Take three different numbers "a,b,c" between "1" and "12". The probability of obtaining 6 on dice is "\\frac{1}{6}", probability of obtaining other number is "1-\\frac{1}{6}=\\frac{5}{6}" . So probability of obtaining "6" for "a"-th, "b"-th and "c"-th times and obtaining other numbers for other times is "\\left(\\frac{1}{6}\\right)^3\\left(\\frac{5}{6}\\right)^{12-3}=\\frac{5^9}{6^{12}}".
The number of choosing "a, b" and "c" is "\\binom{12}{3}", so the probability of obtaining three 6s is "\\binom{12}{3}\\frac{5^9}{6^{12}}"
Answer: "\\binom{12}{3}\\frac{5^9}{6^{12}}"
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