explain xkcd - User contributions [en]

Talk:1159: Countdown

2013-01-17T03:40:52Z

72.71.205.240: Benford's law has no bearing on most of the covered digits.

If you assume (with nothing else known), that large numbers have a probability about reciprocal to themselves to ensure a sum/integral of 1, the digits not being zeroes is extremely unlikely.

Whether black hat guy thinks a supervolcanoe eruption is a favourable event or being spared from one is not made entirely clear. Sebastian --[[Special:Contributions/178.26.121.97|178.26.121.97]] 08:56, 11 January 2013 (UTC)
:I warmly recommend the article {{w|harmonic series (mathematics)}}. ;-) --[[Special:Contributions/131.152.41.173|131.152.41.173]] 13:30, 11 January 2013 (UTC)
::You are right, the harmonic series is divergent. However, the maximal number of digits - which can be possibly displayed - is finite. Which distribution would you suggest? Sebastian --[[Special:Contributions/178.26.121.97|178.26.121.97]] 19:35, 11 January 2013 (UTC)

:Sebastian, do you know the specific name of the statistical principle you're invoking? I agree, but [[User:St.nerol|St.nerol]] does not, and he has a quick tendency to remove things. One part of it is that you don't know the magnitude of a number, exponential distribution is a more appropriate model than linear. Another part is about the unlikelihood of the middle digits being zero. - [[User:Frankie|Frankie]] ([[User talk:Frankie|talk]]) 21:37, 11 January 2013 (UTC)
::{{w|Benford's law}} is about the probability of certain first digit(s). Sebastian --[[Special:Contributions/178.26.121.97|178.26.121.97]] 22:34, 11 January 2013 (UTC)
:::Hmm... "Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution". I missed that the first time I read the article. Okay, that covers the essential parts of the argument. - [[User:Frankie|Frankie]] ([[User talk:Frankie|talk]]) 19:43, 12 January 2013 (UTC)
::::Come on now Frankie, I'm doing my best. I was just too quick to think that the claim was just another of these casual confusions about probability that non-math people have from time to time. (You know, I haven't rolled a 6 for some time, so now the chances must be pretty high...) I hadn't heard about this very counter-intuitive Benson-principle before, but found [http://plus.maths.org/content/looking-out-number-one this page] helpfylly explanatory.
::::So, I trust you on this. What I don't understand is, how do we know that Benfords law can be applied to this particular 14 digit number? The time left to an eruption? Also, how could a calculation of the actual probabiliy of the preciding digits being zero or anything else be made? – [[User:St.nerol|St.nerol]] ([[User talk:St.nerol|talk]]) 22:52, 12 January 2013 (UTC)
:::::What is more important for this comic than the Benford's law itself, is its underlying condition that many naturally existing numbers are lognormally distributed. And not uniformally distributed. Under that premise we can try do hypothesize about the odds of leading zeroes. Sebastian --[[Special:Contributions/178.26.121.97|178.26.121.97]] 00:28, 13 January 2013 (UTC)
:::::The initial timer is a physical quantity, therefore scale invariant, and created by a lognormal distribution (first random experiment). Now there are two possibilities: -- a) BHG specifically got a 14-digit display for the countdown (with the first digit according to Benford's law of course) and the initial timer 14 digits wide. b) The initial timer value possibly was much smaller and it could have been any number which fit on the display. -- Cueball comes in. The shown timer is uniformally distributed within the range below the initial timer (second random experiment). Because of the visible zeroes a) does not seem to be likely and b) would be true, specifically b) with the hidden digits being zero, as the shown zeroes are very unprobable with all large timer values, and the short timer actually is quite probable (lognormal distribution). Is this a valid way to argue for probabilities? Sebastian --[[Special:Contributions/178.26.121.97|178.26.121.97]] 00:55, 13 January 2013 (UTC)
::::::It seems legit, but I can't tell, really. But we have no concrete estimation yet (maybe that's too hard). Do you ''really'' think that this phenomenon is so strong so that (from the 1 in 30000) it makes the probability for four zeroes ''higher'' than for all the other 29999 possibilities together? –[[User:St.nerol|St.nerol]] ([[User talk:St.nerol|talk]]) 08:45, 13 January 2013 (UTC)
:::::::Another effect is that if the initial counter was small to begin with, it is quite unprobable (with only one supervolcanoe eruption) that Cueball comes in during the run of the counter. I will try to do a calculation example to compare the possibilities with reasonable assumptions. Sebastian --[[Special:Contributions/178.26.121.97|178.26.121.97]] 08:52, 13 January 2013 (UTC)
::::::::I restructured the last part somewhat. Hope that I didn't screw anything up, and if so, fix it! And it would be very nice if you could also add some more explanation of the math involved! –[[User:St.nerol|St.nerol]] ([[User talk:St.nerol|talk]]) 19:36, 13 January 2013 (UTC)
::This is a wholly inappropriate accusation to make here. If you have a problem, please put it through appropriate channels. No editor has a perfect score, we all slip up because we're all human. [[User:Lcarsos|lcarsos]]_a ([[User talk:Lcarsos|talk]]) 23:49, 12 January 2013 (UTC)
: Assuming that the middle digits are random, the expected value is 1.53 million years. But: If the display is off-the-shelf, it is probably larger than the largest number actually displayed. Maybe the counter started at 1e8, and the next smaller display had only 8 digits. Maybe we should have a look at the statistical distribution of digits in commercially available LED displays ... [[Special:Contributions/77.88.71.157|77.88.71.157]] 08:42, 14 January 2013 (UTC)

"I forget which one" may be a reference to the 7 known supervolcanoes, or it might be to a list published by the Guardian in 2005 of the top 10 existential threats to life on Earth, which went briefly viral. It included a supervolcano eruption, as well as viral pandemic, meteorite strike, greenhouse gases, superintelligent robots, nuclear war, cosmic rays, terrorism, black holes, and telomere erosion [http://www.guardian.co.uk/science/2005/apr/14/research.science2]

I understand how the hidden numbers could mean that a volcano could either erupt very soon or a very long time. But I don't get why this is a joke. Is there something funnny that I am missing? {{unsigned|72.38.90.50}}

:It's a joke, because a supervolcano eruption would have a major impact on the earth, and Black Hat has a timer that will tell him when one will occur, but he is too lazy to see whether it will happen soon. {{unsigned|76.14.25.84}}

The title-text may be a reference to the line "May the odds be ever in your favor!" in ''The Hunger Games''. I wonder if this might also be a commentary on the foolishness of assuming that a rare event won't happen anytime soon. [[User:gijobarts|gijobarts]] ([[User Talk:gijobarts|talk]]) 19:54, 12 January 2013 (UTC)

The picture could be somewhat symbolic. It could be a sunset or sunrise, like the would could be about to end or not. [[Special:Contributions/67.194.183.127|67.194.183.127]] 06:19, 13 January 2013 (UTC)

Benford's Law has no bearing on what any of the covered digits are except the first, and even then it only weakly applies; it only applies to the FIRST digit of natural numbers, and since we can have leading 0's is really doesn't apply. Furthermore, even if it applied to all the digits, the probability distribution on the covered digits is not affected by the shown digits; that's not how probability works. If I flip a coin 10 times and it's heads all ten times, the probability that the 11th flip is still 50/50. -Mike Powers

Talk:1160: Drop Those Pounds

2013-01-17T03:31:57Z

72.71.205.240: More evidence as to why the 30 pounds is referring to the counterweight.

"Dropping Thirty Pounds Fast"? Is that a reference to the projectile weight being approx 30lb and "dropping" it on someone's walls? [[User:DreamingDaemon|DD]] ([[User talk:DreamingDaemon|talk]]) 10:03, 14 January 2013 (UTC)
:I was thinking more along the lines of thirty pounds of blood and dismembered flesh. '''[[User:Davidy22|Davidy22]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 10:46, 14 January 2013 (UTC)

A trebuchet works by dropping a large weight connected to the swing arm, thereby propelling the projectile in a parabola (hopefully) towards the target. Thus, by dropping 30 lbs fast, you may literally hit your target. {{unsigned|‎62.109.36.140}}

:Anyhow the explanation is a little off. The "subtlety" referred to is not that people tend to ignore weight loss flyers. It is that the flyer ''looks'' like a flyer for a weight loss programme, while it is actually trying to recruit people for something entirely different. Most people would not get this and sign up thinking that they would lose body weight, while they would be signing up for the trebuchet club. The only hint is the drawing, really. I agree with the above comment that the "dropping 30lbs" probably refers to the projectile. [[Special:Contributions/62.25.36.19|62.25.36.19]] 10:52, 14 January 2013 (UTC)

::Actually - I didn't mean that the 30lbs was the projectile but rather the counterweight propelling the projectile. [[Special:Contributions/62.109.36.140|62.109.36.140]] 12:53, 14 January 2013 (UTC)

My vote is that 30lbs stands for the projectile. [[Special:Contributions/70.31.159.230|70.31.159.230]] 15:55, 14 January 2013 (UTC)

30lbs for the projectile is most consistent with the alt-text, which implies that they will be hurling projectiles at the town. A 30lbs counterweight would only be able to fling a projectile an order of magnitude smaller. Also, for medieval trebuchets the "average mass of the projectiles was probably around 50–100 kg" ([http://en.wikipedia.org/wiki/Trebuchet#Counterweight_trebuchet Wikipedia article]) --[[User:Forlackofabettername|Forlackofabettername]] ([[User talk:Forlackofabettername|talk]]) 16:23, 14 January 2013 (UTC)

:A trebuchet club would likely be building smaller models than the original medieval ones, so my vote is the 30lbs is referring to the counterweight, not the projectile. In a trebuchet, the counterweight drops fast, whereas the projectile doesn't initially drop at all, but it rather launches upwards and sideways; it'll be some time before it starts dropping, and even then not very quickly as the vertical speed takes some time to switch from up to zero, and then finally down, eventually building up speed to something that might be considered "FAST". But the "FAST" is mostly in the horizontal direction rather than seen as a "drop". In the meantime, that counterweight had already dropped more directly a long time ago. --boB

::Even the projectiles will take more to drop, it still quite "FAST" compare any weight loss program, so I think it can still refer to the projectile. [[User:Arifsaha|Arifsaha]] ([[User talk:Arifsaha|talk]]) 18:17, 14 January 2013 (UTC)

:::I can just imagine someone from the club saying "Let's drop 30 lbs on the target". Besides, I'd consider the usage of the word "drop" to be more metaphorical because in the operation of a trebuchet, no individual actually drops a counterweight; they simply pull a pin or cut a rope. [[Special:Contributions/70.31.159.230|70.31.159.230]] 20:54, 14 January 2013 (UTC)

A what-if wonder: considering a {{w|trebuchet}} is a {{w|weapon}}, will it be legal to own and place a {{w|trebuchet}} in your own {{w|backyard}}? [[User:Arifsaha|Arifsaha]] ([[User talk:Arifsaha|talk]]) 18:20, 14 January 2013 (UTC)
: The art of [http://www.amazon.com/dp/1613740646 backyard ballistics] is a firmly established niche hobby -- presumably for people with really big backyards. --[[User:Prooffreader|Prooffreader]] ([[User talk:Prooffreader|talk]]) 20:22, 14 January 2013 (UTC)
:: I needed a new hobby since I broke the last one... this is a contender! Thanks! :D [[User:DreamingDaemon|DD]] ([[User talk:DreamingDaemon|talk]]) 16:42, 15 January 2013 (UTC)

This might be example of '''literal''' vs '''figurative''' meaning: for trebuchet it is literally dropping counterweight and literally hitting a target. --[[User:JakubNarebski|JakubNarebski]] ([[User talk:JakubNarebski|talk]]) 17:06, 15 January 2013 (UTC)

Thanks for finally turning my temporary text into a proper explanation :) [[Special:Contributions/62.25.36.19|62.25.36.19]] 17:16, 15 January 2013 (UTC)

Image could also be mistaken for two people taking a walk by a hill to a castle; which would be consistent with mistaking the add for one for weight-loss; The absence of any trebuchet in the picture suggests this is deliberate. [[Special:Contributions/144.124.1.121|144.124.1.121]] 10:04, 16 January 2013 (UTC)

Although there is ambiguity here, I would think that the 30 pounds is referring to the counterweight. This is due to the fact that any device can hurl a projectile (spring catapult, torsion device, and of course trebuchet) but what sets the trebuchet apart from the rest is that it is powered by falling mass. Also, any trebuchet club that is just starting will likely be building small golfball trebuchets which would likely use counterweights on the scale of 30 pounds. I agree the alt-text makes more sense if they are actually hurling 30 pounds, but I think the main joke here is the comic that makes use of the fact that a trebuchet is literally a dropping weight. Lastly, you aren't "dropping" the projectile, you are hurling it.