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May 23

How many experiments are required to establish a probability ?

Let's say I estimate a probability of something happening as 1/10000. How many times do I have to try before I can reasonably say my estimate is valid ? — Preceding unsigned comment added by 2A01:E34:EF5E:4640:889D:27B:ACBF:4AB5 (talk) 12:31, 23 May 2015 (UTC)[reply]

That's a really complex question, and more information is needed before we can give an answer. There are quite different answers for different situations: for example, the probability that intelligent life exists elsewhere in the universe, or that the sun will explode tomorrow, or whether a single jellybean hidden in a box is red, or whether the next jellybean drawn from a pool of 1,000,000 jellybeans will be red or not, are really quite different problems, yet all can be expressed as "probabilities" in some sense. Can you give a more concrete description of the specific scenario you're thinking about? (For example, are your observations statistically independent? Do you know anything about the problem that might give you a prior distribution?) You might want to take a look at the statistical hypothesis testing, Bayesianism and frequentism articles for some pointers about how to frame your problem. -- Impsswoon (talk) 16:59, 23 May 2015 (UTC)[reply]

I believe my problem is closest to the "jellybean" situation. Imagine I have a truly huge number of jellybeans in a box and I suspect that one in ten thousand is red. Is there a general rule of thumb used by statisticians which would allow me to say with x% confidence that after examining y jelly beans and finding z red ones that my 1/10000 ratio is accurate ? — Preceding unsigned comment added by 2A01:E34:EF5E:4640:E9E1:4B51:3817:FB41 (talk) 07:56, 24 May 2015 (UTC)[reply]

The problem as the previous answer said is with your 'I suspect one in a thousand is red'. The number depends on your degree of certainty about your suspicions. If you are absolutely certain you don't need to test any at all. That is your 'prior'. Dmcq (talk) 09:20, 24 May 2015 (UTC)[reply]

In the case of the jellybean case, and when your experiments are independent (i.e. either your box is infinite, or you return the beans and mix after each draw), the distribution of red beans found is governed by the binomial distribution. You might, in particular, want to look at Binomial proportion confidence interval. This is the simple answer with no priors. There is no very simple answer. --Stephan Schulz (talk) 10:03, 24 May 2015 (UTC)[reply]

The probability p that a randomly chosen jellybean is red is estimated from the number k of red jellybeans in a random sample of n jellybeans.

The distribution function of p is a beta distribution.

The mean value of p is

μ = (n+2)⁻¹(k+1).

The standard deviation of p is

σ = (n+2)⁻¹(n+3)^−1/2(k+1)^1/2(n−k+1)^1/2.

To find n and k such that, say

p ≈ 0.00010 ± 0.00001

solve the equations μ = 0.00010, σ = 0.00001.

The answer is approximately n = 1000000, k = 100. Bo Jacoby (talk) 19:45, 24 May 2015 (UTC).[reply]

Excellent, thank you. Although I now realize that my question was incorrect. In fact, what I'm looking for is : "how many experiments are required so that I can be reasonably confident that no more than 1/10000 jellybeans are red".

The *actual* situation relates to scuba decompression profiles: the number of reported accidents (in a very specific yes/no sense) using common profiles is about one per ten thousand dives, so if somebody comes with a radically different profile, and all else being equal how many *accident free* dives would be needed before being able to reasonably start claiming that the profile was safer? 2A01:E34:EF5E:4640:14A0:75F4:548D:EFAD (talk) 20:15, 25 May 2015 (UTC)[reply]

The hypothesis 0 ≤ a ≤ p ≤ b ≤ 1 has the credibility

c={\frac {\int _{a}^{b}p^{k}(1-p)^{n-k}dp}{\int _{0}^{1}p^{k}(1-p)^{n-k}dp}}=(n+1){\binom {n}{k}}\int _{a}^{b}p^{k}(1-p)^{n-k}dp

Setting a = 0 and b = 0.0001 and c ≥ 0.95 and assuming k = 0 you get the condition

0.95\leq (n+1)\int _{0}^{0.0001}(1-p)^{n}dp=1-(1-0.0001)^{n+1}

Note that 1−(1−0.0001)^29954+1 < 0.95 and 1−(1−0.0001)^29955+1 > 0.95.

The number of consecutive accident free dives must be at least 29955 in order for you to be 95% certain that the accident probability is less than 0.0001. Bo Jacoby (talk) 16:35, 26 May 2015 (UTC).[reply]

Fantastic. Thanks a lot. — Preceding unsigned comment added by 2A01:E34:EF5E:4640:5BF:C8E4:804B:7B45 (talk) 19:38, 26 May 2015 (UTC)[reply]

You are welcome! Bo Jacoby (talk) 21:41, 26 May 2015 (UTC).[reply]

(my) formula for normal Latin Squares

My question doesn't require any medical diagnosis or legal advice, and I'm am not sharing an opinion, prediction, or debate, and it is certainly above any homework problem that would ever be assigned to someone. I am simply wanting to post a correction to a wikipedia page.

The page for Latin Squares states an opinion, not fact. I have KNOWN FORMULA for normal Latin Squares, and anyone, yes, anyone,... can appreciate such a mundane answer to such a long-standing problem. Why can't I talk about something that certainly will enhance that webpage? it corrects an unfounded opinion made by the author of Latin Squares wikipedia's webpage. It would be as correct as any, ANY, information that currently resides on that webpage. I could show it to an 8th-grade student, and he/she could enjoy THE FACTS!!! Besides, it is only a talk page. You could allow it on the basis that it is merely a very, very good guess. Please respond. Remember, an 8th-grade student could verify it. Don't be a fool and follow your guidelines as they exist. My idea will make people want to read your webpage and enjoy it for its accuracy! I can supply you with my website where it is already posted. Bill 108.242.169.13 (talk) 20:33, 23 May 2015 (UTC) www.oddperfectnumbers.com/langford-pairings.html read it for yourself. 108.242.169.13 (talk) 22:16, 23 May 2015 (UTC)[reply]

You need to publish your result elsewhere first, Wikipedia doesn't publish original research it only summarizes what is out there in reliable sources. I would advise you to check you actually do get the correct numbers as listed in that article when you evaluate your formula and point that out near the beginning of your paper. Don't worry too much about the difficulty of styling etc, the major first thing is to check your results and then point out that you have done so and then if you stick it ir ArXiv or send it somewhere a person will take the extra time to try and help you through any process. Dmcq (talk) 09:51, 24 May 2015 (UTC)[reply]

Unfortunately that page of yours does not provide a general formula, it gives separate equations for the separate cases and only for the simplest cases. It suggests there might be some such formula but you'd need to give a straightforward algorithm that someone else could follow. That would get you some credit but the main credit would belong to someone who provided an explanation. Dmcq (talk) 17:20, 24 May 2015 (UTC)[reply]

And in addition to a formula, a proof of its validity would be useful. -- Impsswoon (talk) 18:50, 24 May 2015 (UTC)[reply]