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The probability of dice coming up 6 twice in twelve rolls

  1. Jul 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Explain the formulas used to obtain the solution for the question above (What is the probability of two tetrahedral dice landing on a 6 if twelve dice are rolled?)

    1 - (3/4)^12 - 12 * ((3/4)^11) * (1/4) = 0.842

    2. Relevant equations
    If we were looking for one dice landing on 6 (out of 12 dice) it would be

    1 - (3/4)^12

    3. The attempt at a solution

    So, I think that I understand that (3/4)^12 is 12 dice that are NOT 6 (unsuccessful outcomes) and that a dice landing on a six would be 1/4 (a successful outcome). This is because a tetrahedral die has 4 sides (3 out of 4 sides would be unsuccessful, while the 1 out of 4 would be successful/ a 6

    Further in the formula, (3/4)^11 is 11 out of 12 dice that don't land on 6 and 1/4 is that remaining dice that land on a 6. This is multiplied by 12 because there are 12 dice and any one can land on 6 (while the other 11 are a non-6).

    Why is 12 * ((3/4)^11) * (1/4) being subtracted from
    1 - (3/4)^12?


    I hope I'm explaining this right! Thank you
     
    Last edited: Jul 29, 2016
  2. jcsd
  3. Jul 29, 2016 #2

    LCKurtz

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    Most people would assume that when you ask for the probability that 2 dice come up 6, you don't mean that you want the probability that at least two dice come up 6. The calculation you have shown calculates the probability that at least two dice came up 6 by taking 1 - probability none come up 6 - probability 1 comes up 6. That is not the probability you asked for but maybe you just posted the question incorrectly.
     
  4. Jul 29, 2016 #3

    Simon Bridge

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    I think that's fair but could be clearer.
    Try to rework it so you can show the reasoning in words.
    Specifically you need to identify what the target probability is: they just say two 6's... is that at least two, at most two, exactly two, what?

    It helps to have a more precise language:
    ie. If X is the number of successes in 12 trials, and p=1/4 is the probability of success in one trial, then you can look for P(X=x) etc.
    Now you have a mathematical language for talking about the dice.

    ie. P(x=12) = p12 = (1/4)12 right?
    You can write P(x=2) = P(x=1) + P(x=2) or P(x>0)=1-(1-p)12 ... see?

    Also: P(x=1) = P(x<2) - P(x=0) ... which should help you answer your question.
     
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