1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic Probability Question—Rolling Die

  1. Apr 5, 2012 #1
    Basic Probability Question—Rolling Die!!!

    1. The problem statement, all variables and given/known data
    Suppose you roll a fair six-sided die repeatedly until the first time you roll a number that you have rolled before. A) for each r=1,2,... calculate the probability pr that you roll exactly r times.


    2. Relevant equations



    3. The attempt at a solution
    p8=p9=...=pinfinity=0
    p1=0
    p2=1/6
    p3=5/18
    I think I'm doing this the long way and the wrong way. I know these answers are right because they make sense. But it's getting to hard to count. I need insight on how I should think about this problem. It is basic probability from pitman section 1.6 problem 6. This is not homework. I am just reviewing. Please help! Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 5, 2012 #2
    Re: Basic Probability Question—Rolling Die!!!

    Actually I just figured it out. It's similar to the birthday problem. Just create trees for multiplication rule for n events and you will get it.

    p4=(5/6)*(4/6)*(3/6)
    p5=(5/6)*(4/6)*(3/6)*(4/6)
    p6=(5/6)*(4/6)*(3/6)*(2/6)*(5/6)
    p7=(5/6)*(4/6)*(3/6)*(2/6)(1/6)*(6/6)
     
  4. Apr 5, 2012 #3
    Re: Basic Probability Question—Rolling Die!!!

    Why isn't p3 = 5/36?
     
  5. Apr 5, 2012 #4
    Re: Basic Probability Question—Rolling Die!!!

    There's two ways of doing it. Short way: given that you rolled a number, the probability of getting that number on the second roll is 1/6. But since we didn't get that number, we have 5/6. Now, the probability of getting one of those two numbers is 2/6. So we multiply 5/6*2/6=10/36.

    The way I did it first was the dumb way. p(1|12)+...+P(1|16),...P(6|61)+...+P(6|65);P(1|21)+...+P(1|61),...,P(6|16)+...+P(6|56).
    We notice that p3=(5x6+5X6)/6^3=5/18. You're just counting all possible combinations!
     
  6. Apr 6, 2012 #5
    Re: Basic Probability Question—Rolling Die!!!

    Oh, it's until you rolled ANY number that was already rolled. I misread it needs to be the same number as the first one rolled, so I erroneously attributed it a probability of 1/6.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic Probability Question—Rolling Die
  1. Probability in a die (Replies: 0)

Loading...