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Coin Probability

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    If n people flip a coin what is the probaility that one is at odds with the rest?
    (answer is [itex] n2^{1-n} [/itex])
    2. Relevant equations



    3. The attempt at a solution

    I tried (2/2)(1/2)(1/2)(1/2)..... and figure that equaled [itex] \frac{1}{2^{n-1}} [/itex] but that isn't the answer. :(
     
  2. jcsd
  3. Sep 4, 2011 #2
    Try listing the possibilities for n=2 and n=3 explicitly, and see if you can see the pattern. What you have used for the formula doesn't correspond to what the question is asking.

    For n=2, we have two possibilities: TH or HT, and obviously we don't care about the order.

    For n=3, we have six possibilities: THH, HTH, HHT, HTT, THT and TTH and again we don't care about the order.

    Try to convert these explicit cases into a formulaic representation.
     
  4. Sep 4, 2011 #3
    Okay. for two people: HH, HT, TT = 1/3
    three people: HHH, HHT, HTT, TTT = 2/4
    Four people: HHHH, HHHT, HHTT, HTTT, TTTT = 2/5
    five people= HHHHH, HHHHT, HHHTT, HHTTT, HTTTT, TTTTT = 2/6

    So I'm thinking 2/(n+1), except that doesn't work for two people, but hey, I can specify 3 or more peoples easy. Problem is, this is not the answer either. :confused:
     
  5. Sep 4, 2011 #4

    I like Serena

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    You have the pattern XYYYYYY here.
    How many ways to shuffle that?
     
  6. Sep 4, 2011 #5
    Hmm... no that's not right. Sorry, I think you read further into my "we don't care about order" statement than I was intending. I probably shouldn't have written that in the first place. The probability for 2 people, for instance, would be 1/2, since it could happen by getting HT or TH, and the total possibilities are HH, TH, HT, TT.

    I'll let you work with Serena on this, as I see she's helped you before.
     
  7. Sep 4, 2011 #6
    Oh okay, order does matter.

    For two people: (Shuffle XY)(P XY) = (2!/2!)(2/2)(1/2) = 1/2
    Three people: (Shuffle XXY)(P XXY) = (3!/2!)(2/2)(1/2)(1/2) = 3/4

    So for four peeps: (4!/3!)(2/2)(1/2^3)

    So for n peeps: [tex] (\frac{n!}{(n-1)!})(\frac{1}{2^{n-1}}) = n2^{1-n} [/tex] correct?
     
  8. Sep 4, 2011 #7

    I like Serena

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    Yep! :smile:

    Note that the formula breaks down for n=2.
    It seems that the author of your problem has been sloppy.
    (Or was there an extra condition that n>2?)
     
  9. Sep 4, 2011 #8
    No extra condition. Author was sloppy. Book just came out this summer. Sloppy author is actually prof at MY university. Uck. I hate errors.

    Um, on a shuffling side note (I don't see why I still don't get it) take a look at these:

    a) 6,7,8,9
    b) 6,7,8,8
    c) 7,7,8,8
    d) 7,8,8,8

    Now, how many ways can I shuffle each?

    For a) I tried 4!/(1!1!1!1!4!) but that equal 1. What gives?
    For b) 4!/(1!1!2!2!)= 6 but isn't it 10? (by tabling)
    For c) 4!/(2!2!2!)= 3 but isn't it 6?
    For d) I tried 4!/(1!3!) = 4 which I think is correct.
     
  10. Sep 4, 2011 #9

    HallsofIvy

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    If you flip a coin n times the probability of m "heads" and "n- m" tails is
    [tex]\left(\begin{array}{c}n \\ m\end{array}\right)\frac{1}{2^n}[/tex].

    If I read your question correctly, you want the probability that either there are n-1 heads and 1 head or n-1 tails and 1 head.
     
  11. Sep 4, 2011 #10

    I like Serena

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    Well, there's shuffling and shuffling.
    The correcting factor was because you were double-counting certain combinations.
    That's not the case here.
    I'll try to explain using your examples.

    a) The permutions of 4 different specific numbers is simply 4! = 24

    b) In this case you'd be double-counting the 8, so the number is 4! / 2! = 12
    Actually, you'd have 4! / (1!1!2!).
    Where did you get the last 2!? And how did you get 10?

    c) Double-counting 7, and double-counting 8.
    However, 7 and 8 are specific numbers, and not counted once for the one, and another time for the other number.
    So: 4! / (2!2!) = 6.

    d) Yes, this is correct.
     
  12. Sep 4, 2011 #11

    Ray Vickson

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    If b(n,k) = C(n,k)/2^n = probability of getting k heads in n tosses (a binomial probability), then you want b(1,n) + b(n-1,n) = 2*n/2^n.

    RGV
     
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