Dice Probability Over Multiple Rolls

I like your graph Alan.   :)

what alan has done is  

$$\\ ^nC_1\;\left( \frac{1}{6}\right)^1\;\left(\frac{5}{6}\right)^{n-1}\\\\ $you can take this another step$\\\\ $Say you want your particular number to appear exactly m times, where$\;\; m $Then the probability would be$\\\\ ^nC_m\;\left( \frac{1}{6}\right)^m\;\left(\frac{5}{6}\right)^{n-m}\\\\$$

P (NOT rolling a particular number in n trials) = $$\left(\frac{5}{6}\right)^n$$

so

P(Rolling a particular number in n trials ) =      $$1-\left(\frac{5}{6}\right)^n$$

Melody's answer assumes the probability is of the number appearing at least once.

To calculate the probability of it appearing exactly once, use (n/6)(5/6)^n-1.

The two probabilities are compared as a function of n in the graph below:

(The final sentence in the question suggests that Melody's answer is what was being sought.)

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