You throw a fair die until you got 2 sixes. What is the chance of stopping after the 12th throw?

Printable View

- Oct 10th 2008, 01:49 AMmatty888Throw 2 die
You throw a fair die until you got 2 sixes. What is the chance of stopping after the 12th throw?

- Oct 10th 2008, 01:56 AMCaptainBlack
- Oct 10th 2008, 02:13 AMmr fantastic
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:

$\displaystyle ^{11} C_1 \, \left(\frac{1}{6}\right) \, \left(\frac{5}{6}\right)^{10} \, \left(\frac{1}{6}\right) = \left(\frac{11}{36}\right) \, \left(\frac{5}{6}\right)^{10}$ - Oct 10th 2008, 03:52 AMCaptainBlack