You throw a fair die until you got 2 sixes. What is the chance of stopping after the 12th throw?
My interpretation of the question is that 1 die is thrown each time. Then you require exactly one 6 in the first 11 throws and then a 6 on the twelveth throw:
$\displaystyle ^{11} C_1 \, \left(\frac{1}{6}\right) \, \left(\frac{5}{6}\right)^{10} \, \left(\frac{1}{6}\right) = \left(\frac{11}{36}\right) \, \left(\frac{5}{6}\right)^{10}$