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Thread: probability n dice [resolved]

  1. #1

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    Addicted Member dolor's Avatar
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    probability n dice [resolved]

    If three fair standard dice are tossed, what is the probability that the results will be three consecutive integers?
    Last edited by dolor; Jan 26th, 2004 at 01:39 PM.
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  2. #2
    Super Moderator manavo11's Avatar
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    About 32% if this is correct :

    VB Code:
    1. Private Sub Form_Load()
    2. Dim i, j, k, Yes, No As Integer
    3.  
    4. For i = 1 To 6
    5.     For j = 1 To 6
    6.         For k = 1 To 6
    7.             Select Case i + j + k
    8.                 Case 6, 9, 12, 15
    9.                     Yes = Yes + 1
    10.                 Case Else
    11.                     No = No + 1
    12.             End Select
    13.         Next
    14.     Next
    15. Next
    16.  
    17. MsgBox ((Yes) / (Yes + No)) * 100 & " %"
    18. End Sub


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  3. #3
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    sorry but thats wrong, :S
    why ??!
    9 = 2+3+4
    but also
    9 = 2+2+5
    see ??!

    lets not use code,
    use math :P

    the answer will be
    (1/6) ^ 2,
    only in one case:
    if its loop like
    meaning:
    5,6,1
    or
    6,1,2
    is alright,
    thats the only exception that we should exclude

    so its: (1/6)^2 - (1/6)^3 * 2
    =1/54
    this might seem complex but think about it,
    its not.

    in other words:
    the first dice is is free to be anything from 1 to 4
    ie. chance 4/6
    AND the second one should be the one next to it
    ie. 1/6 chance
    AND (hence we should multiply the chances)
    the thrid should be next to the second which has chance
    1/6

    multiply all chances:

    4/6 * 1/6 * 1/6
    =4/216
    =1/54


    a sure answer !!

  4. #4
    Fanatic Member WorkHorse's Avatar
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    But the order of the dice probably doesn't matter. So 3,1,2 would be consecutive integers. So you would have 3! or 6 possibles for each of the 4 integer sets or 6 * 4 = 24 possible combos. 24/216 = 1/9.

  5. #5
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    well, lets see this case ............
    mmmm............


    ok, a little change to our tree,

    adding the other possibilities:

    x , x+1 , x+2
    x , x+2 , x+1

    x+1 , x , x+2
    x+1 , x+2 , x

    x+2 , x , x+1
    x+2 , x+1 , x

    where x ranges from 1-4,
    each line has probability 1/54 (as calculated previously)

    so the answer on this new assumption is
    6/54
    =1/9

    yeah, work horse is right

  6. #6
    Fanatic Member alkatran's Avatar
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    Possibility of getting a certain event with 3 dice 6*6*6 = 1/216

    There are 6 ways to get 1,2,3; 6 ways to get the next, until we hit 6 (4 different series in total) so 6*4/216.
    6*4/216 = 24/216 = 1/9.

    I didn't read workhorse's post before doing this, but we just did probabilities in math and I had to test it out.
    Don't pay attention to this signature, it's contradictory.

  7. #7
    transcendental analytic kedaman's Avatar
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    6,1,2 isn't consecutive thuogh, so there are only 4 permutations out of 216 that are valid.
    Use
    writing software in C++ is like driving rivets into steel beam with a toothpick.
    writing haskell makes your life easier:
    reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
    To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.

  8. #8
    Super Moderator manavo11's Avatar
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    Oh well, it was just an idea... Even if it was wrong


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  9. #9
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    Kedaman said - 4 permutations are possible our of 216.

    No 4 combinations are possible and each can have 3! permutations.

    In short the correct answer is already there 1/9

    Thus easy thread resolved.

  10. #10
    transcendental analytic kedaman's Avatar
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    fundu: depends how you interpret the question
    Use
    writing software in C++ is like driving rivets into steel beam with a toothpick.
    writing haskell makes your life easier:
    reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
    To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.

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