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xlegendx

  • 2 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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  1. hartnn
    • 2 years ago
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    the dreaded riddles are back!

  2. xlegendx
    • 2 years ago
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    these ones are legendary. i don't make puny riddles like the original

  3. xlegendx
    • 2 years ago
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    so @hartnn do you know the answer?

  4. hartnn
    • 2 years ago
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    haven't tried yet...i would always prefer helping than satisfying my own thirst of solving puzzles...i'll try that in free time :)

  5. xlegendx
    • 2 years ago
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    helping huh. haven't tried that in a while

  6. hartnn
    • 2 years ago
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    i 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 \(\large 0.99 =1- \dfrac{\dfrac{366!}{(366-n)!}}{366^n}\) am i on right path ? (wondering how the hell can i solve that for n :O O.o)

  7. xlegendx
    • 2 years ago
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    lol. 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

  8. hartnn
    • 2 years ago
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    rounding to nearest integer i am getting 55 people (infact 55 or more!)

  9. xlegendx
    • 2 years ago
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    ooooh so close

  10. xlegendx
    • 2 years ago
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    but no

  11. hartnn
    • 2 years ago
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    i had my chance, i'll let others try....

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