birth month? 11/12 yielding a probability of 1/12 that they do.



There are 12 ways of picking a birthday for 1 person. There remain only 11 ways to pick a a second person so that you don't have matching birth months. There's a total of 12^2 =144 ways to pick two different birth months.



In general the formula is: $P(k)=frac{12!}{(12-k)!}frac{1}{12^k}$ where $k$



is the number of people in the group. Where $P(k)$ is the probability of not having a matching pair. So the probability of there being a matching pair is $1-P(k).$



So for 10 people the probability of two having the same birth month is 99.6132%.



This assumes all months have the same number of days and all months have the same chances of people being born in them. Both assumptions are slightly off.



The chances hit over 50% for any group over 4 members large.



If $k=2$, $P(2)=12*11/144$, and $1-P(2)=1/12$.



So prob of 2 people no matter what sized group they are chosen from is 1/12 by the same analysis.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 13 at 19:53

























answered Nov 13 at 18:44









TurlocTheRed

60819




60819








  • 1




    I think the question being asked is a little bit different. The question is what is the probability any two randomly chosen people have the same birth month. So given that person 1 has birth month A, there is a $frac{1}{12}$ chance that person B has the same birth month (assuming for simplicity that having a birthday in any month is equally likely).
    – Jack Moody
    Nov 13 at 19:35














  • 1




    I think the question being asked is a little bit different. The question is what is the probability any two randomly chosen people have the same birth month. So given that person 1 has birth month A, there is a $frac{1}{12}$ chance that person B has the same birth month (assuming for simplicity that having a birthday in any month is equally likely).
    – Jack Moody
    Nov 13 at 19:35








1




1




I think the question being asked is a little bit different. The question is what is the probability any two randomly chosen people have the same birth month. So given that person 1 has birth month A, there is a $frac{1}{12}$ chance that person B has the same birth month (assuming for simplicity that having a birthday in any month is equally likely).
– Jack Moody
Nov 13 at 19:35




I think the question being asked is a little bit different. The question is what is the probability any two randomly chosen people have the same birth month. So given that person 1 has birth month A, there is a $frac{1}{12}$ chance that person B has the same birth month (assuming for simplicity that having a birthday in any month is equally likely).
– Jack Moody
Nov 13 at 19:35


















 

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