3
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Could you address which of the concepts is tripping you up instead of just posing the question directly? $\endgroup$ – jonsca Nov 27 '12 at 6:36
  • 1
    $\begingroup$ We need more information! Roll again! $\endgroup$ – Kyle. Nov 27 '12 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 '12 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 '12 at 17:35
  • 1
    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Nov 27 '12 at 17:59
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.