Wikipedia:Reference desk/Archives/Mathematics/2016 October 5

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October 5

Median range

Is there such a thing as the "median range" of a set of numbers? I don't recall encountering the term before, but the article Case Western Reserve University#Undergraduate profile says

Median SAT scores (25% - 75%) were between 1280 and 1450. The median range for ACT scores was 29 to 33.

Whittier Law School#Admission statistics says

Median Range of LSAT: 150-154

Median Range of GPA: 2.75-3.28

Kleinbrook, Texas says

the subdivision had a median price range of $95,000-$151,000.

Traces, Texas says

subdivision had a median price range of $77,000-$124,000 U.S. dollars.

Williamsville, New York says

The median age range was 45–49 years.

Is median range accepted mathematical terminology? Loraof (talk) 01:01, 5 October 2016 (UTC)

I have not come across the term before either, but I'm not a statistician. When I was in school, the statistic in your median SAT scores example was called interquartile range. --100.34.204.4 (talk) 01:47, 5 October 2016 (UTC)

As for Kleinbrook data, you may try asking the author of these edits: Special:Diff/256392404, Special:Diff/256392887.
For Williamsville ask the author of Special:Diff/525445794. Etc... --CiaPan (talk) 12:04, 5 October 2016 (UTC)

Thanks. Our article interquartile range defines it as the difference between the upper and lower quartiles--i.e., as a measure of dispersion. I guess the best way to express the interval is interquartile interval. Even if it's not a technical math term, it's accurate. Loraof (talk) 15:34, 5 October 2016 (UTC)

Geometry Origami problem, take 2

So, now that we have determined that it is theoretically possible to fold a 4-inch square into a 1.38-inch 16-gon (but with at least 10 layers), how exactly can this be done? (Yes, the steel mesh has arrived -- in fact, it arrived quite a while ago -- and it looks and feels flexible enough that even 10 layers of folds might be possible!) Keep in mind the other condition -- one of the faces of the 16-gon (or whatever-gon, as long as the number of sides is 16 or more) must be smooth, so all the folds must be on the other face. 2601:646:8E01:7E0B:240D:7200:4A16:32C2 (talk) 09:41, 5 October 2016 (UTC)

Russian roulette

A couple of questions about Russian roulette: (1) With only 1 bullet in the cylinder, how much less is the actual probability of dying than the theoretical probability of 1/6? (1.1) With this probability, if you play Russian roulette once a day with only 1 bullet, is it even remotely possible to survive for 7 years? (2) Likewise, for any non-symmetric distribution of bullets in the cylinder, how does the actual probability of dying differ from the theoretical probability for that number of bullets? (Question inspired by the movie The Deer Hunter.) 2601:646:8E01:7E0B:240D:7200:4A16:32C2 (talk) 09:55, 5 October 2016 (UTC)

The probability of surviving n days is (1 - 1/6)ⁿ = (5/6)ⁿ. That means e.g. the chance of surviving a week is (5/6)⁷, or about 28%.. The chance of surviving 30 days is (5/6)³⁰ or about 0.4%. Even that seems barely possible. Anything much longer, such as a year or seven, would not seem even remotely possible.

But that is assuming the chance is exactly 1/6. I imagine in reality the chance is somewhat different. A gun is built for reliability not randomness when spinning. Maybe even someone with experience can judge the timing to make sure they always stop the magazine at a safe position, reducing the probability to zero. The maths in this case is probably only an approximate guide.--JohnBlackburne^words_deeds 10:04, 5 October 2016 (UTC)

Well, the American Association of Neurological Surgeons seems to think that head shots are only 95% likely to kill the victim. Of course, this depends on a lot of variables. If you assume an equal likelihood for all 6 chambers, the chance of 7 years without firing a shot is roughly 5*10^-203 or 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000005 (give or take a zero - I'm prone to off-by-one errors). If you stop once you have lost, but still survive, that will dominate your outcome (i.e. you have the 5% survival chance of a head shot - the chance of no shot is negligible). But revolvers with 5 chambers are not uncommon, and the confederates used the 9+1 chamber LeMat Revolver in the US civil war. And apparently Casimir Lefaucheux had a 20 chamber revolver at one time - that would increase your chances significantly (but still not to a good level). --Stephan Schulz (talk) 10:36, 5 October 2016 (UTC)

But the whole point of my asking is, it's NOT an equal likelihood for all 6 chambers -- unless the bullets are distributed symmetrically, the chambers with the bullets will be heavier and would tend to end up on the bottom, thus (for most scenarios, and always if there's only 1 bullet) lowering the odds of shooting yourself. 2601:646:8E01:7E0B:240D:7200:4A16:32C2 (talk) 22:12, 5 October 2016 (UTC)

• Just to expand on the first answer...

(1) is hard to answer without more details of the "randomization" process. But assume said probability is p, and is the same every day (plus, independent tries from day to day). Then the probability to survive n days is P(n,p)=(1 - p)ⁿ. Assume n=2500 (a bit less than 7 years) and define "remotely possible" as P=0.1%, then we must have p=1-exp(ln(1-P)/n)<0.3%.

Basically, you need to cheat, but even with a lot of magazine-spinning practice a 0.3% fail rate is probably hard to achieve.

The somewhat paradoxal lesson is that if you make many tries, an event that is unlikely at each try is virtually guaranteed to happen at some point. The Russian roulette example might seem silly, but there has been real life applications, for instance the Borislav Ivanov case where (to make it simple) an unknown guy skyrocketed in the chess ratings and "coincidentally" his moves matched computers' much more closely than any historical record of world champions. (Bad players routinely match computers' top moves, but even excellent players deviate often enough that this guy was suspicious from day one.) He was eventually excluded without the use of statistics, because it was easier (for PR and/or the legal teams), but many - myself included - hold that the statistics, properly done, are just as proof as catching him red-handed. (Long news story about the statistician hero of the case: here.) Tigraan^{Click here to contact me} 11:12, 5 October 2016 (UTC)

Similarly, see the totalitarian principle in quantum physics. --69.159.61.230 (talk) 22:47, 5 October 2016 (UTC)

Inverse cube expression and spirals

What are the mathematical proof steps of the connection between a inverse-cube central force expression and the form of the trajectory (on spirals) of a (material) mobile point subjected to the central force with the given mathematical expression?--213.233.84.13 (talk) 11:44, 5 October 2016 (UTC)

• Take Newton's second law for the case at hand, express in polar coordinates, solve for r(t) for some initial condition. Anything that decreases monotonically to zero is a spiral in the plain meaning of the term, though you might have a more stringent definition. (Note: I am not sure the result is indeed a spiral; isn't a 1/r^3 force conservative for a 1/r^2 potential?) Tigraan^{Click here to contact me} 16:19, 5 October 2016 (UTC)

The article Cotes spiral says that these spirals are the solutions for the motion of a particle moving under an inverse-cube central force. The article has an external link to a Google book by N. Grossman with a short description of proof that leads to Cotes spiral for inverse cube force. Is that derivation straight forward? I want to add the proof steps into Cotes spiral article. Something may be missing from that source, it seems so. Thoughts ? (Thanks!)--213.233.84.7 (talk) 21:49, 5 October 2016 (UTC)

Binet equation has a derivation, or most of one. But you need to take Binet as a given and its derivation perhaps is not straightforward. --RDBury (talk) 02:43, 6 October 2016 (UTC)