Results 1 to 4 of 4

Math Help - Quick question about probability of success with dice rolls.

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    2

    Quick question about probability of success with dice rolls.

    Hi, this is my first post so bear with me if my formatting is a little unorthodox. I recently was presented with this question by a friend (in relation to a board game, I believe), and I'm not entirely sure I have the correct answer.

    Problem: Imagine that you are rolling an 8 sided die to determine an outcome. If you roll 7 or 8, the outcome succeeds. If you roll 1-6 the outcome fails, but you are allowed to re-roll once. That is to say, if you roll a 2 (for instance) you will then roll the 8 sided die again to give yourself another chance at success (whereas you would not roll again if you got a 7 or 8). The question is, taking the re-roll into account, what is the overall probability of success?

    My understanding is that it works like this: In the initial roll, each possible outcome has a 1/8th probability, giving a total of 2/8ths probability of success and 6/8ths probability of failure. However the re-roll turns each of those 6 failures into 8 possibilities which each have an overall 1/64th chance of happening (1/8th of 1/8th), giving a 2/64 probability of success to each of the 6 initial failures. Overall that would give a 12/64ths probability of success and 36/64ths probability of failure if a 1-6 was rolled initially. Adding in the 2/8ths probability of success from the initial roll, the total probability is 28/64ths chance of success and 36/64ths chance of failure. Is this correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    Quote Originally Posted by Electricblue View Post
    Overall that would give a 12/64ths probability of success and 36/64ths probability of failure if a 1-6 was rolled initially. Adding in the 2/8ths probability of success from the initial roll, the total probability is 28/64ths chance of success and 36/64ths chance of failure. Is this correct?
    I agree with this.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Hi Elictricblue,

    You have it right, but there is a shorter route to the answer.

    Consider the over-all probability of failure; in order to fail, you have to roll 1 through 6 twice. This happens with probability (6/8)^2. So the probability of success is one minus this, 1 - (6/8)^2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2010
    Posts
    2
    Thanks guys, for your quick responses. And you're right awkward, that's a much faster route to get to that answer! Probability was never my strong suit in school, so I'm always happy to learn an easier way to do it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quick Dice /Probability Question
    Posted in the Statistics Forum
    Replies: 4
    Last Post: November 30th 2011, 05:38 PM
  2. Replies: 4
    Last Post: March 10th 2011, 09:09 AM
  3. Varying Dice Rolls in Order
    Posted in the Statistics Forum
    Replies: 6
    Last Post: June 4th 2010, 05:36 PM
  4. Pair of dice question, P(at least 4 rolls)
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: October 5th 2009, 12:53 AM
  5. Math Dice rolls
    Posted in the Statistics Forum
    Replies: 3
    Last Post: November 30th 2006, 03:35 PM

Search Tags


/mathhelpforum @mathhelpforum