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How to calculate probability in dicerolls?

  1. Oct 5, 2008 #1
    How to calculate this certain probability in dicerolls?


    I am writing a game in which AT LEAST a 5 and 4 must be rolled with 2,3,4,5,6 and 7 dice. To calculate this probability I got stuck with the binomial formula nCk*p^k*q^(n-k).
    I am looking for the formula that gives the solution to the question what this probability is.
    I used a Monte Carlo simulation to check my formula but couldn't get close enough.
    I used the formula nPk*p^k*q^(n-k), note the permutational formula, because a 5, then a 4 is different form a 4, then a 5....
    On 7 dice the error was consistent but not to be overlooked.
    How do I get to the right formula?
    thanks for your help!
    Last edited: Oct 5, 2008
  2. jcsd
  3. Oct 5, 2008 #2


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    I'm not sure I understand what you're trying to find. You're rolling some set number of dice (2 to 7; call this number n) which I presume have six sides numbered one through six, and you want to find the probability of getting at least one to land on "4" and at least one other to land on "5"?

    Is that right?

    If so: there are 6^n results, of which 5^n have no 4s, 5^n have no 5s, and 4^n have neither 4s nor 5s. So there are 6^n - 5^n ways of rolling at least one 4, 6^n - 5^n ways of rolling at least one 5, and 6^n - 2 * 5^n + 4^n ways of rolling at least one of each. (The 4^n is added back on because otherwise we'd subtract off the case where neither a 4 nor a 5 is rolled twice.)

    To get the probabilities, divide that by the number of outcomes. My results, to three decimal places:
    2 5.56%
    3 13.9%
    4 23.3%
    5 32.8%
    6 41.8%
    7 50.0%
  4. Oct 5, 2008 #3
    Hi CRGreathouse,

    Yes, you understood the problem correctly and I thank you very much for your solution. This matches my Monte Carlo simulation results. The only thing is, how do I get the formula in "binomial" perspective to this problem. I thought of 1-P(no 5 AND no 4), this translated with a binomial formula would come to 1-(5/6^n)-(5/6^n)+(4/6^n), right?
    thanks Marc
    Last edited: Oct 6, 2008
  5. Feb 26, 2009 #4

    This might not be the right place to add this question but it does involve dice rolls.

    Suppose 2 six-sided dice are rolled in a game where a result of 1,6 is not distinguished from 6,1.

    How would that be incorporated into a probability calculation?

    E.g. a) What is the probability that the sum of the 2 dice is equal or less than 7
    b) Is the probability of getting 2 sixes still 1/36 and if so how has the condition of 1,6 being the same for the game as 6,1 been incorporated into the calculation

  6. Feb 26, 2009 #5


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    If the first roll is a 6, there's only one way for the other die to come up to give a 7 or less. If the first roll is a 5, there are two ways to get a 7 or less. ... If the first roll is a 1, there are six ways to get a 7 or less.

    Add these up and divide by 36.
  7. Feb 26, 2009 #6
    Thanks but I'm not sure you've met all the boundary conditions necessary in this special case.

    The 2 dice dice are rolled simultaneously and not sequentially.

    i.e. your calculation seems identical to the one where a 1 and 6 roll and a 6 and 1 are distinguishable

    but suppose all dice are identical, surely they cannot be distinguished them when they are rolled simulataneously
    and a result of 6,1 will look exactly the same as 1,6
  8. Feb 26, 2009 #7


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    The result is exactly the same.
  9. Feb 27, 2009 #8
    Sorry, I haven't phrased my question very clearly.
    Bear with me while I try again.

    Player A has 2 dice. Player B has 1 dice. Dice are 6 sided,unbiased and identical. Both players record their throws in sets of 2 dice i.e. player B has to roll 2 dice one after another and records the results while player A rolls both dice simultaneously and records the results. The table shows their results. For simplicity I've not included values above 3 and excluded repetitions of the same result.

    Throw..............Throw1; Throw 2

    Whilst (B) is able to distinguish a 2,1 from a 1,2 (A) is unable to, hence his shorter table.

    My query is - what implications are there for this difference
    In which probability calculation would the result be different and in which will the result be the same if data is used from (A) and (B) separately.

    Thanks in advance
  10. Mar 2, 2009 #9


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    If you distinguish "1, 2" from "2, 1", the possibilities are:
    "1, 2": 1/36
    "2, 1": 1/36
    "1, 2" or "2, 1": 1/36 + 1/36 = 1/18

    If you don't, there's a 1/18 chance of rolling a 1 and a 2. It's the same result. The dice don't know if you can tell them apart or not. The dice don't know what order they're rolled in.
  11. Mar 3, 2009 #10
    Yes, I understand all that but what I'm asking is still different.

    I am trying to figure in what probability calculation would the difference in (A) and (B)'s data (as in my previous post) be significant and if so why.
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