Talk:2659: Unreliable Connection

I don’t think this has anything to do with teleconferencing. Am I missing something? 172.70.214.81 22:46, 15 August 2022 (UTC)

Yes. The impliction is that people are expecting you to be available for online communications, and you can use the unreliable Internet connection as an excuse to get out of it. Barmar (talk) 22:51, 15 August 2022 (UTC)

I think it's more about communication in general. He doesn't want anybody calling him or sending him emails, so by saying he has an "unreliable" connection people might assume it will be hard to get in touch with him.

Back in the day, email was usually configured so that it could easily overcome such unreliability, and it's still doable,[1] but today email for most people is a web or local client-server app, as opposed to a local mail store in a peer-to-peer app. Even people in urban areas can suffer unreliable internet, when squirrels or backhoes gnaw through data cables, copper theives strike, or 5G mind control base stations are congested. 172.70.210.143 23:45, 15 August 2022 (UTC)

This could equally cover other instant communication methods where your availability is advertised (e.g. Whatsapp). It could also be about alleviating the social pressure the subject feels to continuously check and immediately respond to messages (including emails), because the immediacy is already hindered by the spotty connection (cf the standard "I will have limited access to email" out of office line, which gives the account owner psychological permission to check it infrequently). 172.70.85.5 09:02, 16 August 2022 (UTC)

According to a PhET simulator (https://phet.colorado.edu/sims/html/plinko-probability/latest/plinko-probability_en.html) for this situation, the ideal standard deviation is 1.732 and ideal mean is 6. I don’t feel like doing the calculations :P 172.70.211.134 23:34, 15 August 2022 (UTC)

If we assume 50-50 for each bounce, the probability that internet is off will be about (11 choose 3)/(2^11), or 8%.--Account (talk) 23:51, 15 August 2022 (UTC)

My first thought was, why so complicated? If each of the twelve switches is equally (and solely) likely to be struck by each ball, it's (100/12)% of the time, or 8⅓%.

Although the equal-chance is wrong, so you're definitely doing "end up with exactly 7 bounce rights and 3 bounce lefts, but in any combination" or similar are you? I'd have summed it differently, though. And not sure where the choose 3 comes in... Just one bounce left off any row-end pin 11 sends to 11 if all others bounce right. Three bounces left hits switch 9, not eight. If I'm counting correctly. Or am I doing telegraph-poles/wires miscounting?

Too early in the morning for me to untangle. The only thing I'm sure about is your division by 2^11 (how many total paths there are to get down). 172.70.91.78 05:00, 16 August 2022 (UTC)

Me again. I hadn't checked that the transcript (which said it was switch #8) was correct. Have now, and found it to be wrong. Have hence also just corrected the Transcript. So I'm gonna assume your 11-choose-3 is entirely correct after all. ;) 172.70.91.78 05:08, 16 August 2022 (UTC)

It's actually 12 switches, not 11, but that doesn't affect the math too much. I originally thought "off" was switch 10, which would have changed the math (to 3%), but that's just the one the current ball hit. The actual "off" switch is switch 9. -boB (talk)

It previously said that there were eleven "on" switches and one "off" switch (which is twelve in total, but it didn't add them up explicitly), and the change to say that there are 12 Ons and 1 Off made it wrong. I corrected/rephrased it (see if you agree with however it looks by the time you get around to reading this) to avoid that reading error (one which happened to me with my own first glance at the phrasing used, but I thought that was just me at the time) without adding any new misinterpretation or easy misinterpretationality.

The maths above is indeed correct enough. The 2^11 relates to the total number of unique paths it can take (assuming a bounce left/right just enough to strike the nearest offset pin below to force a new left/right bounce choice) from the first divider through to any of the 11 final left-right pin-bounces (and onto the 12 switches, at which point we're not bothered with the bouncing - diagram suggests the balls leap outwards and don't hit any other switches).

"11 choose 3" is a way how to ask, given 11 items (possible bounces), how many unique and unordered combinations of exactly 3 of these must exist (leftward-bounces, the rest being right-bounces) to filter onto the off-connected switch. (This is the same as "11 choose 8", if you decide to ask how many right-bounces are necessary, the rest being left-bounces.) That could be layer 1 (the 1-pin), 2 (the 2-pins) and 3 (...), before going consistently right to the final strike of the switch, or layers 9+10+11 (after being pure-right 1..8), but with many intermediate tracks across the pin-spacs (165 in total, as it happens; and it would be 55 to hit switch 10. Or 2, instead of 3, if you orientate things the other way round).

165/2048 (paths hitting the off-switch (at #9) divided by all paths that might happen) is a tad over 8%. On the assumption that it's fair and unbiased and you don't get more rattling around than a simple (single half-step) left/right distribution. 172.70.91.78 03:20, 19 August 2022 (UTC)

To whomever did [2], doesn't [3] prove that symmetrical configurations nearly identical to those shown can produce uniform distributions? They seem to show it's just a matter of horizontal pin spacing. However, I for one can not verify the proof, which uses unusual (novel?) non-Unicode math notation, and a fairly opaque method of proof. 172.70.211.134 00:07, 16 August 2022 (UTC)

Not sure, but this Japanese Wikipedia article is fascinating. 172.70.206.213 01:51, 16 August 2022 (UTC)

Please see section 3.5 on pp. 16-18 of the currently first reference [4]. I am particularly intrigued by, "Open Problem 2: Is every uniform distribution of output probabilities of the form 1/2^k constructible by a 50-50 Pachinko?" on p. 18. However I haven't dived in enough to even know where the parentheses are supposed to be in that expression, yet. Liv2splain (talk) 17:27, 16 August 2022 (UTC)

Good question! https://ibb.co/sRwGwB9 don't look triangular, but it seems the proof might suggest much more triangular solutions. Worth thinking about! 172.69.33.115 21:24, 16 August 2022 (UTC)

What is the chance that the ball will bounce off the first pin, go down the outside of the pins and miss all the switches?

Probably quite high if it's a bouncy ball. With idealized physics though it'd just hit the leftmost/rightmost switch. 172.70.254.127 00:45, 16 August 2022 (UTC)

I would describe the device as a https://en.wikipedia.org/wiki/Galton_board. 172.70.230.109 00:30, 16 August 2022 (UTC)

I was watching the photo and hover-over text and the image disappeared and "Unreliable Connection" showed up in its place. I don't know how often this happens.

Very neat if not a fluke! Can anyone replicate this experience on https://xkcd.com ? 172.70.211.134 14:21, 16 August 2022 (UTC)

"An added source of humour is that Randall could likely achieve the same effect by looking through the router's settings - which most modern ones have a feature to turn on and off at scheduled times - or via purchasing a smart power strip." But by using these other methods, the connection would still be reliable. If it goes out at regular or pre-scheduled intervals then you know when it will be available or not, hence reliable. I think the joke here is that the contraption does in fact make the connection unreliable. 172.70.114.77 14:18, 16 August 2022 (UTC)

Addressed at [5]. Liv2splain (talk) 14:44, 16 August 2022 (UTC)

(Edit conflicted by at least the above, but my answer to the same question...) From a user POV, unless they happen to know that at 11:53 each day (and 12:14, 15:02, 15:07, 16:31, etc...) the scheduler disables tracfic for one (or two, or three) minutes, it is still unreliable, if ultimately predictable once you know the schedule, having seen it go round a few times and taken note. Similarly a timered power-strip could be used (or even several, in serial, the two or three daily interventions by the first also stopping and delaying the subsequent strips' interventions, making their timings uneven, further down the chain) and until you got the pattern it might as well be 'random', not entirely deterministic. (I'm wondering about some OR-gate-like/etc implementation, so power can pass by at least one parallel timer-shut-off to maintain power at the lower levels while some mid-way timers get depowered and thus 'shuffled' in interesting ways, and the resulting single output is governed by an intricate multi-dependent set of routes, but I bet an electrician would be wary about wiring that up...)

You could hack (or patch) the management firmware to be a bit more (pseudo)random about it, though it would still be pseudorandom LFSR/Xorshift with a (long) repetition cycle.

Or make it dependant upon an external factor (if the modulo 12 of the cumulative sum of all observed packet-destination IPs is zero, shut off for the five times the prior modulo 12 test value, in seconds..?), but that's practically the pachinko solution but with software hacking rather than hardware-making/hacking as per the comic.

More effort is needed to make it ultimately unpredictable, but it can still be considered unreliable if it goes out just when you 'want' it.... 172.70.85.5 15:02, 16 August 2022 (UTC)

For real though, isn't this kind of a good idea? Fephisto (talk) 14:34, 16 August 2022 (UTC)

Talk to edtech people in the MOOC space and they will tell you asynchronous is worth it, but talk to people who study educational quality factors like time to receive answers to unanticipated questions, and they will have different ideas. Liv2splain (talk) 14:44, 16 August 2022 (UTC)

Does anyone have an openWRT (or other) implementation of this feature yet?

You can induce it on stock firmware without reflashing, but you need to know the parameters like how often balls come out of the hopper, and what exactly the on/off switches do. As pseudocode:

#!/bin/sh

while true ; do

sleep seconds

if [ `rand100` -le 8 ] ; then

wifictrl off

else

wifictrl on

fi

done

172.70.214.81 00:38, 17 August 2022 (UTC)

There are spaces between the button that the balls can fall into, and this could complicate the stuff a bit. However if the ratio between probability of hitting ON and probability of hitting OFF remain the same (1883:165), the average OFF time will still be the same (165/2048 of the time). The behavior that the network is switching between ON and OFF will probably be changed though. Lamty101 (talk) 04:44, 17 August 2022 (UTC)

I would have expected the negative reviews to have mentioned all the balls on the floor and perhaps the need to periodically refill the hopper. Philhower (talk) 16:18, 17 August 2022 (UTC)