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Homework Help: (Probability) The birthday problem P(at least 2) DIRECT APPROACH

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the probability that given a group of 5 people, at least 2 will share the same birthday?

    2. Relevant equations

    I know that 1-P(0 matches) = answer, but there is a reason I am going about this the head on approach. I am trying to develop my combinatoric skills.

    3. The attempt at a solution
    EDIT: I think I am missing some cases. What about the case where we have exactly 1 triplet and 1 pair? Hmm..

    OKAY so IF I am correct, then here is what happened in my original work:
    When I did the case of 1 triplet, I REALLY did the case of 1 triplet and 2 different birthdays. Which means I was missing the case of 1 triplet and 2 same birthdays!

    So there are C(5,2) ways to choose the people to be part of the pair = 10 ways. And similar logic yields the final piece of the puzzle. My solutions are equivalent now. Anybody want to verify?

    I am calculating the probabilities of the following to solve this:
    0 matches
    1 pair
    2 pairs
    1 triplet
    1 quadruplet
    1 quintuplet

    Here is my table of work:
    http://img826.imageshack.us/img826/8428/birthday5peopleproblem.jpg [Broken]

    The method is as follows:
    Code (Text):
    No matches has 1 way to assign the lack of matches.
     The # of days for first person is 365, 2nd is 364,etc..
    Multiply everything together and divide by (365^5) to get the probability

    Code (Text):
    1 pair has C(5,2) ways to assign the pairs.
    The # of days for first person is 365. 2nd is 1 way.
     3rd is 364, 4th 363, 5th 362. Multiply all together and divide by 365^5

    Code (Text):
    2 pairs has 3*5 =15 ways of combinations. this is derived by observing:
    A = pair 1
    B = pair 2
    C = standalone
    There are 5 ways of assigning the standalone(no match) for each group, hence 15 ways.

    The way of choosing days is as follows: 365*1*364*1*363
    Multiply that with 15, divide by 365^5
    Code (Text):
    For a triplet, there are C(5,3) ways to assign the
    matching day. The # of days are 365*1*1*364*363

    Code (Text):
    For a quadruplet, there are C(5,4) ways to
    assign the matching day. the # of days are 365*1*1*1*364

    Code (Text):

    For a quintuplet, there is only 1 way to
    assign the matching day, and there is 365 ways to choose the day.
    Please see my chart, and note that the probabilities differ by an extremely small #. I can't figure out where the error is?!?!
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 25, 2012 #2
    Have you considered the possibility of one pair and one triplet?
  4. Feb 25, 2012 #3
    Yes! I realized last night I was missing that one! (and now my solution for the 5 person case is correct, thank you)

    Okay guys, so I thought all was well, but to really test my combinatoric skills I decided to try the 8 person case. Well...once again there is a discrepancy! I will attach my chart.
    Edit: Please ignore "quadrupleruple" lol...I used find/replace to change words around and that happened...oops
    Edit 2: I didn't actually sum rows 21-23...oops. Answer is still off by roughly the same margin.

    Attached Files:

    Last edited: Feb 25, 2012
  5. Feb 25, 2012 #4


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    Science Advisor

    Are the people in your group indistinguishable from each other?

    I mean, if you label them 1,2,3,4,5 , are you distinguishing between ,say,

    1,2 having the same birthday or 2,4, etc?

    Also, in case you're interested in a realistic model --and not just practicing

    your counting--there is data that strongly suggests that birthdays are not

    distributed uniformly.
  6. Feb 25, 2012 #5
    Okay for example:
    If 1,2 have the same birthday as 3,4 , then that is considered a quadruple, not 2 pairs.
    For a group of 4 people, we can have the following cases:
    (0 same): 1 <> 2 <> 3 <> 4
    (1 pair) 1=2 and 3<>4 , 1 = 3 and 2<>4, 1=4 and 2<>3. 2=3 and 1<>4, 2=4 and 1<>3, 3=4 and 1<>2
    2 pairs 1=2 and 3=4, 1=3 and 2=4, 1=4 and 2=3
    (1 triplet) 1=2=3 and 4 is different, 1=2=4 and 3 is different, 1=3=4 and 2 is different, 2=3=4 and 1 is different
    (1 quadruplet) 1=2=3=4

    And for example, suppose we have 2 birthdays: A and B

    _ _ _ _ = 1,2,3,4

    There are only 3 ways to have 2 pairs of birthdays:

    Note that BABA is identical to ABAB (because A can be B and B can be A, but not at the same time).

    Does that answer your question?

    And yes I am aware, but I am solving this to work on my counting skills really.
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