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How do you calculate the chance of rolling a certain number on a six sided dice once (assuming the rolls are completely random) in a certain number of rolls? Ex: rolling a dice 6 times won't get you a 100% chance of rolling a six.

Guest Feb 19, 2015

#3**+10 **

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}\\\\$$

Melody Feb 19, 2015

#1**+5 **

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 Feb 19, 2015

#2**+5 **

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)

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.)

.

Alan Feb 19, 2015

#3**+10 **

Best Answer

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}\\\\$$

Melody Feb 19, 2015