Results 1 to 4 of 4

Math Help - Throw 2 die

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    132

    Throw 2 die

    You throw a fair die until you got 2 sixes. What is the chance of stopping after the 12th throw?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by matty888 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by matty888 View Post
    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 View Post
    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:

    ^{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}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by mr fantastic View Post
    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:

    ^{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

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. probability of dice throw
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 21st 2010, 10:05 AM
  2. probability: free throw shooter
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: April 11th 2009, 04:54 PM
  3. Prob distribution of a throw of the die
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: December 22nd 2007, 03:16 AM
  4. Proving Convarience of a Dice Throw
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 26th 2007, 12:40 PM
  5. Throw me some Formulars to work out
    Posted in the Algebra Forum
    Replies: 6
    Last Post: August 20th 2007, 01:33 PM

Search Tags


/mathhelpforum @mathhelpforum