Talk:Segal's law

2007-02-9 Automated pywikipediabot message
--CopyToWiktionaryBot 06:00, 9 February 2007 (UTC)


 * FYI, the page is not actually available on Wiktionary. –Blue Hoopy Frood (talk) 17:47, 16 June 2011 (UTC)

Relevance
Is it necessary to have an entire article about a relatively obscure adage? I think not. If there is some cultural relevance to this expression here that I am not aware of, could someone please add information about it? —Preceding unsigned comment added by Augurar (talk • contribs) 07:09, 29 June 2010 (UTC)

Origins?
I have come across this quote several times (most recently at the US Navy site), and am interested in its origins. Was it coined by someone named Lee Segal (as suggested on Quoteland)? Was it attributed to Einstein (as implied on aeshom.com)? Does it derive from a French proverb (also Quoteland)? Does anyone know? –Blue Hoopy Frood (talk) 17:57, 16 June 2011 (UTC)

It's actually a commentary on the illusion of right and wrong. That 2 independent, morally guided individuals may come to completely differing conclusions of what is just, yields the conclusion that there is no absolute right or wrong. Seeker of truth, you just found it!106.69.191.177 (talk) 05:11, 23 June 2014 (UTC)

Name
Why is it called Segal's Law? Who was Segal, and what else is he noted for?

Furthermore, there is no citation for Segal's statement of his law. Or for the person who named it.

--  Solo Owl   04:59, 15 August 2014 (UTC)

This is embarrassing.
The article says $$ P = p^3 + 3p^2q + 3pq^2 = p + p^2 (1-p) $$. Is this really correct? Is there a citation for this equation?

When I do it, I get $$ P = p^3 - 3p^2 + 3p = p (p^2 - 3p + 3) $$.

On the interval $$ 0 \le p \le 1 $$, the parabola $$ y = p^2 - 3p + 3 $$ decreases from $$ y = 3 $$ to $$ y = 1 $$, that is, $$ y \ge 1 $$for all permissible values of $$ p $$. Therefore $$ P \ge p $$on this interval. And the rest of the paragraph is valid.

(Note that $$ P = p $$ only at the endpoints $$ 0 $$ and $$ 1 $$. That is, the probability $$ P > p $$ except for the extreme cases when all the watches are known to be broken or all the watches are known to be good.)

No doubt I will be accused of original research, even though this is just high school elementary algebra. Even if the incorrect formula is Wikipedia-verifiable in the sense that it was copied from some “reputable” (but incorrect) publication, it behooves Wikipedia to get it right. Wikipedia should never ever repeat mathematically invalid statements as if they were true.

--  Solo Owl   07:09, 15 August 2014 (UTC) I corrected my math code several times since posting--  Solo Owl   15:10, 15 August 2014 (UTC)


 * The coefficient of the final term should be 1, not 3,


 * $$ P = p^3 + 3p^2q + pq^2$$


 * Simple typo. It should work out now. SpinningSpark 08:02, 15 August 2014 (UTC)


 * I think not. The probability that at least one watch works is $$ P = 1 - q^3 = (p + q)^3 - q^3 = p^3 + 3 p^2 q + 3 p q^2 $$. --  Solo Owl   08:11, 15 August 2014 (UTC)
 * See my comment below. SpinningSpark 08:12, 15 August 2014 (UTC)

Proof that n watches are better than 1 watch.
Let $$ P = 1 - q^n $$ be the probability that not all the watches are broken, that is, that at least one watch works. Since $$ 0 \le q \le 1 $$, we have
 * $$ \begin{align} 0 \le q^{n-1} &\le 1 \\ q^n &\le q \\ q^n - 1 &\le q - 1 \\ 1 - q^n &\ge 1 - q \\ P &\ge p \end{align} $$

This works for all integers $$ n \ge 2 $$.

This proof is simple, elegant, and general. I think it should replace the proof in the article. It should not be that hard to find $$ x^m \le 1 $$for $$ 0 \le x \le 1 $$and $$ m \ge 2 $$in a standard textbook, but I am too lazy and it is almost dawn.

--  Solo Owl   07:54, 15 August 2014 (UTC) I Fixed math code several times. --  Solo Owl   15:15, 15 August 2014 (UTC)


 * Except that one working watch is not sufficient to be able to tell the correct time. It is not possible to tell which watch is right.
 * By the way, you need to view your posts with MathJax turned off. They are a complete broken mess for anyone not using MathJax. SpinningSpark 08:08, 15 August 2014 (UTC)


 * That's because I don't know how to get the >, <, ≥, ≤, etc., except by using unicode. What do you do for that? --  Solo Owl   08:19, 15 August 2014 (UTC)
 * It's >, <, \ge, \le respectively. See Help:Displaying a formula. SpinningSpark 08:26, 15 August 2014 (UTC)
 * Thanks Spark, I needed the hint. It looks better now.--  Solo Owl   15:17, 15 August 2014 (UTC)

Maths doesn't make sense
Some logical thought can show that the maths in this article doesn't make sense:


 * 1) No watch can show the correct time, they can only show ever more accurate estimates of the correct time which asymptotically converges towards the correct time.
 * 2) The degree to which all watches are "broken" varies. This is not taken into account in the article.

— Nicholas (reply) @ 15:52, 3 August 2015 (UTC)


 * There an implicit assumption that "working" means showing the right time to within some prescribed limits. Does not the phrase in the article "yield the correct time to within some accuracy depending on the specification of the timepieces" not say just that?  Note that the degree of accuracy specified does not make any difference to the number of watches required to judge whether one has the right time with greater probability than with one watch.  It does of course make a difference to the total probability of having the right time. <b style="background:#FAFAD2;color:#C08000">Spinning</b><b style="color:#4840A0">Spark</b> 14:09, 8 January 2016 (UTC)

I think that the entire section under "In Reality" is arguably Original Research, and in some deep ways missing the point of the quote. Points of time exist on a continuum, so that even in the context of an absolute correct reference time, every other estimate is going to be "wrong" except for the infinitesimal chance that the estimate coincides with the exact time. Doing a flat categorical analysis of "right time" and "wrong time" misses this element completely. The concept is more closely aligned with the famous George E. P. Box quote, that "All models are wrong but some are useful." While we have only one watch, we can temporarily forget this complexity, but as soon as we have two watches, then with probability one (in the mathematical sense), the two will differ and we will be forced to recall that of course the two pieces of evidence are each, themselves, not exact. If we're going to apply probability theory to help to explain the quote, it should be using distributions over the Reals (possibly two Normals) and talk about things like Bayesian inference, empirical confidence intervals, etc. Categorical distributions are utterly inapt. But really, citations to the history and explanation of the naming would be much better than unreferenced mathematical explanations by you and me. 68.148.98.88 (talk) 19:49, 17 February 2016 (UTC)]]
 * This is exactly the logic that was used by the Admiralty in the issue of chronometers to ships. That much at least is citable.  What exactly the true meaning of the adage is, no source is ever going to tell you that definitively because such adages are not real scientific laws. <b style="background:#FAFAD2;color:#C08000">Spinning</b><b style="color:#4840A0">Spark</b> 23:30, 17 February 2016 (UTC)