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Math Help - Please verify: a twist on the Birthday Problem

  1. #1
    Gui
    Gui is offline
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    Please verify: a twist on the Birthday Problem

    Hello,

    I was assigned some optional practice on the birthday question. I got all the questions on the birthday problem itself correct, but although I got an answer for this last one (n >= 612.257), it seems VERY unintuitive to me. Can you please confirm whether it's right or what I did wrong? Here is the problem:

    Given n people randomly chosen, find the smallest value of n so that the probability at least one of them were born on the same day of the year that you are born is at least 50%.

    Now, here's what I thought:

    P(X >= 1) >= 0.5
    X~Binomial(n, 1/365)

    P(X >= 1) = 1 - P(X=0) - P(X=1)

    1 - P(X=0) - P(X=1) >= 0.5

    0.5 >= P(X=0) + P(X=1)

    0.5 >= (nCr(n,0) * (1/365)^0 * (364/365)^n) + (nCr(n,1) * (1/365) * (364/365)^(n-1))

    Now, nCr(anything, 0) is always 1, and nCr(n, 1) is always n. So:

    0.5 >= (364/365)^n + (n/365)*(364/365)^(n-1)

    Plugging this into wolfram alpha, I get that n >= 612.257. This result seems very unintuitive to me. Is it correct?
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  2. #2
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    Re: Please verify: a twist on the Birthday Problem

    Hello, Gui!

    Given n people randomly chosen, find the smallest value of n so that the probability that
    at least one of them were born on the same day of the year that you are born is at least 50%.

    We will find the probability that none of them have your borthday.

    The probability that none of the n people have your birthday is:. \left(\frac{364}{365}\right)^n
    . . We want this to be less than 50%.

    We have: . \left(\frac{364}{365}\right)^n \,<\:0.5

    Take logs: . \ln\!\left(\frac{364}{365}\right)^n \,<\;\ln(0.5) \quad\Rightarrow\quad n\!\cdot\!\ln\!\left(\frac{364}{365}\right) \,<\:\ln(0.5)

    Therefore: . n \:<\:\dfrac{\ln(0.5)}{\ln\left(\frac{364}{365} \right)} \:<\:252.6519836

    . . It will take 253 people.
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