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Thread: Probability of throwing n different numbers in m throws of a dice

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    Probability of throwing n different numbers in m throws of a dice




    Folks

    I know how to calculate the probability of throwing the same number in every throw of a dice (die, if you prefer!), and I know how to calculate the probability of throwing a different number in each throw of a dice. But how do I calculate the probability of throwing n different numbers in m throws of a dice?

    Just to be clear, it doesn't matter what the numbers are or how many times each number is thrown so two ones and three twos in five throws will be equivalent to four threes and one four in the same number of throws, for example.

    Thanks

    Remster

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    Re: Probability of throwing n different numbers in m throws of a dice


    I don't know of a single formula, and my guess would be that there isn't one. However, the probabilities are easy to calculate using an iterative formula. If p(n,m) is the probability of getting exactly n different numbers after m throws, then just before the last throw there are either n numbers and you get another of the ones you already have, or there are n-1 numbers and you get a different one. So p(n,m) = p(n,m-1)*n/6 + p(n-1,m-1)*(6-(n-1))/6.
    To start off, p(0,m) = 0, p(n,0) = 0 and p(1,1) = 1. Set it up in Excel, put in the formula, and copy in all directions. Post back if you need to see an actual sheet. kat

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