# Non-fair die - College Probability

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.

• Could you address which of the concepts is tripping you up instead of just posing the question directly? – jonsca Nov 27 '12 at 6:36
• We need more information! Roll again! – Kyle. Nov 27 '12 at 17:31
• @kyle. This is for a midterm I have later today.. It seems quite a bit ambiguous, but that is how it's been stated. – RAF Nov 27 '12 at 17:33
• @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! – Kyle. Nov 27 '12 at 17:35
• 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. – Dilip Sarwate Nov 27 '12 at 17:59

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.