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anonymous
 3 years ago
LEGEN...waitforit...DARIDDLE:
How many people should be in a room so that there is a 99% probability that two or more people share the same birthday?
Hint: whoever gets it right with solutions gets a medal
anonymous
 3 years ago
LEGEN...waitforit...DARIDDLE: How many people should be in a room so that there is a 99% probability that two or more people share the same birthday? Hint: whoever gets it right with solutions gets a medal

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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1the dreaded riddles are back!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0these ones are legendary. i don't make puny riddles like the original

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so @hartnn do you know the answer?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1haven't tried yet...i would always prefer helping than satisfying my own thirst of solving puzzles...i'll try that in free time :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0helping huh. haven't tried that in a while

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1i try these problems by taking small numbers first let there be just 2 people in that room then the probability "that two or more people share the same birthday" will be 1 probability that BOTH will not have same birthday (assuming 366 days a year) so, \(\large 1 \dfrac{366}{366}\times \dfrac{365}{366}\) now extending this for 'n' people \(\large 0.99 =1 \dfrac{\dfrac{366!}{(366n)!}}{366^n}\) am i on right path ? (wondering how the hell can i solve that for n :O O.o)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol. am not saying if you're on the right path or not. just tell me the answer and i'll tell you if it's right. that's how riddles work

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1rounding to nearest integer i am getting 55 people (infact 55 or more!)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1i had my chance, i'll let others try....
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