What is the probability somebody's birthday is the day before mine?












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What is the probability that someone's birthday is the day before my birthday? For example, my birthday is Feb 28, what is the probability that my mom's birthday is Feb 27? Is it just $frac1{365}$? That seems too simple to me but maybe I'm just complicating things unnecessarily.






probability statistics







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  • 6




    $begingroup$
    Well, obviously the probability of her her birthday being Feb. 27 is 1/365 no matter what your birthday is.
    $endgroup$
    – fleablood
    9 hours ago






  • 3




    $begingroup$
    What if your birthday is the 1st of March? Do the 29th and 28th of February both count then? So then there's the probability your birthday is the 1st of March, and the probability they're the 28th or 29th.
    $endgroup$
    – James
    8 hours ago








  • 3




    $begingroup$
    @fleablood The word "obviously" should be discouraged here. Although it was not your intention to be rude, it doesn't really help to say the answer is obvious in response to a question about that very thing.
    $endgroup$
    – 6005
    7 hours ago










  • $begingroup$
    @6005. The question was about the probability of a birthday the day before and the OP gave an example of "if my birthday is 2/28" and her brthday is 2/27". I was pointing out that the probability of her birthday being 2/27 is the same regardless of his birthday, hence the reason it isn't "too simplistic".
    $endgroup$
    – fleablood
    6 hours ago










  • $begingroup$
    @fleablood I know what you meant! But what I'm pointing out is that the word "obviously" is almost guaranteed to be interpreted differently than you intended. If I ask, "what is X?" and you answer, "obviously, Y", I am likely to interpret that my question was stupid -- I should have known the answer myself.
    $endgroup$
    – 6005
    6 hours ago


















8












$begingroup$


What is the probability that someone's birthday is the day before my birthday? For example, my birthday is Feb 28, what is the probability that my mom's birthday is Feb 27? Is it just $frac1{365}$? That seems too simple to me but maybe I'm just complicating things unnecessarily.






probability statistics







share|cite|improve this question









New contributor




Alana Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 6




    $begingroup$
    Well, obviously the probability of her her birthday being Feb. 27 is 1/365 no matter what your birthday is.
    $endgroup$
    – fleablood
    9 hours ago






  • 3




    $begingroup$
    What if your birthday is the 1st of March? Do the 29th and 28th of February both count then? So then there's the probability your birthday is the 1st of March, and the probability they're the 28th or 29th.
    $endgroup$
    – James
    8 hours ago








  • 3




    $begingroup$
    @fleablood The word "obviously" should be discouraged here. Although it was not your intention to be rude, it doesn't really help to say the answer is obvious in response to a question about that very thing.
    $endgroup$
    – 6005
    7 hours ago










  • $begingroup$
    @6005. The question was about the probability of a birthday the day before and the OP gave an example of "if my birthday is 2/28" and her brthday is 2/27". I was pointing out that the probability of her birthday being 2/27 is the same regardless of his birthday, hence the reason it isn't "too simplistic".
    $endgroup$
    – fleablood
    6 hours ago










  • $begingroup$
    @fleablood I know what you meant! But what I'm pointing out is that the word "obviously" is almost guaranteed to be interpreted differently than you intended. If I ask, "what is X?" and you answer, "obviously, Y", I am likely to interpret that my question was stupid -- I should have known the answer myself.
    $endgroup$
    – 6005
    6 hours ago
















8












8








8


1



$begingroup$


What is the probability that someone's birthday is the day before my birthday? For example, my birthday is Feb 28, what is the probability that my mom's birthday is Feb 27? Is it just $frac1{365}$? That seems too simple to me but maybe I'm just complicating things unnecessarily.






probability statistics







share|cite|improve this question









New contributor




Alana Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




What is the probability that someone's birthday is the day before my birthday? For example, my birthday is Feb 28, what is the probability that my mom's birthday is Feb 27? Is it just $frac1{365}$? That seems too simple to me but maybe I'm just complicating things unnecessarily.






probability statistics




probability statistics






share|cite|improve this question









New contributor




Alana Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Alana Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









Parcly Taxel

44.1k1375105




44.1k1375105






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asked 10 hours ago









Alana DuffyAlana Duffy

442




442




New contributor




Alana Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Alana Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alana Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 6




    $begingroup$
    Well, obviously the probability of her her birthday being Feb. 27 is 1/365 no matter what your birthday is.
    $endgroup$
    – fleablood
    9 hours ago






  • 3




    $begingroup$
    What if your birthday is the 1st of March? Do the 29th and 28th of February both count then? So then there's the probability your birthday is the 1st of March, and the probability they're the 28th or 29th.
    $endgroup$
    – James
    8 hours ago








  • 3




    $begingroup$
    @fleablood The word "obviously" should be discouraged here. Although it was not your intention to be rude, it doesn't really help to say the answer is obvious in response to a question about that very thing.
    $endgroup$
    – 6005
    7 hours ago










  • $begingroup$
    @6005. The question was about the probability of a birthday the day before and the OP gave an example of "if my birthday is 2/28" and her brthday is 2/27". I was pointing out that the probability of her birthday being 2/27 is the same regardless of his birthday, hence the reason it isn't "too simplistic".
    $endgroup$
    – fleablood
    6 hours ago










  • $begingroup$
    @fleablood I know what you meant! But what I'm pointing out is that the word "obviously" is almost guaranteed to be interpreted differently than you intended. If I ask, "what is X?" and you answer, "obviously, Y", I am likely to interpret that my question was stupid -- I should have known the answer myself.
    $endgroup$
    – 6005
    6 hours ago
















  • 6




    $begingroup$
    Well, obviously the probability of her her birthday being Feb. 27 is 1/365 no matter what your birthday is.
    $endgroup$
    – fleablood
    9 hours ago






  • 3




    $begingroup$
    What if your birthday is the 1st of March? Do the 29th and 28th of February both count then? So then there's the probability your birthday is the 1st of March, and the probability they're the 28th or 29th.
    $endgroup$
    – James
    8 hours ago








  • 3




    $begingroup$
    @fleablood The word "obviously" should be discouraged here. Although it was not your intention to be rude, it doesn't really help to say the answer is obvious in response to a question about that very thing.
    $endgroup$
    – 6005
    7 hours ago










  • $begingroup$
    @6005. The question was about the probability of a birthday the day before and the OP gave an example of "if my birthday is 2/28" and her brthday is 2/27". I was pointing out that the probability of her birthday being 2/27 is the same regardless of his birthday, hence the reason it isn't "too simplistic".
    $endgroup$
    – fleablood
    6 hours ago










  • $begingroup$
    @fleablood I know what you meant! But what I'm pointing out is that the word "obviously" is almost guaranteed to be interpreted differently than you intended. If I ask, "what is X?" and you answer, "obviously, Y", I am likely to interpret that my question was stupid -- I should have known the answer myself.
    $endgroup$
    – 6005
    6 hours ago










6




6




$begingroup$
Well, obviously the probability of her her birthday being Feb. 27 is 1/365 no matter what your birthday is.
$endgroup$
– fleablood
9 hours ago




$begingroup$
Well, obviously the probability of her her birthday being Feb. 27 is 1/365 no matter what your birthday is.
$endgroup$
– fleablood
9 hours ago




3




3




$begingroup$
What if your birthday is the 1st of March? Do the 29th and 28th of February both count then? So then there's the probability your birthday is the 1st of March, and the probability they're the 28th or 29th.
$endgroup$
– James
8 hours ago






$begingroup$
What if your birthday is the 1st of March? Do the 29th and 28th of February both count then? So then there's the probability your birthday is the 1st of March, and the probability they're the 28th or 29th.
$endgroup$
– James
8 hours ago






3




3




$begingroup$
@fleablood The word "obviously" should be discouraged here. Although it was not your intention to be rude, it doesn't really help to say the answer is obvious in response to a question about that very thing.
$endgroup$
– 6005
7 hours ago




$begingroup$
@fleablood The word "obviously" should be discouraged here. Although it was not your intention to be rude, it doesn't really help to say the answer is obvious in response to a question about that very thing.
$endgroup$
– 6005
7 hours ago












$begingroup$
@6005. The question was about the probability of a birthday the day before and the OP gave an example of "if my birthday is 2/28" and her brthday is 2/27". I was pointing out that the probability of her birthday being 2/27 is the same regardless of his birthday, hence the reason it isn't "too simplistic".
$endgroup$
– fleablood
6 hours ago




$begingroup$
@6005. The question was about the probability of a birthday the day before and the OP gave an example of "if my birthday is 2/28" and her brthday is 2/27". I was pointing out that the probability of her birthday being 2/27 is the same regardless of his birthday, hence the reason it isn't "too simplistic".
$endgroup$
– fleablood
6 hours ago












$begingroup$
@fleablood I know what you meant! But what I'm pointing out is that the word "obviously" is almost guaranteed to be interpreted differently than you intended. If I ask, "what is X?" and you answer, "obviously, Y", I am likely to interpret that my question was stupid -- I should have known the answer myself.
$endgroup$
– 6005
6 hours ago






$begingroup$
@fleablood I know what you meant! But what I'm pointing out is that the word "obviously" is almost guaranteed to be interpreted differently than you intended. If I ask, "what is X?" and you answer, "obviously, Y", I am likely to interpret that my question was stupid -- I should have known the answer myself.
$endgroup$
– 6005
6 hours ago












3 Answers
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oldest

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$begingroup$

Assuming a uniform distribution of birthdays throughout a 365-day year, and that "somebody" means "a fixed person chosen at random from the human population", $frac1{365}$ is correct, since there is only one day that is immediately before a given day. If "somebody" is a person we can choose arbitrarily, the answer is $1$ since at least one person has been born on each day of the year, including 29 February.



In reality more people are born in the summer months, so the distribution of birthdays is variable and not at all uniform.




When considering 29 February as well as the usual 365 days in the first formulation of "somebody", the year also needs to be known in order to completely determine which day is "the day before my birthday". Once again, assume birthdays are uniformly distributed by day (not date). Since there are 97 leap years in the 400-year Gregorian cycle, the probability that a randomly chosen person has a specified birthday is $frac{400}{400cdot365+97}$ if that birthday is not 29 February and $frac{97}{400cdot365+97}$ otherwise.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29.
    $endgroup$
    – Chase Ryan Taylor
    9 hours ago












  • $begingroup$
    @ChaseRyanTaylor That'd just be making a distribution of all birthdays on the planet with 29 February taken into account - theoretically possible, since only a finite number of people have been born - and then, given your birthday, look at the day before it and look up the corresponding probability.
    $endgroup$
    – Parcly Taxel
    9 hours ago










  • $begingroup$
    Wouldn't that just be 1/365.25 ? I mean approximately, obviously the actual length of a year is slightly off and we cover it with the 100 yr. and 400 yr. rules for leap years. Mind you, those exceptions being as far apart as they are, they're longer than the average human lifespan, so assuming two people who are both alive at the same time, 1/365.25 is probably good enough unless you're within one lifespan a century boundary that was not leap year.
    $endgroup$
    – Darrel Hoffman
    8 hours ago










  • $begingroup$
    @DarrelHoffman I said 365-day year.
    $endgroup$
    – Parcly Taxel
    8 hours ago






  • 2




    $begingroup$
    "It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29." There's no accounting for taste but I'd have utterly no interest in that. It's enough to know some days are more common that others and if you told me summer dates are more common I'd take you word for it, but the exact data and numbers is of absolutely no interest to me other than theoretical.
    $endgroup$
    – fleablood
    8 hours ago



















4












$begingroup$

If you want to make it hard:



There are $365$ choices her birthday (H) could be, and $365$ choices yours(Y) could be. So there are are $365^2$ pairs of birthdays (H,Y).



Of those $365^2$ pairs there are $365$ cases where $Y = H+1$. (Case 1: Y= Jan. 1-H=Dec 31. Case 2: Y= Jan. 2-H= Jan 1. Case 3: Y=Jan. 3-H= Jan. 2..... Case 365: Y= Dec. 31, H= Dec. 30$.



So the probability is $frac {365}{365^2} = frac 1{365}$.



If you want to do it the easy way.



There are $365$ days she could be born. The day before your birthday is precisely one of them and as likely as any other. So probability is $frac 1{365}$.



.....



As for your birthday being Feb. 28.... No matter what your birthday is, the probability of her birthday being Feb. 27 is $frac 1{365}$.



...



Finally your title is "What is the probability somebody's birthday is the day before mine?"



The answer to that is 100%. There are tens of millions of people with birthdays the day before you.



.....



" That seems too simple to me" Why? Why should it be complicated?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If you want to make it really hard (and not assume any knowledge about people's birthdays), the probability of somebody's birthday being the day before the asker's is not 100%. I asked that very question a few years ago, and the answer for 7 billion people would be extremely close to 100%, but not quite.
    $endgroup$
    – maxb
    9 hours ago










  • $begingroup$
    @maxb I thought about that but didn't want to confuse the issue. It is, of course, $1 - (frac {364.25}{365.25})^{7,000,000,000}$ which looks significantly less than $1$ than it actually is. Thing is, for there not to be anybody having a birthday the day before there must be a date in which no-one on the planet was born and no one was born on that date in any year. I think the statement "Every day millions of people are born" is a safe assumption in a "Junior Jumbo Book of Facts" way.
    $endgroup$
    – fleablood
    8 hours ago






  • 1




    $begingroup$
    Once we get into real-world questions, probabilities become much more complicated. The fraction $left(frac{364.25}{365.25}right)^{7,000,000,000}$ is much smaller than any actual uncertainty you deserve to have about pretty much anything: there is an incredibly tiny but much higher probability that for some mysterious reason you're not aware of nobody is ever born on March 12th.
    $endgroup$
    – Misha Lavrov
    6 hours ago












  • $begingroup$
    @fleablood ""Every day millions of people are born" isn't a safe assumption at all. For a population of 7bn, ignoring population growth, and assuming a global life expectancy of 50 years, there are only about 400,000 births per day. You would need make some creative assumptions to change that estimate to "millions" (plural).
    $endgroup$
    – alephzero
    5 hours ago












  • $begingroup$
    ... or be off be a factor of 50.....
    $endgroup$
    – fleablood
    2 hours ago



















0












$begingroup$

Considering February 29 I say it is



1 in 365.25



as every four year we have an extra day.
The chance of another person being born on the day before is equally big as any other day of the year (except for February 29).






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

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    active

    oldest

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    13












    $begingroup$

    Assuming a uniform distribution of birthdays throughout a 365-day year, and that "somebody" means "a fixed person chosen at random from the human population", $frac1{365}$ is correct, since there is only one day that is immediately before a given day. If "somebody" is a person we can choose arbitrarily, the answer is $1$ since at least one person has been born on each day of the year, including 29 February.



    In reality more people are born in the summer months, so the distribution of birthdays is variable and not at all uniform.




    When considering 29 February as well as the usual 365 days in the first formulation of "somebody", the year also needs to be known in order to completely determine which day is "the day before my birthday". Once again, assume birthdays are uniformly distributed by day (not date). Since there are 97 leap years in the 400-year Gregorian cycle, the probability that a randomly chosen person has a specified birthday is $frac{400}{400cdot365+97}$ if that birthday is not 29 February and $frac{97}{400cdot365+97}$ otherwise.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29.
      $endgroup$
      – Chase Ryan Taylor
      9 hours ago












    • $begingroup$
      @ChaseRyanTaylor That'd just be making a distribution of all birthdays on the planet with 29 February taken into account - theoretically possible, since only a finite number of people have been born - and then, given your birthday, look at the day before it and look up the corresponding probability.
      $endgroup$
      – Parcly Taxel
      9 hours ago










    • $begingroup$
      Wouldn't that just be 1/365.25 ? I mean approximately, obviously the actual length of a year is slightly off and we cover it with the 100 yr. and 400 yr. rules for leap years. Mind you, those exceptions being as far apart as they are, they're longer than the average human lifespan, so assuming two people who are both alive at the same time, 1/365.25 is probably good enough unless you're within one lifespan a century boundary that was not leap year.
      $endgroup$
      – Darrel Hoffman
      8 hours ago










    • $begingroup$
      @DarrelHoffman I said 365-day year.
      $endgroup$
      – Parcly Taxel
      8 hours ago






    • 2




      $begingroup$
      "It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29." There's no accounting for taste but I'd have utterly no interest in that. It's enough to know some days are more common that others and if you told me summer dates are more common I'd take you word for it, but the exact data and numbers is of absolutely no interest to me other than theoretical.
      $endgroup$
      – fleablood
      8 hours ago
















    13












    $begingroup$

    Assuming a uniform distribution of birthdays throughout a 365-day year, and that "somebody" means "a fixed person chosen at random from the human population", $frac1{365}$ is correct, since there is only one day that is immediately before a given day. If "somebody" is a person we can choose arbitrarily, the answer is $1$ since at least one person has been born on each day of the year, including 29 February.



    In reality more people are born in the summer months, so the distribution of birthdays is variable and not at all uniform.




    When considering 29 February as well as the usual 365 days in the first formulation of "somebody", the year also needs to be known in order to completely determine which day is "the day before my birthday". Once again, assume birthdays are uniformly distributed by day (not date). Since there are 97 leap years in the 400-year Gregorian cycle, the probability that a randomly chosen person has a specified birthday is $frac{400}{400cdot365+97}$ if that birthday is not 29 February and $frac{97}{400cdot365+97}$ otherwise.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29.
      $endgroup$
      – Chase Ryan Taylor
      9 hours ago












    • $begingroup$
      @ChaseRyanTaylor That'd just be making a distribution of all birthdays on the planet with 29 February taken into account - theoretically possible, since only a finite number of people have been born - and then, given your birthday, look at the day before it and look up the corresponding probability.
      $endgroup$
      – Parcly Taxel
      9 hours ago










    • $begingroup$
      Wouldn't that just be 1/365.25 ? I mean approximately, obviously the actual length of a year is slightly off and we cover it with the 100 yr. and 400 yr. rules for leap years. Mind you, those exceptions being as far apart as they are, they're longer than the average human lifespan, so assuming two people who are both alive at the same time, 1/365.25 is probably good enough unless you're within one lifespan a century boundary that was not leap year.
      $endgroup$
      – Darrel Hoffman
      8 hours ago










    • $begingroup$
      @DarrelHoffman I said 365-day year.
      $endgroup$
      – Parcly Taxel
      8 hours ago






    • 2




      $begingroup$
      "It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29." There's no accounting for taste but I'd have utterly no interest in that. It's enough to know some days are more common that others and if you told me summer dates are more common I'd take you word for it, but the exact data and numbers is of absolutely no interest to me other than theoretical.
      $endgroup$
      – fleablood
      8 hours ago














    13












    13








    13





    $begingroup$

    Assuming a uniform distribution of birthdays throughout a 365-day year, and that "somebody" means "a fixed person chosen at random from the human population", $frac1{365}$ is correct, since there is only one day that is immediately before a given day. If "somebody" is a person we can choose arbitrarily, the answer is $1$ since at least one person has been born on each day of the year, including 29 February.



    In reality more people are born in the summer months, so the distribution of birthdays is variable and not at all uniform.




    When considering 29 February as well as the usual 365 days in the first formulation of "somebody", the year also needs to be known in order to completely determine which day is "the day before my birthday". Once again, assume birthdays are uniformly distributed by day (not date). Since there are 97 leap years in the 400-year Gregorian cycle, the probability that a randomly chosen person has a specified birthday is $frac{400}{400cdot365+97}$ if that birthday is not 29 February and $frac{97}{400cdot365+97}$ otherwise.






    share|cite|improve this answer











    $endgroup$



    Assuming a uniform distribution of birthdays throughout a 365-day year, and that "somebody" means "a fixed person chosen at random from the human population", $frac1{365}$ is correct, since there is only one day that is immediately before a given day. If "somebody" is a person we can choose arbitrarily, the answer is $1$ since at least one person has been born on each day of the year, including 29 February.



    In reality more people are born in the summer months, so the distribution of birthdays is variable and not at all uniform.




    When considering 29 February as well as the usual 365 days in the first formulation of "somebody", the year also needs to be known in order to completely determine which day is "the day before my birthday". Once again, assume birthdays are uniformly distributed by day (not date). Since there are 97 leap years in the 400-year Gregorian cycle, the probability that a randomly chosen person has a specified birthday is $frac{400}{400cdot365+97}$ if that birthday is not 29 February and $frac{97}{400cdot365+97}$ otherwise.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 8 hours ago

























    answered 10 hours ago









    Parcly TaxelParcly Taxel

    44.1k1375105




    44.1k1375105












    • $begingroup$
      It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29.
      $endgroup$
      – Chase Ryan Taylor
      9 hours ago












    • $begingroup$
      @ChaseRyanTaylor That'd just be making a distribution of all birthdays on the planet with 29 February taken into account - theoretically possible, since only a finite number of people have been born - and then, given your birthday, look at the day before it and look up the corresponding probability.
      $endgroup$
      – Parcly Taxel
      9 hours ago










    • $begingroup$
      Wouldn't that just be 1/365.25 ? I mean approximately, obviously the actual length of a year is slightly off and we cover it with the 100 yr. and 400 yr. rules for leap years. Mind you, those exceptions being as far apart as they are, they're longer than the average human lifespan, so assuming two people who are both alive at the same time, 1/365.25 is probably good enough unless you're within one lifespan a century boundary that was not leap year.
      $endgroup$
      – Darrel Hoffman
      8 hours ago










    • $begingroup$
      @DarrelHoffman I said 365-day year.
      $endgroup$
      – Parcly Taxel
      8 hours ago






    • 2




      $begingroup$
      "It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29." There's no accounting for taste but I'd have utterly no interest in that. It's enough to know some days are more common that others and if you told me summer dates are more common I'd take you word for it, but the exact data and numbers is of absolutely no interest to me other than theoretical.
      $endgroup$
      – fleablood
      8 hours ago


















    • $begingroup$
      It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29.
      $endgroup$
      – Chase Ryan Taylor
      9 hours ago












    • $begingroup$
      @ChaseRyanTaylor That'd just be making a distribution of all birthdays on the planet with 29 February taken into account - theoretically possible, since only a finite number of people have been born - and then, given your birthday, look at the day before it and look up the corresponding probability.
      $endgroup$
      – Parcly Taxel
      9 hours ago










    • $begingroup$
      Wouldn't that just be 1/365.25 ? I mean approximately, obviously the actual length of a year is slightly off and we cover it with the 100 yr. and 400 yr. rules for leap years. Mind you, those exceptions being as far apart as they are, they're longer than the average human lifespan, so assuming two people who are both alive at the same time, 1/365.25 is probably good enough unless you're within one lifespan a century boundary that was not leap year.
      $endgroup$
      – Darrel Hoffman
      8 hours ago










    • $begingroup$
      @DarrelHoffman I said 365-day year.
      $endgroup$
      – Parcly Taxel
      8 hours ago






    • 2




      $begingroup$
      "It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29." There's no accounting for taste but I'd have utterly no interest in that. It's enough to know some days are more common that others and if you told me summer dates are more common I'd take you word for it, but the exact data and numbers is of absolutely no interest to me other than theoretical.
      $endgroup$
      – fleablood
      8 hours ago
















    $begingroup$
    It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29.
    $endgroup$
    – Chase Ryan Taylor
    9 hours ago






    $begingroup$
    It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29.
    $endgroup$
    – Chase Ryan Taylor
    9 hours ago














    $begingroup$
    @ChaseRyanTaylor That'd just be making a distribution of all birthdays on the planet with 29 February taken into account - theoretically possible, since only a finite number of people have been born - and then, given your birthday, look at the day before it and look up the corresponding probability.
    $endgroup$
    – Parcly Taxel
    9 hours ago




    $begingroup$
    @ChaseRyanTaylor That'd just be making a distribution of all birthdays on the planet with 29 February taken into account - theoretically possible, since only a finite number of people have been born - and then, given your birthday, look at the day before it and look up the corresponding probability.
    $endgroup$
    – Parcly Taxel
    9 hours ago












    $begingroup$
    Wouldn't that just be 1/365.25 ? I mean approximately, obviously the actual length of a year is slightly off and we cover it with the 100 yr. and 400 yr. rules for leap years. Mind you, those exceptions being as far apart as they are, they're longer than the average human lifespan, so assuming two people who are both alive at the same time, 1/365.25 is probably good enough unless you're within one lifespan a century boundary that was not leap year.
    $endgroup$
    – Darrel Hoffman
    8 hours ago




    $begingroup$
    Wouldn't that just be 1/365.25 ? I mean approximately, obviously the actual length of a year is slightly off and we cover it with the 100 yr. and 400 yr. rules for leap years. Mind you, those exceptions being as far apart as they are, they're longer than the average human lifespan, so assuming two people who are both alive at the same time, 1/365.25 is probably good enough unless you're within one lifespan a century boundary that was not leap year.
    $endgroup$
    – Darrel Hoffman
    8 hours ago












    $begingroup$
    @DarrelHoffman I said 365-day year.
    $endgroup$
    – Parcly Taxel
    8 hours ago




    $begingroup$
    @DarrelHoffman I said 365-day year.
    $endgroup$
    – Parcly Taxel
    8 hours ago




    2




    2




    $begingroup$
    "It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29." There's no accounting for taste but I'd have utterly no interest in that. It's enough to know some days are more common that others and if you told me summer dates are more common I'd take you word for it, but the exact data and numbers is of absolutely no interest to me other than theoretical.
    $endgroup$
    – fleablood
    8 hours ago




    $begingroup$
    "It would be very interesting to see the solution without assuming a uniform distribution, including considering Feb 29." There's no accounting for taste but I'd have utterly no interest in that. It's enough to know some days are more common that others and if you told me summer dates are more common I'd take you word for it, but the exact data and numbers is of absolutely no interest to me other than theoretical.
    $endgroup$
    – fleablood
    8 hours ago











    4












    $begingroup$

    If you want to make it hard:



    There are $365$ choices her birthday (H) could be, and $365$ choices yours(Y) could be. So there are are $365^2$ pairs of birthdays (H,Y).



    Of those $365^2$ pairs there are $365$ cases where $Y = H+1$. (Case 1: Y= Jan. 1-H=Dec 31. Case 2: Y= Jan. 2-H= Jan 1. Case 3: Y=Jan. 3-H= Jan. 2..... Case 365: Y= Dec. 31, H= Dec. 30$.



    So the probability is $frac {365}{365^2} = frac 1{365}$.



    If you want to do it the easy way.



    There are $365$ days she could be born. The day before your birthday is precisely one of them and as likely as any other. So probability is $frac 1{365}$.



    .....



    As for your birthday being Feb. 28.... No matter what your birthday is, the probability of her birthday being Feb. 27 is $frac 1{365}$.



    ...



    Finally your title is "What is the probability somebody's birthday is the day before mine?"



    The answer to that is 100%. There are tens of millions of people with birthdays the day before you.



    .....



    " That seems too simple to me" Why? Why should it be complicated?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If you want to make it really hard (and not assume any knowledge about people's birthdays), the probability of somebody's birthday being the day before the asker's is not 100%. I asked that very question a few years ago, and the answer for 7 billion people would be extremely close to 100%, but not quite.
      $endgroup$
      – maxb
      9 hours ago










    • $begingroup$
      @maxb I thought about that but didn't want to confuse the issue. It is, of course, $1 - (frac {364.25}{365.25})^{7,000,000,000}$ which looks significantly less than $1$ than it actually is. Thing is, for there not to be anybody having a birthday the day before there must be a date in which no-one on the planet was born and no one was born on that date in any year. I think the statement "Every day millions of people are born" is a safe assumption in a "Junior Jumbo Book of Facts" way.
      $endgroup$
      – fleablood
      8 hours ago






    • 1




      $begingroup$
      Once we get into real-world questions, probabilities become much more complicated. The fraction $left(frac{364.25}{365.25}right)^{7,000,000,000}$ is much smaller than any actual uncertainty you deserve to have about pretty much anything: there is an incredibly tiny but much higher probability that for some mysterious reason you're not aware of nobody is ever born on March 12th.
      $endgroup$
      – Misha Lavrov
      6 hours ago












    • $begingroup$
      @fleablood ""Every day millions of people are born" isn't a safe assumption at all. For a population of 7bn, ignoring population growth, and assuming a global life expectancy of 50 years, there are only about 400,000 births per day. You would need make some creative assumptions to change that estimate to "millions" (plural).
      $endgroup$
      – alephzero
      5 hours ago












    • $begingroup$
      ... or be off be a factor of 50.....
      $endgroup$
      – fleablood
      2 hours ago
















    4












    $begingroup$

    If you want to make it hard:



    There are $365$ choices her birthday (H) could be, and $365$ choices yours(Y) could be. So there are are $365^2$ pairs of birthdays (H,Y).



    Of those $365^2$ pairs there are $365$ cases where $Y = H+1$. (Case 1: Y= Jan. 1-H=Dec 31. Case 2: Y= Jan. 2-H= Jan 1. Case 3: Y=Jan. 3-H= Jan. 2..... Case 365: Y= Dec. 31, H= Dec. 30$.



    So the probability is $frac {365}{365^2} = frac 1{365}$.



    If you want to do it the easy way.



    There are $365$ days she could be born. The day before your birthday is precisely one of them and as likely as any other. So probability is $frac 1{365}$.



    .....



    As for your birthday being Feb. 28.... No matter what your birthday is, the probability of her birthday being Feb. 27 is $frac 1{365}$.



    ...



    Finally your title is "What is the probability somebody's birthday is the day before mine?"



    The answer to that is 100%. There are tens of millions of people with birthdays the day before you.



    .....



    " That seems too simple to me" Why? Why should it be complicated?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If you want to make it really hard (and not assume any knowledge about people's birthdays), the probability of somebody's birthday being the day before the asker's is not 100%. I asked that very question a few years ago, and the answer for 7 billion people would be extremely close to 100%, but not quite.
      $endgroup$
      – maxb
      9 hours ago










    • $begingroup$
      @maxb I thought about that but didn't want to confuse the issue. It is, of course, $1 - (frac {364.25}{365.25})^{7,000,000,000}$ which looks significantly less than $1$ than it actually is. Thing is, for there not to be anybody having a birthday the day before there must be a date in which no-one on the planet was born and no one was born on that date in any year. I think the statement "Every day millions of people are born" is a safe assumption in a "Junior Jumbo Book of Facts" way.
      $endgroup$
      – fleablood
      8 hours ago






    • 1




      $begingroup$
      Once we get into real-world questions, probabilities become much more complicated. The fraction $left(frac{364.25}{365.25}right)^{7,000,000,000}$ is much smaller than any actual uncertainty you deserve to have about pretty much anything: there is an incredibly tiny but much higher probability that for some mysterious reason you're not aware of nobody is ever born on March 12th.
      $endgroup$
      – Misha Lavrov
      6 hours ago












    • $begingroup$
      @fleablood ""Every day millions of people are born" isn't a safe assumption at all. For a population of 7bn, ignoring population growth, and assuming a global life expectancy of 50 years, there are only about 400,000 births per day. You would need make some creative assumptions to change that estimate to "millions" (plural).
      $endgroup$
      – alephzero
      5 hours ago












    • $begingroup$
      ... or be off be a factor of 50.....
      $endgroup$
      – fleablood
      2 hours ago














    4












    4








    4





    $begingroup$

    If you want to make it hard:



    There are $365$ choices her birthday (H) could be, and $365$ choices yours(Y) could be. So there are are $365^2$ pairs of birthdays (H,Y).



    Of those $365^2$ pairs there are $365$ cases where $Y = H+1$. (Case 1: Y= Jan. 1-H=Dec 31. Case 2: Y= Jan. 2-H= Jan 1. Case 3: Y=Jan. 3-H= Jan. 2..... Case 365: Y= Dec. 31, H= Dec. 30$.



    So the probability is $frac {365}{365^2} = frac 1{365}$.



    If you want to do it the easy way.



    There are $365$ days she could be born. The day before your birthday is precisely one of them and as likely as any other. So probability is $frac 1{365}$.



    .....



    As for your birthday being Feb. 28.... No matter what your birthday is, the probability of her birthday being Feb. 27 is $frac 1{365}$.



    ...



    Finally your title is "What is the probability somebody's birthday is the day before mine?"



    The answer to that is 100%. There are tens of millions of people with birthdays the day before you.



    .....



    " That seems too simple to me" Why? Why should it be complicated?






    share|cite|improve this answer









    $endgroup$



    If you want to make it hard:



    There are $365$ choices her birthday (H) could be, and $365$ choices yours(Y) could be. So there are are $365^2$ pairs of birthdays (H,Y).



    Of those $365^2$ pairs there are $365$ cases where $Y = H+1$. (Case 1: Y= Jan. 1-H=Dec 31. Case 2: Y= Jan. 2-H= Jan 1. Case 3: Y=Jan. 3-H= Jan. 2..... Case 365: Y= Dec. 31, H= Dec. 30$.



    So the probability is $frac {365}{365^2} = frac 1{365}$.



    If you want to do it the easy way.



    There are $365$ days she could be born. The day before your birthday is precisely one of them and as likely as any other. So probability is $frac 1{365}$.



    .....



    As for your birthday being Feb. 28.... No matter what your birthday is, the probability of her birthday being Feb. 27 is $frac 1{365}$.



    ...



    Finally your title is "What is the probability somebody's birthday is the day before mine?"



    The answer to that is 100%. There are tens of millions of people with birthdays the day before you.



    .....



    " That seems too simple to me" Why? Why should it be complicated?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 9 hours ago









    fleabloodfleablood

    72.2k22687




    72.2k22687












    • $begingroup$
      If you want to make it really hard (and not assume any knowledge about people's birthdays), the probability of somebody's birthday being the day before the asker's is not 100%. I asked that very question a few years ago, and the answer for 7 billion people would be extremely close to 100%, but not quite.
      $endgroup$
      – maxb
      9 hours ago










    • $begingroup$
      @maxb I thought about that but didn't want to confuse the issue. It is, of course, $1 - (frac {364.25}{365.25})^{7,000,000,000}$ which looks significantly less than $1$ than it actually is. Thing is, for there not to be anybody having a birthday the day before there must be a date in which no-one on the planet was born and no one was born on that date in any year. I think the statement "Every day millions of people are born" is a safe assumption in a "Junior Jumbo Book of Facts" way.
      $endgroup$
      – fleablood
      8 hours ago






    • 1




      $begingroup$
      Once we get into real-world questions, probabilities become much more complicated. The fraction $left(frac{364.25}{365.25}right)^{7,000,000,000}$ is much smaller than any actual uncertainty you deserve to have about pretty much anything: there is an incredibly tiny but much higher probability that for some mysterious reason you're not aware of nobody is ever born on March 12th.
      $endgroup$
      – Misha Lavrov
      6 hours ago












    • $begingroup$
      @fleablood ""Every day millions of people are born" isn't a safe assumption at all. For a population of 7bn, ignoring population growth, and assuming a global life expectancy of 50 years, there are only about 400,000 births per day. You would need make some creative assumptions to change that estimate to "millions" (plural).
      $endgroup$
      – alephzero
      5 hours ago












    • $begingroup$
      ... or be off be a factor of 50.....
      $endgroup$
      – fleablood
      2 hours ago


















    • $begingroup$
      If you want to make it really hard (and not assume any knowledge about people's birthdays), the probability of somebody's birthday being the day before the asker's is not 100%. I asked that very question a few years ago, and the answer for 7 billion people would be extremely close to 100%, but not quite.
      $endgroup$
      – maxb
      9 hours ago










    • $begingroup$
      @maxb I thought about that but didn't want to confuse the issue. It is, of course, $1 - (frac {364.25}{365.25})^{7,000,000,000}$ which looks significantly less than $1$ than it actually is. Thing is, for there not to be anybody having a birthday the day before there must be a date in which no-one on the planet was born and no one was born on that date in any year. I think the statement "Every day millions of people are born" is a safe assumption in a "Junior Jumbo Book of Facts" way.
      $endgroup$
      – fleablood
      8 hours ago






    • 1




      $begingroup$
      Once we get into real-world questions, probabilities become much more complicated. The fraction $left(frac{364.25}{365.25}right)^{7,000,000,000}$ is much smaller than any actual uncertainty you deserve to have about pretty much anything: there is an incredibly tiny but much higher probability that for some mysterious reason you're not aware of nobody is ever born on March 12th.
      $endgroup$
      – Misha Lavrov
      6 hours ago












    • $begingroup$
      @fleablood ""Every day millions of people are born" isn't a safe assumption at all. For a population of 7bn, ignoring population growth, and assuming a global life expectancy of 50 years, there are only about 400,000 births per day. You would need make some creative assumptions to change that estimate to "millions" (plural).
      $endgroup$
      – alephzero
      5 hours ago












    • $begingroup$
      ... or be off be a factor of 50.....
      $endgroup$
      – fleablood
      2 hours ago
















    $begingroup$
    If you want to make it really hard (and not assume any knowledge about people's birthdays), the probability of somebody's birthday being the day before the asker's is not 100%. I asked that very question a few years ago, and the answer for 7 billion people would be extremely close to 100%, but not quite.
    $endgroup$
    – maxb
    9 hours ago




    $begingroup$
    If you want to make it really hard (and not assume any knowledge about people's birthdays), the probability of somebody's birthday being the day before the asker's is not 100%. I asked that very question a few years ago, and the answer for 7 billion people would be extremely close to 100%, but not quite.
    $endgroup$
    – maxb
    9 hours ago












    $begingroup$
    @maxb I thought about that but didn't want to confuse the issue. It is, of course, $1 - (frac {364.25}{365.25})^{7,000,000,000}$ which looks significantly less than $1$ than it actually is. Thing is, for there not to be anybody having a birthday the day before there must be a date in which no-one on the planet was born and no one was born on that date in any year. I think the statement "Every day millions of people are born" is a safe assumption in a "Junior Jumbo Book of Facts" way.
    $endgroup$
    – fleablood
    8 hours ago




    $begingroup$
    @maxb I thought about that but didn't want to confuse the issue. It is, of course, $1 - (frac {364.25}{365.25})^{7,000,000,000}$ which looks significantly less than $1$ than it actually is. Thing is, for there not to be anybody having a birthday the day before there must be a date in which no-one on the planet was born and no one was born on that date in any year. I think the statement "Every day millions of people are born" is a safe assumption in a "Junior Jumbo Book of Facts" way.
    $endgroup$
    – fleablood
    8 hours ago




    1




    1




    $begingroup$
    Once we get into real-world questions, probabilities become much more complicated. The fraction $left(frac{364.25}{365.25}right)^{7,000,000,000}$ is much smaller than any actual uncertainty you deserve to have about pretty much anything: there is an incredibly tiny but much higher probability that for some mysterious reason you're not aware of nobody is ever born on March 12th.
    $endgroup$
    – Misha Lavrov
    6 hours ago






    $begingroup$
    Once we get into real-world questions, probabilities become much more complicated. The fraction $left(frac{364.25}{365.25}right)^{7,000,000,000}$ is much smaller than any actual uncertainty you deserve to have about pretty much anything: there is an incredibly tiny but much higher probability that for some mysterious reason you're not aware of nobody is ever born on March 12th.
    $endgroup$
    – Misha Lavrov
    6 hours ago














    $begingroup$
    @fleablood ""Every day millions of people are born" isn't a safe assumption at all. For a population of 7bn, ignoring population growth, and assuming a global life expectancy of 50 years, there are only about 400,000 births per day. You would need make some creative assumptions to change that estimate to "millions" (plural).
    $endgroup$
    – alephzero
    5 hours ago






    $begingroup$
    @fleablood ""Every day millions of people are born" isn't a safe assumption at all. For a population of 7bn, ignoring population growth, and assuming a global life expectancy of 50 years, there are only about 400,000 births per day. You would need make some creative assumptions to change that estimate to "millions" (plural).
    $endgroup$
    – alephzero
    5 hours ago














    $begingroup$
    ... or be off be a factor of 50.....
    $endgroup$
    – fleablood
    2 hours ago




    $begingroup$
    ... or be off be a factor of 50.....
    $endgroup$
    – fleablood
    2 hours ago











    0












    $begingroup$

    Considering February 29 I say it is



    1 in 365.25



    as every four year we have an extra day.
    The chance of another person being born on the day before is equally big as any other day of the year (except for February 29).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Considering February 29 I say it is



      1 in 365.25



      as every four year we have an extra day.
      The chance of another person being born on the day before is equally big as any other day of the year (except for February 29).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Considering February 29 I say it is



        1 in 365.25



        as every four year we have an extra day.
        The chance of another person being born on the day before is equally big as any other day of the year (except for February 29).






        share|cite|improve this answer









        $endgroup$



        Considering February 29 I say it is



        1 in 365.25



        as every four year we have an extra day.
        The chance of another person being born on the day before is equally big as any other day of the year (except for February 29).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 9 hours ago









        Sven van den BoogaartSven van den Boogaart

        125111




        125111






















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            迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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            南乌拉尔铁路局

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            南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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