I saw some similar questions to this floating around, but I don't think they are quite the same as this one.

If you ask two people what month they were born in, and they respond February, what is the chance that they will have the same birthday? I was talking through this problem with a friend of mine, and we thought the chance was 1/(28.25^2), due to the expected number of days in but we're not completely sure. To make the problem complete, all normal leap year operations are true, ie every 4 years except multiples of 100, but yes to multiples of 4&100 (400).

  • $\begingroup$ Forgetting the complication involving February 29, you shouldn't square. The first person can be any date...there is then a (roughly) $\frac {1}{28.25}$ chance the second one matches. As to leap years, happily 2000 was one, so I think it's reasonable to treat it as .25 of a day. $\endgroup$ – lulu Jul 24 '15 at 2:26
  • $\begingroup$ Oh, I see. That makes sense. I wonder if anyone has taken the time to compute the expected number of days in February (from negative to positive infinity) $\endgroup$ – diminishedprime Jul 24 '15 at 2:34
  • $\begingroup$ Oh, it's very close to 28.25 . Easy to get the correction...you wrote down the formula already. $\endgroup$ – lulu Jul 24 '15 at 2:35
  • $\begingroup$ Of course, you won't meet many people whose birth year was a multiple of $100$ but not $400$. $\endgroup$ – Robert Israel Jul 24 '15 at 4:17
  • $\begingroup$ @RobertIsrael There are only four such living people in the world, and anyway none of them were born in February: en.wikipedia.org/wiki/List_of_the_verified_oldest_people $\endgroup$ – Travis Jul 24 '15 at 4:42

There are $4*28+1=113$ days in four years of February.
Out of $113^2$ possible $(a,b)$, there are $28*4^2+1^2=449$ ways they can coincide. So the chance is $449/113^2=0.03516...$


What about this we have two people A and B both born in February. I will make four cases:

Case1:Both born in a year with February with 28 day. this year has the probability 3/4 Hence we have the set which present their birthday day $\{(1,1) , (1, 2) , \cdots (28 , 28)\}$ This set has $28 \cdot 28$ days and $28$ element with same coordinates. This case has the probability ${9 \over 16}$. So to have the same birthday day we get $\frac{28}{28\cdot 28} \times \frac{9}{16}$.

Case 2 : A born in a 28 Feb year. B in a 29 Feb year. This has the probability ${3\over 16}$ again we make the set $\{ (1,1) , \cdots , (28,28), (1, 29) ,\cdots , (28 , 29)\}$ This set has $28\cdot 29$ elements with $28$ element with similar coordinates hence we have the probability $\frac{3}{16} \times \frac{28}{28 \cdot 29}$.

Case 3: A in 29 , B in 28 Feb year. This has a similar probability for case 2.

Case4: A and B in a year with 29 day in Feb. This will have the probability $\frac{1}{16} \times \frac{29}{29 \cdot 29}$

Thus the probability is the sum if the cases. $ \frac{9}{16} \cdot \frac{1}{28} + 2 \left(\frac{3}{16} \cdot \frac{1}{29} \right)+ \frac{1}{16} \cdot \frac{1}{29}$

  • 2
    $\begingroup$ People are more likely to be born in a leap year (366/365 more likely). $\endgroup$ – Empy2 Jul 24 '15 at 8:20
  • $\begingroup$ I guess my mistake is I calculated the probability to have same day and month without the same year. Right? $\endgroup$ – AmerYR Jul 24 '15 at 9:19

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