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I am not sure how to phrase this so feel free to ask questions in comment, edit and help me clarify my question.

If you throw a dice 100 times you get a certain deviation from the "ideal" outcome (1/6 for each side). If you throw the same dice 1000 times you will with a certain probability get closer to the ideal distribution.

First question: what is the relation between how close you get to the ideal distribution when you toll the dice ten times as many times?

Second question: image that you want the distribution for any possible outcome to be within 1/5 and 1/7 with p=0,95, how do you calculate how many times you need to roll the dice to achieve this?

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  • $\begingroup$ How will you measure how far from "the ideal" the outcome of the experiment is? Are you considering a single side chosen in advance, or the largest deviation among all sides, or something else? $\endgroup$ – saulspatz Mar 31 '18 at 23:31
  • $\begingroup$ Since it is a dice, an ideal outcome is 1/6 for each side. What do you mean with "measure"? "Just" count the result an infinite number of times. $\endgroup$ – d-b Apr 1 '18 at 0:05
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(2) To be 95% sure for any one face, you need $n$ such that $1.96\sqrt{\frac{(1/6)(5/6)}{n}} < \frac{1}{42},$ using a pretty good normal approximation. Something like $n = 980$ should suffice. This is for greater than $1/7.$ Less than $1/5$ takes only about 500.

Sorry, I'm not sure what you're asking in (1).

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