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Math Help - Birthday Paradox

  1. #1
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    Birthday Paradox

    Apparently, there are two methods of solving this probability problem. The first, using p(n) = 1 - (365^n falling / 365^n) makes sense to me; however, there is an alternate solution using the Taylor series of e^x.

    Here's the full solution on Wikipedia:
    Birthday problem - Wikipedia, the free encyclopedia

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  2. #2
    Super Member Random Variable's Avatar
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    let x=-x

    then e^{-x} \approx 1 -x

    so  1 - \frac{1}{365} \approx e^{-1/365}

     1-\frac{2}{365} \approx e^{-2/365}

    and so on
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by jrkevinfang1234 View Post
    Apparently, there are two methods of solving this probability problem. The first, using p(n) = 1 - (365^n falling / 365^n) makes sense to me; however, there is an alternate solution using the Taylor series of e^x.

    Here's the full solution on Wikipedia:
    Birthday problem - Wikipedia, the free encyclopedia

    I got lost at the transition from


    to



    What's happening here?
    e^x\approx 1+x\implies e^-x\approx 1-x and so \left(1-x\right)(1-y)(1-z)\cdots(1-t)\approx e^{-x}\cdot e^{-y}\cdot e^{-z}\cdots e^{-t}
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  4. #4
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    Interesting birthday paradox.

    I have an interesting situation. My friend and I have the same birthday. One of her siblings and one of my siblings have the same birthday. She has 2 sibs and I have 6. What are the odds and probability of that occuring? Thanks for responding.
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