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Probability question related to dice

  1. Oct 20, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A die is thrown n times(n being odd). The probability that even face turns odd number of times is

    2. Relevant equations

    3. The attempt at a solution
    The probability that even face appears is 1/2 and odd face appears is 1/2. From any of the n throws, the even face can appear 1, 3, 5 ..... or n times. Thus the required probability is
    [itex]\dfrac{1}{2^n} [^nC_1+^nC_3+^nC_5+............+^nC_n][/itex]

    But I don't know how to sum this series.
     
  2. jcsd
  3. Oct 20, 2013 #2
    Multiply and divide by 2. Then investigate the expansion of ##(1+x)^n## and ##(1-x)^n##.
     
  4. Oct 21, 2013 #3

    haruspex

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    What relationships can you find between that probability and the probability that even faces appear an even number of times?
     
  5. Oct 21, 2013 #4

    utkarshakash

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    They both add up to 1. That's what I can think.
     
  6. Oct 21, 2013 #5

    haruspex

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    Yes, they add up to 1, but there's another relationship. Try rephrasing one of the two conditions in terms of probability of a number of odd faces. Can you see a useful symmetry?
     
  7. Oct 21, 2013 #6

    tiny-tim

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    hi utkarshakash! :smile:

    as a general rule, if you're stuck, try a simple case and see if you can see a pattern

    eg what is the answer for n = 3? :wink:
     
  8. Oct 22, 2013 #7

    utkarshakash

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    I am not able to make out what is wrong in my original solution. For eg- If I take n=3 (as suggested by tiny-tim) then I have to find the probability that even face appears either 1 or 3 times.If it appears 1 times out of 3 throws I can write 3C1(1/2)^3. SUppose instead that if it appears 3 times out of 3 throws, the probability will be 3C3(1/2)^3. Adding the 2 probabilities should give the correct answer.
     
  9. Oct 22, 2013 #8

    tiny-tim

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    hi utkarshakash! :smile:

    ah, but your original question was …
    … and that's what we've all been talking about!
    and that answer is … ? :wink:
     
  10. Oct 22, 2013 #9

    utkarshakash

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    [itex](1+x)^n-(1-x)^n=2(C_1x+C_3x^2......) \\
    2^{n-1}=C_1+C_3+C_7......[/itex]
    So the answer should be [itex]2^{n-1}/2^n[/itex] which unfortunately is incorrect.

    The correct answer is [itex]\dfrac{1.3.5.7.........(2n-1)}{2^n n!}[/itex]
     
  11. Oct 22, 2013 #10

    haruspex

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    If that's the answer then I don't think I have understood the question.
    The reasoning I was trying to lead you towards is:
    - prob of odd number of even throws equals prob of odd number of odd throws, by symmetry
    - prob of odd number of odd throws equals prob of even number of even throws since they are the same event
    - prob of odd number of even throws and prob of even number of even throws add to 1
    Putting those three together we get 1/2.
    So I strongly suspect the problem has been misinterpreted somewhere along the line. Is it the exact wording? A translation? Should it be 2n throws?
     
  12. Oct 22, 2013 #11

    utkarshakash

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    Oh! I'm really sorry as I've made a mistake in posting the correct answer. Actually the correct answer for this question is 1/2 and the answer posted here is the answer of some other question. I apologize for any inconvenience that may have occurred to you.
     
  13. Oct 22, 2013 #12

    Ray Vickson

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    Since people have already told you the answer, I might as well explain one of the other approaches that was suggested to you, but which you seemed to ignore. Let ##S_E = \sum{k} C(n,2k)## and ##S_O = \sum_{k} C(n,2k+1)## be the even and odd sums. We have
    [tex]0 = (1-1)^n = C(n,0)-C(n,1)+C(n,2)- \cdots = S_E - S_O [/tex]
    and
    [tex]2^n = (1+1)^n = C(n,0) + C(n,1) + C(n,2) + \cdots = S_E + S_O.[/tex]
    Therefore, ##S_E = S_O = 2^{n-1}##.
     
  14. Oct 22, 2013 #13

    utkarshakash

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    I did not ignore it. I followed this approach to reach at the answer. You should read my previous post.
     
  15. Oct 22, 2013 #14

    tiny-tim

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    another way to show that the odd sum equals the even sum … which is clearer if you use 2n+1 instead of n … is by using the symmetry of nCi:

    2n+1C2i+1 = ∑ 2n+1C2(n-i) :wink:
     
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