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Math Help - Advanced Birthday Problem

  1. #1
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    Question Advanced Birthday Problem

    In a town of 42,077 people 12 people run for local office.

    Of those 12 only 9 are elected.

    What is the probability that 2 of the 9 elected have the same birthday?





    I don't even know where to start and I think I broke my calculator. And please, I'd love to see the work/equation.
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  2. #2
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    Am I misreading something? I don't see why the majority of the information is being given. Why don't you just need to calculate the probability of matching birthdays in a group of 9 people, assuming a 365 day year with each day equally likely?

    Also, are you looking for the probability of at least one match, or one and only one match?
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  3. #3
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    You're not misreading. The extra information might be there to throw me off and from your reaction it sounds like it is.

    I am so horrible at trying to decide what numbers matter in these types of problems.

    And the problem is for ONLY 2 of 9 having the same birthday. So...one and only one match.
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  4. #4
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    Do you know combinational mathematics? I don't know your school system

    Perhaps this could be something in the right direction? I dont know

    P(2 people same)= (365/365)*(364/365)*(363/365)*(362/365)*(361/365)*(360/365)*(359/365)*(358/365)*(8/365)
    P(2 people same)=365!/(357!*365^8)*(8/365)
    P(2 people same)=(365!*8!)/(357!*8!*365^8)*(8/365)
    P(2 people same)=(365 choose 357)*(8!*8)/(365^9) which gives an answer about 0.020288... Even if it's wrong, I hope that you might get any idea or something out of this!
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  5. #5
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    In that case, let's tackle this one step at a time. Hopefully I'll do this right, since I'm not very good at these - if this is wrong hopefully the forum will forgive me. The set of all possible outcomes is \{1, 2, ... , 365\}^9 (using ordered 9-tuples), and each possibility is equally likely. In this situation, you just take the number of ways to have one and only one matching birthday, and divide that number by the number of possibilities. The number of possibilities is easy: 365^9.

    So the majority of the problem is counting the number of vectors in \{1, 2, ... , 365\}^9 that have one and only one matching number. First, we will pick the two of the nine that match, and there are 9C2 ways to do that. Then we assign those two a number: 365 ways to do that. Then, we fill in the remaining 7 numbers, and order matters for the purpose of counting up our possibilities, so 364 P 7 ways to do that. What you end up with at the end is

    <br />
\frac{<br />
\left( <br />
\begin{array}{c} 9 \\ 2 \end{array} <br />
\right)<br />
P \left( <br />
\begin{array}{c} 365 \\ 8 \end{array} <br />
\right)<br />
}<br />
{<br />
365^9<br />
}<br />

    Since it's a little ambiguous, the top is 9C2 * 365P8. It's hard to make things particularly intuitive in counting the number of things you have in the numerator. The key, I think, is to make sure the way you break up "doing the jobs" is one-one and onto the set you are trying to count. Fail at the one-one step, and you end up double counting some possibilities, and if it isn't onto then you are leaving some out. Basically, if you look at a particular possibility you should be able to reconstruct exactly the things you did to get there. For example, for (3, 17, 22, 150, 153, 17, 240, 270, 340) using my method of counting I

    1) Picked the 2nd and 6th place to match (9C2 ways)
    2) Filled those both with 17 (365 ways)
    3) Filled the 1st position with 3 (364 ways)
    4) Filled the 3rd position with 22 (363 ways)
    .
    .
    .
    9) Filled the last position with 340 (358 ways)

    and this is the only way to do this using the method I outlined.
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