I have a nuclear family consisting of dad, mom, me and my younger sister. Our birth week i.e. week in which we were born are linear according to our date of birth i.e. my father who is oldest was born on wednesday, my mom on thursday, me on friday and finally my sister(who is youngest) on saturday.

When i was small i used to think this was always the case with any family (this was my definition of being a family) but this is surely not right but actually quite rare to happen. So, i was just wondering given a size of nuclear family consisting of n members (mine is 4) what is the probability that it will have exactly this condition as mine given known number of families in the world(hypothetically).

  • $\begingroup$ you might want to look at this post which also discusses coincidences around birthdates. $\endgroup$ – TooTone Mar 12 '14 at 19:00
  • $\begingroup$ i know that problem... its a old birthday problem... i am not talking about birthdate but brithweek over here... $\endgroup$ – Sumit Shrestha Mar 13 '14 at 5:20
  • $\begingroup$ The details don't matter: this question is a duplicate because it concerns a post hoc formulation of a probability question. The issues surrounding that are addressed in the duplicate answers. $\endgroup$ – whuber Apr 22 '16 at 14:27

Three assumptions:

  1. Each person can be born on any day with equal chance - eg. a Monday birth is not more likely than a Saturday birth.

  2. The birth day are "independent" - knowing which day one family member was born does not change the chances for another. eg. Wednesday birth for dad does not make a Sunday birth for mum more likely

  3. There are $365$ days in a year

These assumptions mean that the chance for any day is $\frac {1}{365} $. So we have that for a sequence with $4$ family members, the day of the oldest can be any day, but then we have just one day the other three can be born on. So the probability is $\frac {1}{365} \times\frac {1}{365} \times\frac {1}{365} =\frac {1}{48627125} $ . Extremely unlikely!!


If we are talking about consecutive days of the week, but not consecutive dates, then we simply replace $365$ with $7$ in the above and we have $\frac {1}{7} \times\frac {1}{7} \times\frac {1}{7} =\frac {1}{343} $ - much more likely!

| cite | improve this answer | |
  • $\begingroup$ It is not one day per year but is one per day of week. So, 1/7 should come somewhere $\endgroup$ – Sumit Shrestha Apr 21 '16 at 20:23
  • $\begingroup$ @sumit - I have updated my answer accordingly. $\endgroup$ – probabilityislogic Apr 22 '16 at 9:52
  • $\begingroup$ But, you dont include the fact that first can be born on any of the 7 days but other following has to be only next day of the one first was born. This applies in series as its a family. $\endgroup$ – Sumit Shrestha Apr 22 '16 at 17:17
  • $\begingroup$ @sumit - it does take account of this - that's why there are only three terms multiplied. Another way to think of it is there are $7^4$ combinations of days born for a family of size $4$ (all equally likely given the assumptions). Of these, there are $7$ combinations which are in sequence (one for each day it starts). Hence the probability is $\frac {7}{7^4} =\frac {1}{343} $ as my answer says. $\endgroup$ – probabilityislogic Apr 23 '16 at 3:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.