# Throw 2 die

• Oct 10th 2008, 01:49 AM
matty888
Throw 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 AM
CaptainBlack
Quote:

Originally Posted by matty888
You throw a fair die until you got 2 sixes. What is the chance of stopping after the 12th throw?

The chance of 2 sixes on a single throw is 1/36.

To stop exactly after 12 throws requires that the first 11 throws are not double 6 (probability per throw 35/36) and that the last throw is double six (probability 1/36)

RonL
• Oct 10th 2008, 02:13 AM
mr fantastic
Quote:

Originally Posted by matty888
You throw a fair die until you got 2 sixes. What is the chance of stopping after the 12th throw?

Quote:

Originally Posted by CaptainBlack
The chance of 2 sixes on a single throw is 1/36.

To stop exactly after 12 throws requires that the first 11 throws are not double 6 (probability per throw 35/36) and that the last throw is double six (probability 1/36)

RonL

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 AM
CaptainBlack
Quote:

Originally Posted by mr 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:

$\displaystyle ^{11} C_1 \, \left(\frac{1}{6}\right) \, \left(\frac{5}{6}\right)^{10} \, \left(\frac{1}{6}\right) = \left(\frac{11}{36}\right) \, \left(\frac{5}{6}\right)^{10}$

Sorry I was reading the question in a parallel reality(Tongueout)

CB