Probability Son, Father and Grandfather have same birthday.

Here is a quick request for comments for Probability 1 students. Here are two answers saying that the probability that a grandfather, father and son are all born on the same day.

The first answer is sort of wrong because it assumes you specify in advance date of birth. The second answer is right because we assume in advance we are given three generations and we assume we deal with a first born son.

Please leave comments below and will forward them on to BBC.


12 thoughts on “Probability Son, Father and Grandfather have same birthday.”

  1. I think the first answer is right, because three generations are independent , so the probability of this is 1\365 *1\365*1\365 =1\48627125. it just like throw coins , if we want get 10 times heads ,the probability of this is 0.5^10 . they are independent . it is not a conditional probabilities. it is my personal perspective.


    1. Yes, but here we have to specify that the 1st date is 6th Oct. The probability of Oct 6, Oct 6 and Oct 6 (namely \frac{1}{48627125} is less than the probability of a grandfather being born on some day, then the son on same day, and the grandson on the same day (namely \frac{1}{365} \times \frac{1}{365}=1/133225).


  2. Surely this would not be right as there isn’t 1/365 chance the baby was born on October 6th as that is dependent on when it was conceived, if they were all conceived on the same day then the chance of all 3 being born on October 6th isn’t 1/48,000,000 as it is much more likely than that. For example if the baby was conceived in January, there is hardly any chance of it being born in December or November so there are not 365 possibilities of the birthday, there are less.


    1. It is a very good point. Reading the article more carefully there are various dependencies introduced. We can only multiply the probabilities together \frac{1}{365} \times \frac{1}{365} if the outcomes are independent of on another. Also see Sean’s answer above we see even amongst the nation probabilities of different days are not uniform, so 6th Oct has a higher probability too.


    2. This is a good point, a close reading of the mirror article shows that there are various dependencies being introduced, where maybe on parent being born of 6th Oct might effect the sons Birthday. So we maybe can’t assume that the events are independent. From a maths perspective it is hard to account for this so multiplying probabilities acts as an approximation to multiply the probabilities together.


  3. 1/133590, probability of the grand father sharing a birthday with his son and grand-son if exactly one was born in a leap year as (1/365)x(1/366). 1/133956, if both were born in a leap year as (1/366)x(1/366). 1/133225, if neither were born in a leap year. But this can be extended to include the likelihood of 0/1/2 births being during a leap year.


  4. If we assume that each family has two children (1 grandfather, 2 fathers, 4 sons) then the probability that one person from each generation = (1x2x4)/(365×365). This is because the grandfather can be born on any day so the probability is 1. The second generation has to be born on the same day as the grandfather, but there are two of them so there are two chances. Then the third generation has four chances of being born on the same day. So the final probability in this case is 8/133225.


    1. Yes that the BBC article is more specific it states that we are dealing with first born sons for a grandfather, farther and son, but if we say that we can do this for any farther and his son then the answer changes. If you have a family, with two sons from each generation and ask what is the probability that a farther and son have the same birthday as the grandfather, then you essentially get 4 attempts, the answer is then roughly 4 times higher than \frac{1}{133225}. (But because we have the possibility of having 4 or more with the same birthday the \frac{4}{133225} isn’t quite right, we have to use something like the inclusion-exclusion formula from you exercise sheets e.g. \mathbb P (A \cup B\ cup C ) = \mathbb P(A) + \mathbb P (B) + \mathbb P(C) - \mathbb P (A\cap B) - \mathbb P(A \cap B) - \mathbb P(B \cap C) + \mathbb P ( A \cap C) - \mathbb P (A \cap B \cap C) but now for 4 sets. I didn’t bother to do the calculation because that answer should not be too far off \frac{4}{133225}).


  5. The ONS says that October 6 is the 45th most popular birthday in the UK, with 0.3% more babies being born on that date than on an average day. So the probability of the father and son both being born on October 6 given that the grandfather was born on October 6 is 8.000×10^-6 which is the same as 1 in ~125,000. This is slightly better odds than the assumption that the odds of being born on a particular date are 1/365 which gives 1 in ~130,000.


    1. Great answer! Yes, it seems October 6th in on the cusp of a popular time of year (late September):

      Plus if we couple this with the fact there are 8 Million families with children (many of these will have a paternal grandfather and half should at least one son) there should be a good number with same birthdays. Approximately

      4,000,000 \times \frac{1}{365} \times \frac{1}{365} \approx 300.

      (Obviously there are a lot of approximations going on here but it seems reasonable that there are a few hundred families in the UK where the grandfather, father and son share the same birthday)


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