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Math Help - Simple probability question. Involves drawing name for gift.

  1. #1
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    Simple probability question. Involves drawing name for gift.

    We had a dinner with my family where we draw names to give gift this christmas. The only rule was to not pick our name.

    We did it a couple of time because we kept picking our name.

    Finally we succeded and i started to think about what was the probability of someone picking his own name thus making this fail.

    I built an excel simulation program which simulate drawing and got a consistent 50 % failure chance for 2 persons, about 66 % for 3 persons...and so on...but it's like there is a somekind of a maximum of about 68 % as i add people to my simulation.

    For 6 persons i get 68 % failure and for 100 persons, it's about the same. I don't understand why that is.

    So i started thinking...and the figures that i think it should be is about :

    success rate should be 1/number of person, so 1/6 for 6 persons so a failure rate of 5/6....

    Can anyone explain me what i should be getting ?
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  2. #2
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    Quote Originally Posted by moris7 View Post
    We had a dinner with my family where we draw names to give gift this christmas. The only rule was to not pick our name
    We did it a couple of time because we kept picking our name.
    Finally we succeded and i started to think about what was the probability of someone picking his own name thus making this fail.
    Just to be clear on the question: “Six people draw names from the hat.
    What is the probability that no one draws his/her own name?
    If that is the question, there is a well-known answer. These are derangements.
    The number of derangements on six objects is D(6)=265.
    So the probability no one has a match is \frac{265}{6!}=0.3681.
    Subtracting that number from 1, gives the probability of at least one match.

    BTW: If n\ge 3 then a very close approximation of D(n) \approx \frac{{n!}}{e}.
    \frac{6!}{e}=264.873197643438
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  3. #3
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    Thank you very much for your answer...this seems like the right answer...i will try to see why my simulation does not point exactly toward this number.
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    It's because of excel random function which is clearly not random.
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  5. #5
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    a descriptive title

    I solved it once without derangements... think of this name-drawing as a permutation where each spot represents a person... and then those spots get filled in with their names (ie 1 through 6) in some order.... 123456 143265 654312 etc... being in the correct spot represents drawing your own name.


    • Number of permutations where all 6 draw their name: 1


    • Number of permutations where exactly five draw their own name: 0 (it's impossible)


    • Number of permutations where exactly four draw their own name: 6C4*1 = 15


    • Number of permutations where exactly three draw their own name: 6C3*2=40 (6C3 ways to choose which three draw their own names and 2 ways for the remaining people to not draw their names)


    • Number of permutations where exactly two draw their own name: 6C2*Q where Q=the number of ways for four people to not draw their own name.

    To get Q you have to recursively use the previous steps... think permutations like 1243 1234 etc.

    permutations (length 4) where all 4 draw their own name: 1
    permutations where exactly 3 draw their own name: 0
    permutations where exactly 2 draw their own name: 4C2*1=6
    permutations where exactly 1 draws their own name: 4C1*2=8
    permutations where none draw their own name: (4x3x2x1)-(1+6+8)=9
    therefore Q=9
    6C2*Q=6C2*9= 135 and therefore there are 135 permutations (length 6) where exactly 2 people draw their own name.
    • number of permutations where exactly 1 person draws their own name: 6C1*L where L is the number of ways for 5 people to not draw their own names

    permutations (length 5) where all 5 draw their own name: 1
    permutations where exactly 4 draw their own name: 0
    permutations where exactly 3 draw their own name: 5C3*1= 10
    permutations where exactly 2 draw their own name: 5C2*2= 20
    permutations where exactly 1 draws their own name: 5C1*Q=5C1*9= 45
    permutations where none draw their own name: (5x4x3x2x1)-(1+10+20+45)=44
    therefore L=44
    6C1*L=6C1*44= 264 and therefore there are 264 permutations (length 6) where exactly 1 person draws their own name.
    Therefore, the probability of no one drawing their own name in a group of six people is:
    1-P(1 or 2 or 3 or 4 or 5 or 6 people draw their own name)
    =1- (1+15+40+135+264)/(6x5x4x3x2x1)
    =0.36805555555
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