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I have been reading about the birthday paradox & was trying to take it further & apply it to an example from my own life:-

I am a Brit married to an American. My British sister-in-law shares the same birthday, 19th January, as an American friend that I know through my husband. Both of them gave birth to a son on the same day, 26th June 2011.

I couldn't even get past the probabilities of just knowing two people sharing birthdays whose sons were born on the same day, let alone including the two continents! Anyone fancy having a stab at it?

EDITED.... OK - clarification asked for.

1) What is the probability of knowing two people who share a birthday and who also have a child born on the same day. 2) Whether/How that probability is affected by the fact that number of American friends of my husband that I know directly is about 12. 3) Whether/How that probability is affected by the specific month/day of birth. 4) Whether/How that probability is affected by the two people being in different countries/continents (UK & USA).

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closed as too broad by whuber Mar 25 '18 at 13:02

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Almost 100%. The event in question could be characterized as "what is the chance that a random person might find two pairs of people within her extended acquaintances who share the same birthday." The point is that your question is so vague about the event it wishes to characterize that it permits a huge range of possible answers. For instance, it does not say whether the continents must be America and Europe or whether any two continents will do; it does not stipulate what population should be considered; and too much more is left to the imagination for this to be answerable. $\endgroup$ – whuber Jan 19 '15 at 15:59
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    $\begingroup$ OK. (2) through (4) can now be answered, but (1) is still unanswerable (as the answers to (2) through (4) should be able to demonstrate). (3) has been addressed in other Birthday Paradox-related problems on this site. $\endgroup$ – whuber Jan 19 '15 at 18:22