# 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? Commented Nov 27, 2012 at 6:36
• We need more information! Roll again! Commented Nov 27, 2012 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
Commented Nov 27, 2012 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! Commented Nov 27, 2012 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. Commented Nov 27, 2012 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.