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Dice probability

  1. Sep 23, 2013 #1
    Say you roll a dice twice.You want to calculate the probablity of getting both dice to land on 3. Using the formula for combinations: the total number of combinations is 21. so is the probabilty of getting a 3 on both dice 1 /21 ?
     
  2. jcsd
  3. Sep 23, 2013 #2

    pwsnafu

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    The probability of rolling a three on a six sided die is ##\frac16##. Doing it twice is ##\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}##
     
  4. Sep 23, 2013 #3
    explain

    I don't see the logic
    how is it permuation formula(combination formula ).
    What is wrong with my method?
     
  5. Sep 23, 2013 #4
    Whatever formula you have in mind, but there are 36 possible outcomes: [itex]\{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}[/itex]
    If each of these seems equally likely to you, then the answer is [itex]\frac1{36}[/itex].
     
  6. Sep 23, 2013 #5
    I'm guessing your method was to plug in "6 choose 2", which is the formula that tells you how many ways to pick a pair of people from a collection of 6 people. That doesn't describe the situation you named.
     
  7. Sep 23, 2013 #6

    tiny-tim

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    hi new_at_math! :smile:
    no, you're completely misunderstanding what combinations are for :redface:

    21 is the number of different results you can get from two seven-sided dice if you're not allowed doubles …

    12 13 14 15 16 17 23 24 25 26 27 34 35 36 37 45 46 47 56 57 67 …

    start again: write out the possible combinations for a 3 (you did mean 3-total?) :smile:
     
  8. Sep 23, 2013 #7
    I get it now it was a permutation with repetition;my bad.
     
  9. Sep 23, 2013 #8
    I don't believe any formula with the word "permutation" or the word "combination" is an effective way to approach this problem.
     
  10. Sep 23, 2013 #9

    pwsnafu

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    No. Permutation is a rearrangement of a collection of objects.
    Your example is a Bernoulli trial.
     
  11. Sep 24, 2013 #10

    HallsofIvy

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    You roll a die twice. (Or you roll two dice.) "Dice" is the plural of "die".


    Another way of looking at this is that if "A" and "B" are independent events, then the probability of "A and B" is the probability of A times the probability of B.

    There are 6 faces on a die, one of which is a "3". As long as the faces are all equally likely to come up, the probability of a "3" is 1/6. The second roll of the die is independent of the first so the probability that both will come up "3" is (1/6)(1/6)= 1/36.
     
    Last edited: Sep 24, 2013
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