How many times should I roll this dice?

Discussion in 'General Math' started by AngleWyrm, Oct 19, 2008.

  1. AngleWyrm

    AngleWyrm Guest

    I want to get an Ace on the roll of a single six-sided die. I've played with
    the die for quite some time, and over 6000 rolls, it came up about 1000 on
    each face, so I'm willing to say that it is a fair dice.

    Also, I am willing to be wrong once in twenty.

    So how few times can I throw this die, and be assured that I will see at
    least one Ace come up?
     
    AngleWyrm, Oct 19, 2008
    #1
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  2. AngleWyrm

    Nick Guest

    17
     
    Nick, Oct 19, 2008
    #2
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  3. AngleWyrm

    fishfry Guest

    Assured? As in absolutely certain? No finite number of rolls will do. No
    matter how many times you roll a fair die -- a thousand, a million, a
    zillion -- there is always some small positive probability that you
    never see a 1. The only way to answer your question is to qualify the
    meaning of "assured" to mean that you want 99% certainty or 99.99999%
    certainty of eventually seeing a 1. You can never get 100%.
     
    fishfry, Oct 19, 2008
    #3
  4. AngleWyrm

    se16 Guest

    AngleWyrm did qualify the question by saying "I am willing to be wrong
    once in twenty"
     
    se16, Oct 20, 2008
    #4
  5. AngleWyrm

    hagman Guest

    If the die is fair, the probability of *not* seeing an ace in n rolls
    is (5/6)^n.
    To make this probability less than 1/20, you need 17 rolls because
    log(1/20)/log(5/6) = 16.431...

    Even if the die is not totally fair and we have to replace 5/6 with a
    bigger number,
    17 rolls are still enough as long as that bigger number is at most
    0.92632...
    But under that circumstances it is extremely unlikely to have seen
    about 1000 aces in 6000 rolls before (as that would have been about
    26sigma more than
    the expected 442)
     
    hagman, Oct 24, 2008
    #5
  6. AngleWyrm

    Rob Johnson Guest


    The probability of not rolling any particular number after n rolls is
    (5/6)^n and there are C(6,1) ways of choosing that face.

    The probability of not rolling two given numbers after n rolls is
    (4/6)^n and there are C(6,2) ways of choosing those faces.

    Continuing like this and using the Principle of Inclusion-Exclusion,
    we get that the probability of not rolling some face after n rolls is

    6
    --- k-1 6-k n
    --- 6
    k=1

    n = 26 gives 5.20% chance of missing a face
    n = 27 gives 4.34% chance of missing a face

    Rob Johnson <>
    take out the trash before replying
    to view any ASCII art, display article in a monospaced font
     
    Rob Johnson, Oct 24, 2008
    #6
  7. AngleWyrm

    Rob Johnson Guest

    I did the wrong problem. I answered the question: "How many rolls
    until we see all the faces at least once with 95% probability." You
    only wanted to see an ace.

    The probability of not seeing an ace after n rolls is (5/6)^n, so

    n = 16 gives 5.41% chance of not seeing an ace
    n = 17 gives 4.51% chance of not seeing an ace

    After 17 rolls, there is a 95.49% probability that an ace will show.

    Rob Johnson <>
    take out the trash before replying
    to view any ASCII art, display article in a monospaced font
     
    Rob Johnson, Oct 24, 2008
    #7
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