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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            數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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            When can things happen in Etherscan, such as the picture below?

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            2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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            格利澤436b

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            格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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