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Dice probability

  1. Apr 21, 2009 #1
    Not sure if I should ask this here but

    I'm trying to find the probabilty of consecutive numbers on tossing three dice. eg 1 2 3 , 4 5 6 etc

    My workings so is

    nPr = 3P3 (how many permutaions of 3 numbers in order)

    = n!/(n-r)! = 3!/0! = 6

    What
    I did next is calulate all possible outcomes of three dice = 6 x 6 x 6 = 216

    Therfore the probabilty of consecutive numbers on three dice = 6/216 = 1/36

    Does this look right ?
    I'm not quite sure if the permutaion calculation is right or if i have to multiply it by six...
    regards
    Brendan
     
  2. jcsd
  3. Apr 21, 2009 #2

    CRGreathouse

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    Science Advisor
    Homework Helper

    Yep.

    Going up: 3 ways of choosing the first die, other two are fixed: 3/216
    Going down: 3 ways of choosing the first die, other two are fixed: 3/216
     
  4. Apr 21, 2009 #3
    Thanks for your reply.
    I've created an function using R to simulate rolling the dice 1000 times and I needed to compare it with the calculated probability.

    Here' my code

    RollDie=function(n) sample(1:6,n,replace=T) #function to roll 1 dice

    result=0 #results hold the number of successive throws


    for (x in c(1:1000) ) #loop 1000 times
    {
    die1 = RollDie(1) #roll first die

    die2 = RollDie(1) #roll second die

    die3 = RollDie(1) #roll third die


    if (((die1 == die2-1) & (die2 == die3-1)) || ((die1 == die2+1) & (die2 == die3+1))) #if dice in consecutive order add to result
    {
    result = result+1

    }
    else
    {}
    }

    print(result) #print number of successive values

    P3succesivenumbers=(result/1000) #calculate probability

    print(P3consecutivenumbers) #print probability



    And the Probabilty came out as:

    print(P3consecutivesivenumbers) #print probability
    [1] 0.029


    Which is pretty close to 1/36 = .0277

    thanks
    Brendan
     
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