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This is related to the Doomsday rule. Certain days of the year, such as 4/4 and 6/6 are always on the same day of the week as each other in a given year.

A first approximation would be $\frac17$.

But if one person was born in January or February, and the second person was born after February, then there is no way that they would always have their birthdays on the same day of the week on both leap years, and non-leap years.

If they are both in the January to February range, or they are both in the March to December range, then it's OK.

This restriction brings the odds down a little lower than $\frac17$ but how much lower?

I'm not sure what to do with people who were born on leap day. So I guess we can leave them out. Solve for two people where neither was born on leap day.

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If it is impossible to be born on 29 February and other dates are equally likely then in January and February there are $59=8\times 4+9 \times 3$ days and in the other ten months there are $306=43\times 2+44 \times 5$ days

so the probability would be $\dfrac{8^2\times4+9^2\times3+43^2\times2+44^2\times5}{365^2} \approx 0.10416$, substantially less than $\frac17 \approx 0.14286$

If it is possible to be born on 29 February (say with $\frac14$ the probability of other dates) and other dates are equally likely then perhaps the probability would be $\frac{8^2\times4+9^2\times3+0.25^2\times1 +43^2\times2+44^2\times5}{365.25^2}\approx 0.10402$, which is barely changed

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  • $\begingroup$ The February 29 calculation seems to be missing cases where one person was born on February 22 (for example) and the other on February 29. Accounting for those cases I get about $0.10405$ ... so not really much better. Note that the probability will get worse as the ratio between the two parts of the year gets closer to $1,$ so adding days to the shorter half (without making it the longer half) will always reduce the probability. $\endgroup$ – David K Apr 3 at 2:17
  • $\begingroup$ @DavidK - your first point may be a matter of interpretation: I took it to be that if you were born on 29 February and I was born on 22 February (or 7 March), then our birthdays are not the same day of the week in 2019 as you do not have a birthday this year - otherwise sameness loses transitivity $\endgroup$ – Henry Apr 3 at 7:52
  • $\begingroup$ My understanding is that people with birthdays on February 29 celebrate their birthdays on the day after Februrary 28, whether that's Februrary 29 or March 1. Otherwise you don't even have the $0.25^2$ term, since not having a birthday means you can't have the same birthday. But you're right, that's only one interpretation. In The Pirates of Penzance it's interpreted the other way, one birthday every four years. $\endgroup$ – David K Apr 3 at 11:52
  • $\begingroup$ @Henry - Ignoring leap day babies, you got the correct answer, 0.10416. And you got there a lot faster than me! But I don't understand how you got there. Can you add more detail? $\endgroup$ – Dan Cron May 2 at 19:36
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    $\begingroup$ @DanCron: The probability both were born on any of the eight days 7 Jan, 14 Jan, ... or 25 Feb is $\dfrac{8^2}{365^2}$. There are four sets like this so $\dfrac{8^2\times 4}{365^2}$ and there are other possibilities leading to my expression $\endgroup$ – Henry May 2 at 22:19
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A back of the envelope estimate, ignoring the fact that months have different lengths, is as follows.

The chance that both people were born in January or February is $(1/6)^2$ and the chance that both people were born between March and December is $(5/6)^2$. So you should get something like

$$ {1 \over 7} \left( \left( {1 \over 6} \right)^2 + \left( {5 \over 6} \right)^2 \right) = {26 \over 252} \approx 0.103$$

as your answer. The exact answers people have come up with are very close to this.

Depending on why you are trying to answer this question, you may not be answering the right question. My birthday is December 9 and my mother's birthday is January 13. Are our birthdays on the same day of the week? If we're talking about in the same year, no. If we're comparing December 9, year $N$ to January 13, year $N+1$, the two days are 35 days apart and the answer is yes. (If you don't care about my family: are Christmas and New Year's Day on the same day of the week?)

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    $\begingroup$ Interesting point. Yes. I meant within one calendar year, do the two people's birthdays fall on the same day of the week. Your point of view makes sense too, it's just a different question. I think the decimal point is off by one. Did you mean 0.103? $\endgroup$ – Dan Cron Apr 3 at 17:01
  • $\begingroup$ Why am I asking the question? I read about someone who said that they and their two siblings always had their birthdays on the same day of the week every year. I was wondering, with a family of 3 kids, what are the chances of that? (About 1/100) If two kids had their birthday on Christmas and one was on New Years Day, the way I stated the problem, they would not be on the same day of the week. But your way of looking at things appeals to my common sense more. How exactly would you specify the question to get Christmas and New Year considered together? $\endgroup$ – Dan Cron Apr 3 at 17:19
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    $\begingroup$ You're right, it's 0.103 - I fixed it. As for how to specify it, I can think of two reasonable ways: (1) force the two birthdays to be separated by less than half a year; (2) redefine the "year" to start on March 1, which means you don't have to worry about the leap-year special cases. $\endgroup$ – Michael Lugo Apr 3 at 17:58
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Assuming that we are ignoring people born on February 29.

The first person could be born on any of the 365 days. the second person could be born on any of the 365 days. The two people were not necessarily born in the same year.

There are 59^2 combinations where they were both born in January or February. 1/7 of these pairs will have the same birthday.

There are 306^2 combinations where they were both born in March to December. 1/7 of these pairs will have the same birthday.

There are 365^2 combinations in total.

((((59^2)/7) + ((306^2)/7))/(365^2) which is about 0.10414


Edit:

There is an issue with the solution above. In order to illustrate the issue, lets consider a different question. What are the chances that two people born from January 1 to January 8 have their birthday on the same day of the week every year? Using the method above, you would see that there are 64 possible combinations, and each of the combinations has about 1/7 chance of being on the same day of the week, so an approximate answer to the January 1 to January 8 question is ((8^2)/7)/(8^2) = 1/7 which is about 0.14286.

However, if you actually drew out the 8 by 8 grid, and checked off the boxes on the same day of the week, you would check off the 8 boxes on the diagonal, and the two other corners. So you would check off a total of 10 boxes. So the exact correct answer to the January 1 to January 8 question is 10/64 which is 0.15625.


Temporarily ignoring leap year completely... If we are considering a grid with x days, and we want to count the number of combinations "checked off" with the same day of the week, we can use (1+x^7)/(1-x)^2/(1-x^7), which is listed as A008814 in the On-line Encyclopedia of Integer Sequences.

This exact count replaces the approximation of (x^2)/7 in my original solution.

Hmm... the integer sequence is correct, but that expression does not look correct.


Edit:

The formula from OEIS does not look correct. I calculated the values for the 59th and 306th values of the integer sequence with a spreadsheet instead, and I got the results 499 and 13,378 So, the exact correct answer is (499 + 13378)/(365^2) which is about 0.10416

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