1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Die tossing problem

  1. Jul 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A fair die is cast until a 6 appears. What is the probability that it must be cast more than five times.

    2. Relevant equations
    The die is fair, hence like most of the problems I can assume equally likely outcomes.
    [itex]P(A^c)=1-P(A)[/itex] for any event A

    3. The attempt at a solution
    Theoretically, a 6 may never come up. It should be better to calculate the complement of the event.
    The complement, I think, is the event in at least 5 tosses, a 6 occurs. So a 6 may occur in the 1st toss or the 2nd toss or... or the 5th toss.

    So I have [itex] \frac 16 + \frac 56 \cdot \frac 16 + .... + (\frac 56)^4 \cdot \frac 16 [/itex]
    Then factoring out [itex]\frac 16[/itex] and writing the probabilities as a summation, I have
    [itex] \frac 16 \cdot \sum_{k=1}^5 (\frac 56)^{5-i} [/itex]

    Is this correct? I don't have the answer. I suspect this can be derived from a probability distribution. If my suspicions are correct, which probability distribution?
  2. jcsd
  3. Jul 6, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Isn't this just the probability that none of the first five is a 6?
  4. Jul 6, 2014 #3
    Right you are. In my originial post, I should have wrote [itex] 1 - (\frac 16 \cdot \sum_{i=1}^5(\frac 56)^{5-i})[/itex].

    Which is approximately 0.4.

    What you suggested (and I agree with) is [itex]P(Event)= (\frac 56)^5[/itex]. Which is the same number.

    This seems to much like a coincidence. Any thoughts?
  5. Jul 6, 2014 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I didn't really look at the LaTeX in your post. I just looked at the problem statement and immediately thought that the problem should be equivalent to finding the probability of five consecutive results in the 1-5 range. Then I read enough of your post to see that you appeared to be solving a more complicated problem. For a moment I thought that I must have misunderstood the problem, so I thought about it some more, but I couldn't find anything wrong with my first thought.

    My interpretation of the problem is this: Suppose that we roll the die over and over until we get a six. Then we write down the sequence we got. Repeat this indefinitely. The problem is asking what fraction of the sequences we're writing down will not have a 6 among the first five numbers. If there's less than five numbers in the sequence, we can just pad it with more sixes so that we have a 5-digit sequence. The frequency of 5-digit initial sequences without sixes has to be equal to the probability that an arbitrary 5-digit sequence doesn't contain any sixes.
  6. Jul 6, 2014 #5


    User Avatar
    Science Advisor
    Homework Helper

    Your sum is the first few terms of a geometric series. It's not hard to find the sum as a fraction.
  7. Jul 6, 2014 #6


    User Avatar
    Homework Helper

    I just want to add that your summation is unnecessarily more complicated than it should be. Given
    [tex] \frac 16 + \frac 56 \cdot \frac 16 + .... + (\frac 56)^4 \cdot \frac 16 [/tex]

    This summation expression that you gave
    [tex]\frac 16\cdot \sum_{k=1}^5 (\frac 56)^{5-i} [/tex]
    (by the way, you used k and i as your dummy variable accidentally)

    is equivalent to
    [tex]\frac 16\cdot \sum_{k=0}^4 (\frac 56)^{k} [/tex]

    which I'm sure you would agree is easier to read.
  8. Jul 7, 2014 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    There is no coincidence here; you are just computing the right-tail of the geometric distribution. Google is your friend. See, eg., http://www.math.uah.edu/stat/bernoulli/Geometric.html .
    Last edited: Jul 7, 2014
  9. Jul 7, 2014 #8


    User Avatar
    Science Advisor

    The probability you do NOT get a 6 on any one throw is 5/6. What is the probability you will NOT get a 6 on any of 5 consecutive throws?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted