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Math Help - Birthday Coincidence Probability (w/ leap years)

  1. #1
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    Cool Birthday Coincidence Probability (w/ leap years)

    What is the probability of k people having a coincidence of birthdays, taking leap-years into account? Hint: Consider all the days in a 4-year period, in which the leap-day 2/29 occurs once.


    Help tonight/before tomorrow afternoon would be lovely.

    Thanks!!
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  2. #2
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    Quote Originally Posted by cassie00 View Post
    What is the probability of k people having a coincidence of birthdays, taking leap-years into account? Hint: Consider all the days in a 4-year period, in which the leap-day 2/29 occurs once.


    Help tonight/before tomorrow afternoon would be lovely.

    Thanks!!
    It's easier to find the probability that among k people there are no coincidences and subtract this probability from 1.
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    I originally tried splitting up the years (3 "normal years" for every 1 leap-year) and weighting them as 3/4 and 1/4.

    Would you recommend using a larger sample of days? 3(365)+366?

    Thanks!
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  4. #4
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    Quote Originally Posted by cassie00 View Post
    I originally tried splitting up the years (3 "normal years" for every 1 leap-year) and weighting them as 3/4 and 1/4.

    Would you recommend using a larger sample of days? 3(365)+366?

    Thanks!
    Yes, this is what the hint suggests. The probability of having a birthday other than Feb. 29 is going to be \frac{4}{3(365) + 366}, whereas the probability of having Feb. 29 will be \frac{1}{3(365) + 366}.
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    Yes, this is what the hint suggests. The probability of having a birthday other than Feb. 29 is going to be \frac{4}{3(365) + 366}, whereas the probability of having Feb. 29 will be \frac{1}{3(365) + 366}.

    Why isn't the numerator of the first probability 1460?
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  6. #6
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    I'm taking the hint :)

    In the course of 4 years (i.e. 3 regular years and 1 leap year), each regular day occurs 4 times while February 29 occurs once. There are 365 regular days. The total number of days in the course of 4 years is:

    365 x 4 + 1 = 1461

    During that period, a person celebrates his/her birthday 4 times if it falls on a regular day and 1 time if it falls on February 29 (of course, he/she can choose to celebrate it on February 28 or March 1, but that's not the point of the problem).

    The probability that his/her birthday falls on January 1 is 4/1461. The probability is the same for each regular day. Meanwhile, the probability that his/her birthday falls on February 29 is 1/1461.

    The probability that k people have their birthdays on January 1 is:

    <br />
( \frac {4}{1461} )^k<br />

    The probability is the same for the other 365 regular days. Meanwhile, the probability that k people have their birthdays on February 29 is:

    <br />
( \frac {1}{1461} )^k<br />

    Hence, the total probability that k people have the same birthdays is:

    <br />
365 ( \frac {4}{1461} )^k + ( \frac {1}{1461} )^k<br />
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  7. #7
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    Quote Originally Posted by cassie00 View Post
    What is the probability of k people having a coincidence of birthdays, taking leap-years into account? Hint: Consider all the days in a 4-year period, in which the leap-day 2/29 occurs once.

    Thanks!!
    Of related interest: The Birthday Paradox
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