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Dice Probability

  1. Sep 22, 2009 #1
    Hello, I have a dice (or just general probability) question.

    Its kind of a question of odds of odds.

    If you roll a 17 sided die 12500 times (sorry for the non rounded number) the probability with the highest odds is that it will come up on one particular number (pick one) 735 times.

    What are the odds that that number would come up 850 times?

    Simple right?

    Thanks.
     
  2. jcsd
  3. Sep 22, 2009 #2

    HallsofIvy

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    The probability that a number will come up on any one throw is 1/17. The probability it will NOT come up is 16/17. The probability it will come up on the first 850 tosses but not on any of the 12500- 850= 11650 tosses is [itex](1/17)^{850}(16/17)^{11650}[/itex].
    Now we can rearrange 850 "S" (for "success" in throwing that particular number) and 11650 "F" (for "failure" in throwing that particular number)
    [tex]\left(\begin{array}{c}12500 \\ 850\end{array}\right)= \frac{12500!}{850!11650!}[/tex]
    which can also be written 12500C850.

    So the probability of rolling exactly 850 of a specific number is
    [tex]_{12500}C_{850}\left(\frac{1}{17}\right)^{850}\left(\frac{16}{17}\right)^{11650}[/tex]
     
  4. Sep 22, 2009 #3
    Sorry, my fault, I didn't quite state my question correctly.

    What I meant to say was, what is the probability that the number will come up 850 times or more.

    i.e. probability of it coming up 850 times plus all other options to 12500 combined.

    Thanks for your help already.
     
  5. Sep 22, 2009 #4
    Sorry, I was also just wondering if you get the same answer for your calculation shown of approximately 1.6209 X 10-6?

    Thanks.
     
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