Whats the probability of 3 brothers being born on the same day of the month in 3 different months? I ask because I'm one of them. Its April, May, and October 12th. What's the difference including whether or not we'd be male/female?

  • $\begingroup$ Do you have any other siblings? $\endgroup$
    – whuber
    Mar 15, 2014 at 14:50
  • $\begingroup$ nope, only 2 other brothers. Intersting fact is my father had only males, too: four brothers. $\endgroup$
    – Mazura
    Mar 15, 2014 at 20:53

1 Answer 1


I didn't really get if you want to know the probability of being born on the same day in 3 different months or precisely on the 12th on those months (in this case the probability is obviously lower).

Assume we are not in a leap year.

The probability of being born on the same day in 3 different months is equal to the one of being born from the first to the 28th + the one of being born in the 29th or 30th + the one of being born on the 31th:

$P(B) = \frac1{365^3} \cdot (336\cdot11\cdot10 + 22\cdot10\cdot9 + 7\cdot6\cdot5) = \frac{39150}{48627125} = 0.0008$

Let's assume we have a proportion of $P(F)$ females and $1 - P(F)$ males and that the fact of being born on a specific day does not affect the gender, thus $P(F|B) = P(F)$.

So $P(B|F) = \frac{P(F|B)\cdot P(B)}{P(F)} = \frac{P(F)\cdot P(B)}{P(F)} = P(B)$

If I got the point, there is no need to consider the gender. If there are mistakes in my reasoning, please show them to me. Thank you.

  • $\begingroup$ Why is the probability to be born any of April 1? E.g. none of my siblings is born in April. $\endgroup$
    – Verena
    Mar 15, 2014 at 12:25
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    $\begingroup$ Also I don't think the OP wants 3 specific months, he likes to know all possible combinations of 3 different months. $\endgroup$
    – Verena
    Mar 15, 2014 at 12:29
  • $\begingroup$ @ExpectoPatronum You are completely right. I edited my answer. Check it. $\endgroup$ Mar 15, 2014 at 13:24
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    $\begingroup$ Your answer would be much clearer if it were to explain where the values 336, 22, etc. come from in your formula. You make other important assumptions worth stating, of which the key ones are (1) there are only three boys under consideration (rather than, say, some three brothers out of a family of 12 children); (2) their dates of birth are independent; and (3) each day of the year is equally likely. The first is crucial. The next two are incorrect but the errors made in (3) are known to be small. The second is questionable, so it's worth pointing out and discussing. $\endgroup$
    – whuber
    Mar 15, 2014 at 14:39
  • $\begingroup$ If you ignore leap days .. and consider 365^3 combinations: For how many combinations are all three born on the 1st? 12 x 12 x 12 = 1728. Same for 2nd through 28th. How many where they are born on the 29th? 11 x 11 x 11 = 1331. Same for 30th. How many for the 31st? 7 x 7 x 7 = 343. Total is 28 x 1728 + 2 x 1331 + 343 = 51389. Resulting in a probability of 51389 / 48627125 = 0.00105679700372991. How do you account for the difference in your count? $\endgroup$ Mar 15, 2014 at 16:30

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