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!

All six faces of a die

  1. Apr 9, 2005 #1
    Does anyone know how to determine the probability that the sixth side of a fair die will appear on the Nth consecutive throw.
     
  2. jcsd
  3. Apr 9, 2005 #2

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I suggest you sit down and try to reason this through, but it's just

    [tex]P(N)=\frac{1}{6^N}[/tex]
     
  4. Apr 10, 2005 #3
    The problem I was trying to ask was " You throw a die until all six faces have shown up and then you stop." What is the probability that you stop on the Nth throw. I believe that P(N=6)=6!/(6^6) and P(N<6)=0 but I am confused about the probability of stopping for N>1.
     
  5. Apr 10, 2005 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't expect that to have a simple answer. It's easy enough to answer:

    "What is the probability of getting u 1's, v 2's, w 3's, x 4's, y 5's, and z 6's in N (= u+v+w+x+y+z) throws"

    But then to get the answer to your problem, you have to add up a lot of individual cases. :frown:
     
  6. Apr 10, 2005 #5
    Counting the number of ways that all the faces can come up on n rolls is the same as counting the number of onto functions from {1, 2, ... n} to {1, 2, 3, 4, 5, 6}, since each time you roll the die you effectively map the number of the roll to one of the faces on the die.
     
    Last edited: Apr 10, 2005
  7. Apr 10, 2005 #6
    There is a formula for counting the number of onto functions.
     
  8. Apr 13, 2005 #7

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    I think a good strategy would be counting the number of ways to get at least 1 of each of 5 numbers in N-1 throws and exactly 0 of the last. This way order doesn't matter, which simplifies the problem (I think/hope). This should be exactly six times P(N).
     
    Last edited: Apr 13, 2005
  9. Apr 13, 2005 #8

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Musing aloud

    Let [tex]F_k(n)[/tex] represent the number of ways a die with [tex]k[/tex] equiprobible faces can land once on all but one faces at least once, with the last face never coming up, out of [tex]n+k-1[/tex] throws. Clearly the probability of the desired event is [tex]P(n)=\frac{F_6(n-6)}{6^n}[/tex] as long as n>=6.

    [tex]F_6(0)=6\cdot5![/tex], since there are 5! ways for the 5 objects to be chosen once each, and 6 ways to choose which element is last. This brings us to P(6)=6!/6^6=5/324, as LittleWolf pointed out.

    In fact, in general [tex]F_k(0)=\frac{k!}{k^k}[/tex], since each of the faces other than the last must be landed on exactly once.

    [tex]F_2(n)=2[/tex], since either all the results are 1 or all the results are 2.

    [tex]F_3(n)=3\left(\sum_{l=1}^{n+1}{n+2\choose l}\right)=3\left(\sum_{l=0}^{n+2}{n+2\choose l}-{n+2\choose0}-{n+2\choose n+2}\right)=3\left(2^{n+1}-2\right)[/tex]

    Is there some good pattern, or some Bell-related OEIS sequence I'm missing here?
     
  10. Apr 13, 2005 #9
    I said there was already a formula for counting the number of onto functions. Given that LittleWolf was asked the problem, he probably already has that formula. I looked it up and this is it, using ^ for exponentiation and C(a, b) for "a choose b":

    The number of onto functions from a set of size m to a set of size n is equal to the sum for i = 0 to n of ((-1)^i * C(n, i) * (n - i)^m).

    So to find the answer to LittleWolf's question, you substitute N for m and 6 for n in that expression.
     
  11. Apr 14, 2005 #10

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    First: Thanks for the formula, I couldn't think of that one. Actually, I'm not sure if I even knew of it before... it looks vaguely familliar, but no more. If you don't mind I'll [tex]\TeX[/tex] it so I can follow it a little better....

    [tex]\sum_{i=0}^{n}(-1)^i{n\choose i}(n-i)^m=\sum_{i=0}^{n-1}(-1)^i{n\choose i}(n-i)^m[/tex]

    [tex]\sum_{i=0}^{6}(-1)^i{6\choose i}(6-i)^N= {6\choose 0}6^N-{6\choose 1}5^N+{6\choose 2}4^N-{6\choose 3}3^N+{6\choose 4}2^N-{6\choose 5}= 6^N-6\cdot5^N+15\cdot4^N-20\cdot3^N+15\cdot2^N-6[/tex]

    OK... this is the number of ways to have all 6 faces show up by the Nth turn. The probability of it turning up on the Nth turn is then
    [tex]P(N)=6^{-N}\left((6^N-6\cdot5^N+15\cdot4^N-20\cdot3^N+15\cdot2^N-6)-(6^{N-1}-6\cdot5^{N-1}+15\cdot4^{N-1}-20\cdot3^{N-1}+15\cdot2^{N-1}-6)\right)[/tex]

    Right?
     
  12. Apr 14, 2005 #11
    Looks right to me.

    I think the reason it might be familiar to you is that (so says my book) the summation is a Stirling number of the second kind, S(m, n), multiplied by n!.
    http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html
    The Stirling numbers count the number of ways to divide a set of size m into n nonempty sets, so to count functions you want to have the order of the n sets important so you multiply by n!.
     
    Last edited: Apr 14, 2005
  13. Apr 14, 2005 #12

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    I thought that Stirling numbers (of the second kind; I've never used Stirling numbers of the first kind) should come up, but I didn't actually follow through with that.

    Thanks for the formula!
     
  14. Apr 21, 2005 #13
    Modifying your notation, the probability P(N,6) is the cumulative distribution function. The probability that the last face shows up on the Nth throw is 6*P(N,5). If you consider the recursion formula for Stirling#s of the 2nd kind, you can see why P(N,6) is the CDF. Thanks to everyone for clearing up this problem.
     
  15. Apr 21, 2005 #14
    a related problem is
    what is the the expected number of throws to get a six on a fair dice

    this was raised at
    https://www.physicsforums.com/archive/topic/t-45262_The_expected_value_of_a_Geometric_Series.html
    which seems to be closed now.

    yes the answer is 6 but how do you prove it.

    if p=1/6 and q=1-p
    x is the number of throws

    then E(x) is sum(x.P(X=x)) = sum(x.p.q^(x-1))
    the sums are for x=1 to infinity
    i.e. E(x) is p(1 +2q +3q^2 +4q^3+...)

    let S= E(x)/p = 1 +2q +3q^2 +4q^3+...

    rearranging the sum as
    S=
    1 + q + q^2 + q^3 + ...
    + q + q^2 + q^3 + ..
    + q^2 + q^3 +...
    then rearrange these lines as

    S= 1 ( 1 + q + q^2 + q^3 + ...)
    +q (1 + q + q^2 + q^3 + ...)
    +q^2 ( 1 + q + q^2 + q^3 + ...)
    ....

    so S= (1 + q + q^2 + q^3 + ...)*(1 + q + q^2 + q^3 + ...)

    i.e S= (1 + q + q^2 + q^3 + ...)^2
    this geometric series sums to 1/(1-q)

    i.e. S= (1/(1-q))^2
    i.e. = 1/p^2
    So E(x) =pS = 1/p
    in the case of six sided die, p=1/6, E(x) =6


    this solutiuon found at http://math.berkeley.edu/~mchrist/Math55/Lectures/L28.pdf

    probably a good source several problems in this forum
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: All six faces of a die
  1. A Six Dice puzzle (Replies: 20)

  2. Sum of a die (Replies: 4)

Loading...