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# Dice Probability Over Multiple Rolls

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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
+91510
+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
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#1
+91510
+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
+26412
+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)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.)

.

Alan  Feb 19, 2015
#3
+91510
+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

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