Please Help with Probabilities Formula for the following problem....

Discussion in 'After Hours Lounge (Off Topic)' started by Mark Giles, Apr 10, 2007.

  1. Mark Giles

    Mark Giles Second Unit

    Joined:
    Aug 30, 2002
    Messages:
    272
    Likes Received:
    0
    Simple....I have three 6-sided dice. I want to know the total number of unique results I can obtain by repeatedly throwing all three dice.

    111
    112
    113
    114
    115
    116
    121
    etc, etc....

    I'm guessing it's 6^3, but can someone confirm and prove this?

    Thanks for you help! [​IMG]
     
  2. betooz

    betooz Agent

    Joined:
    Dec 11, 2006
    Messages:
    41
    Likes Received:
    0
    6^3 is correct
     
  3. Mark Giles

    Mark Giles Second Unit

    Joined:
    Aug 30, 2002
    Messages:
    272
    Likes Received:
    0
    Do you have anything I can use to support this? I want to be able to explain it to my friend too.
     
  4. Mark Giles

    Mark Giles Second Unit

    Joined:
    Aug 30, 2002
    Messages:
    272
    Likes Received:
    0
    Ok, nevermind. I figured it out. Every dice increases the power.

    One die would give you 6 possibilities or 6^1

    Two would give you 6^2

    and so on...
     
  5. Brian D H

    Brian D H Second Unit

    Joined:
    Sep 2, 2004
    Messages:
    453
    Likes Received:
    0
    This one's easy. Any aging role-playing gamers could explain this one (and draw the bell curve) in their sleep.

    Range 3 to 18
    Center of bell curve 10.5
    etc.
    etc.
    etc.

    (God, I'm a geek.)
     
  6. andrew markworthy

    Joined:
    Sep 30, 1999
    Messages:
    4,762
    Likes Received:
    11
    The formula is simply n1 x n2 x ..... where n is the number of choices on each die (this allows for difference sided dice being used in the same process). The fun sets in when you have different biases in individual dice. And this is a major reason why I was so glad to give up probability and stick to statistics.
     
  7. Mark Giles

    Mark Giles Second Unit

    Joined:
    Aug 30, 2002
    Messages:
    272
    Likes Received:
    0


    So let me ask this then. Lets say there were six 6-sided dice and you can toss any number of dice you want, as long as you tossed at least one.

    Based on your formula, the amount of possible outcomes would this be....

    n=6

    (n) + (n1 x n2) + (n1 x n2 x n3) + (n1 x n2 x n3 x n4) + (n1 x n2 x n3 x n4 x n5) + (n1 x n2 x n3 x n4 x n5 x n6)

    ?? Is that correct?


    And no, Im not cheating on a test.... [​IMG]
     
  8. Holadem

    Holadem Lead Actor

    Joined:
    Nov 4, 2000
    Messages:
    8,967
    Likes Received:
    0
    Yes but your notations are a bit sloppy there and mostly pointless with your n=6 condition.

    --
    H
     
  9. Sami Kallio

    Sami Kallio Screenwriter

    Joined:
    Jan 6, 2004
    Messages:
    1,035
    Likes Received:
    0
    x[n] = 6^n + x[n-1]
     
  10. Gregg Loewen

    Gregg Loewen Video Standards Instructor, THX Ltd.
    Insider

    Joined:
    Nov 9, 1999
    Messages:
    6,373
    Likes Received:
    32
    Location:
    New England
    Real Name:
    Gregg Loewen
    sounds like you need help with your home work ??
     
  11. Mark Giles

    Mark Giles Second Unit

    Joined:
    Aug 30, 2002
    Messages:
    272
    Likes Received:
    0
    No, actually a co-worker asked me the original question earlier in the day and I wanted to be able to explain it to her. The second one was asked because that's the way she first presented the question to me. My curiosity was taking over. I took Prob. & Stats years and years ago and remember it as well as I remember the one semester of Japanese as a freshman.

    Not to mention, I'm bored today.... [​IMG]
     
  12. Chris Lockwood

    Chris Lockwood Producer

    Joined:
    Apr 21, 1999
    Messages:
    3,215
    Likes Received:
    0
    It's actually not 6 to the 3rd power, since the original question said "unique".
     
  13. Sami Kallio

    Sami Kallio Screenwriter

    Joined:
    Jan 6, 2004
    Messages:
    1,035
    Likes Received:
    0
    It's 6^3 since 1,1,2 and 2,1,1 are unique.
     
  14. DaveF

    DaveF Moderator
    Moderator

    Joined:
    Mar 4, 2001
    Messages:
    17,493
    Likes Received:
    1,382
    Location:
    One Loudoun, Ashburn, VA
    Real Name:
    David Fischer

    What's your formula for, Kami?

    x(n) = 6^n + x(n-1) = (3 dice, 6 sides)
    x(n) = 6^3 + 6^2 + 6^1 + 6^0 + 6^(-1) + ...

    or maybe you just meant
    x(n) = 6^3 + 6^2 + 6^1 + 6^0 = 259

    ???

    As for unique: is 1,1,2 = 1,2,1 = 2,1,1 or are these unique? If I'm rolling identical dice, I can't distinguish these outcomes. They are not unique in typical gameplaying situations.
     
  15. Al.Anderson

    Al.Anderson Cinematographer

    Joined:
    Jul 2, 2002
    Messages:
    2,535
    Likes Received:
    68
    Real Name:
    Al
    The "unique" in the orginal question threw me off too. Now that we get the question (which I think is better phrased as non-unique), I think what you want is a Permutation: n!/(n-r)!. Or in this case 6!/(6-3)! = 120.
     
  16. cafink

    cafink Producer

    Joined:
    Apr 19, 1999
    Messages:
    3,043
    Likes Received:
    36
    Real Name:
    Carl Fink
    Although Mark asked for the number of "unique results" in his original post, he also listed both 112 and 121 in his list, so I imagine that 6^3 is the answer he's looking for.
     
  17. Sami Kallio

    Sami Kallio Screenwriter

    Joined:
    Jan 6, 2004
    Messages:
    1,035
    Likes Received:
    0
    As I understood, they are unique. Just like a lock on your suitcase has three numbers you roll and the unique combinations are 10^3. Dice #1 and dice #2 for example are not interchangeable. If they are then 6^3 is not the answer.
     
  18. Haggai

    Haggai Producer

    Joined:
    Nov 3, 2003
    Messages:
    3,883
    Likes Received:
    0

    Yes, if the three dice are distinguishable from each other, the answer is 6^3 = 216.

    If the dice are NOT distinguishable, i.e. if the outcomes of 112 and 121 and 211 are all considered to be the same, then you need to account for those repeat possibilities in the right way, which I think can be done as follows:

    EDIT: Screwed this up the first time. Corrections below.

    In the total of 216, multiple-counting of the same possibility can happen in two different ways.

    One is if two different numbers show up on the three dice, e.g. the 112/121/211 possibility mentioned before. We also have to take into account the 122/212/221 option here, since we are classifying this as "what if the numbers 1 and 2 are the only ones that appear." As we can see from this example, this will result in one possibility counting six times (in the 216 total), when we only want it to count twice.

    The other type of multiple counting is if three different numbers show up on the three dice, e.g. 123/132/213/231/312/321. This will result in a sextuple counting, when we only want it to count once.

    So the multiple counted possibilities that we want to throw away are:

    "6 choose 2" (binomial coefficient) ways of getting two different numbers, each of which contributes 4 excess possibilities (e.g. we'll keep 112 and 122, but we want to discard 121/211 and 212/221).

    "6 choose 3" ways of getting three different numbers, each of which contributes 5 excess possibilities (e.g. we'll keep 123, but discard the other 5 permutations)

    The total we want then turns out to be (see the link above for how to calculate the binomial coefficient):

    6^3 - (6 choose 2)*4 - (6 choose 3)*5 = 216 - 15*4 - 20*5 = 56.
     
  19. Mark Giles

    Mark Giles Second Unit

    Joined:
    Aug 30, 2002
    Messages:
    272
    Likes Received:
    0


    Ok, I'm sorry about the confusion in the beginning. 112 and 121 shoudnt be counted twice. So 216 isnt correct. I'll take some time and let your formula sink in Haggai. Thanks a lot for the breakdown!
     
  20. Sami Kallio

    Sami Kallio Screenwriter

    Joined:
    Jan 6, 2004
    Messages:
    1,035
    Likes Received:
    0
    In that case Al's solution is correct, which is basically the same as 6*5*4 = 120

    For the six dice you get 6! = 720
     

Share This Page