Handyman wrote:A question:
How many people are needed on a birthday party to have a 50% chance that it is someone else birthday too?
In 23 people group it is very likely to have two people with same birthday than everyone having different days.
Choose one person. He may be born any day of the year. Take another person. He may be born any day except on the day of the first person was born, meaning a day from the rest 364 days. Select the third person, who may be born any day except those in which 1 and 2 person was born, that is, on any date of the year 363. Selected as the fourth, fifth, and so on up to till 23rd person.
Check-in principle, the dates can be selected in
365 x 364 x 363 x 362 x ... x 343
different ways.
Of all the possible ways there are (365^23) pieces because
1 - 365 x 364 x 363 x 362 x ... x 343 / 365^23 ≈ 0,5073 > 0,5
And the problem is addressed.
Age si quid agis.