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Math Help - IRV Probability Question - check my work?

  1. #1
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    IRV Probability Question - check my work?

    Hi folks. My class (not a probability course) just covered a bunch of probability in one class period. Last prob course I had was in early high school so I'm a little fuzzy on it.

    Here's an example we have to work on:
    Everyone goes to a party with a gift and gives it to the secret santa, who passes them back to everyone randomly (in a random order). If n people go to the party, what is the expected number of party-goers that will get their own gifts back?

    I need to use Indicator Random Variables for this. So here's what I have.

    \mbox {S is a sample space } \\ \mbox {E is an event of S}

    So each person either gets the gift or does not.

    SO, first thing's first, I need to identify my IRV C_i

    C_i = \left\{\begin{array}{cc}0, & \mbox{ if they get a real gift } \\1, & \mbox{ if they get back their own gift } \end{array}\right.

    Next, since C is the random variable that represents # of people who got back their own gifts...
     C = \begin{array}{cc}n\\ \sum \\ n=1\end{array} {C_i}

    Then, getting a good gift back is:
    Pr[{C_i}=0] = \frac{1}{2}
    And getting your own gift back is:
    Pr[{C_i}=1] = \frac{1}{2}

    So our expected value of C_i is:
    E[{C_i}] = 0 * Pr[{C_i =0}] + 1 * Pr[{C_i =0}] = 0 * 1 * \frac{1}{2} = \frac{1}{2}

    And finally, the expected value of C is...
    E[C] = E[ (\begin{array}{cc}n\\ \sum \\ i=1\end{array}) {C_i} ]
    E[C] = \begin{array}{cc}n\\ \sum \\ i=1\end{array} E[{C_i}]
    E[C] = \begin{array}{cc}n\\ \sum \\ i=1\end{array} \frac{1}{2} = \frac{n}{2}

    (I hope you like my latex, I really tried to learn that for this post. )

    So is this right? Any advice? Thanks!
    Last edited by mander; February 23rd 2009 at 05:48 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I'm not sure what you want me to check.
    These 0,1 rvs are called Bernoullis and their sum is a Binomial.
    You have a few typos.
    C runs from 1 to n.
    You accidentally are starting at 0.
    O,1,...,n are n+1 terms.
    But n/2 is correct.
    You want 1,...,n in your sums.
    Also that 2^n is just weird.
    It can't be a lower value since it's greater than n, your upper limit.
    You want 1,..,n in your series, all of them.
    Last edited by matheagle; February 23rd 2009 at 06:00 PM.
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  3. #3
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    Yeah I was working out the problem a different way and then tried to change it and... the 2^n is from my original solution which I don't think works, sorry about that! (So, it should be n=1 instead of the random 2^n right?)

    I'm not sure what you mean that I need 1,...,n in my series though. I'll fix my C starting at 0, thanks! I understand what you mean about that.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Your definition of C is still off, It's i=1,...,n.
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  5. #5
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    Thanks again, Matheagle. I think I get it now. You're awesome.
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