How many times must you roll a non-fair die to be at least 84% sure that the sample probability will be within 3% from the actual probability.

Since the die is not-fair, we do not know p. My question is, can I assume it to be around 1/6, or not?

And so far, I found the Z-scores for 84% of the probability from the mean to be + & - 1.4. But, I do not know how to move forward with an unknown probability.

  • 4
    $\begingroup$ Could you address which of the concepts is tripping you up instead of just posing the question directly? $\endgroup$
    – jonsca
    Nov 27, 2012 at 6:36
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    $\begingroup$ We need more information! Roll again! $\endgroup$
    – Kyle.
    Nov 27, 2012 at 17:31
  • $\begingroup$ @kyle. This is for a midterm I have later today.. It seems quite a bit ambiguous, but that is how it's been stated. $\endgroup$
    – RAF
    Nov 27, 2012 at 17:33
  • $\begingroup$ @RAF Thanks for the edits to your question. I think, since this is a homework problem, people wanted to help you with the underlying concepts, rather than just answer the question, which will help you more on your midterm anyway! $\endgroup$
    – Kyle.
    Nov 27, 2012 at 17:35
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    $\begingroup$ The maximum value of the variance of the value obtained from a die roll is $(6-1)^2/4$ and occurs when the die is so biased that only a $1$ and a $6$ show up (with equal probability $\frac{1}{2}$) and $2,3,4,5$ never show up. Work out a $z$-score from this and determine the number of rolls needed. This will be an upper bound on the actual number of rolls needed with a die that is less strongly biased. $\endgroup$ Nov 27, 2012 at 17:59

1 Answer 1


As others say in the comments, the question makes no sense because it does not state the probability of what. The probability of the die falling to a surface is presumably one, for example, regardless of how loaded it is. I will assume it means the probability of rolling a six (I know this is by no means the only possibile interpretation of this ambiguous question).

The wording of the question asks for you to be "at least 84% sure that the sample probability will be within 3% from the actual probability." The "at least" is surely significant. You can answer the question by assuming the worst case scenario - the die is loaded so six shows up 50 percent of the time so the variance of the number of sixes is $n(\frac{1}{2})^2$. Solve from there using usual methods.

So my answer is no, you cannot assume p is 1/6. As you have been asked to be conservative ("at least") you must assume the worst case scenario, which is p=1/2.


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