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For want of a better word, there's a trope used in the depiction of creatures (humanoid or other) depicted in video games where the characters are constantly jiggling about. I can guess several reasons why this might be the case. Is there a name for this sort of depiction? Does it have an interesting history perhaps? --jpgordon^{𝄢𝄆𝄐𝄇} 00:58, 17 November 2024 (UTC)[reply]

That's rather vague, can you give an example? Are you thinking of the spasms sometimes afflicting puppets due to ragdoll physics? Or something simpler, like the jerkiness of two-frame animation?  Card Zero  (talk) 09:52, 17 November 2024 (UTC)[reply]

If you've played any video games in last decade, you'll have seen it; here is an example. Has nothing to do death throes (unless that's where the trope started.) It doesn't seem to have any purpose other than visual; it's not denoting actual motion or anything vaguely realistic. --jpgordon^{𝄢𝄆𝄐𝄇} 16:05, 17 November 2024 (UTC)[reply]

The name of that type of depiction is idle or idling animation, as indicated on that page. Shantavira|^{feed me} 17:48, 17 November 2024 (UTC)[reply]

Idle animation. Thank you, that's what I was looking for. --jpgordon^{𝄢𝄆𝄐𝄇} 18:34, 17 November 2024 (UTC)[reply]

One form of two-factor authentication (or 2FA for short) uses six-digit codes. So, how likely will one encounter a code that one has already seen before? If a six-digit code is generated 1,000,001 times, then the pigeonhole principle guarantees that at least one of them must be repeated. So, if a six-digit code is generated every 30 seconds starting from the beginning of a year, then there must inevitably be a repeated code by the end of the year. GTrang (talk) 15:34, 17 November 2024 (UTC)[reply]

Yes, and? The codes don't need to be unique. --jpgordon^{𝄢𝄆𝄐𝄇} 16:40, 17 November 2024 (UTC)[reply]

The likelihood of encountering a code that one has already seen before (which appears to be your question) depends entirely on how many you have seen before. Shantavira|^{feed me} 17:52, 17 November 2024 (UTC)[reply]

Let ${\displaystyle n}$ stand for the number of possible outcomes of a discrete random variable with a uniform distribution. For a fair standard die, ${\displaystyle n=6;}$ For the six-digit codes with range 000000–9999999, ${\displaystyle n=1,000,000.}$ Assume that each next turn is independent of the history. Let ${\displaystyle q(t)}$ denote the probability that the first ${\displaystyle t}$ turns gave ${\displaystyle t}$ different outcomes – no repeats (yet). Obviously, ${\displaystyle q(0)=1.}$ For turn ${\displaystyle t+1}$ to be different from the earlier ${\displaystyle t}$ turns, the outcome has to be one of the still remaining ${\displaystyle n-t}$ outcomes that have not yet occurred. The probability, independent of the past, is ${\displaystyle {\frac {n-t}{n}},}$ so

${\displaystyle q(t+1)=q(t){\frac {n-t}{n}}={\frac {n}{n}}{\frac {n-1}{n}}{\frac {n-2}{n}}\cdots {\frac {n-t}{n}}.}$

For ${\displaystyle t>n}$ this has a factor ${\displaystyle {\frac {n-n}{n}},}$ so then ${\displaystyle q(t)=0.}$ Otherwise,

${\displaystyle q(t)={\frac {n!}{(n-t)!\,n^{t}}}.}$

When ${\displaystyle n=1,000,000,}$ you already have ${\displaystyle q(1178)\approx 0.49981\,,}$ less than 50% chance of repeat-free survival. See also Birthday paradox.  --Lambiam 19:19, 17 November 2024 (UTC)[reply]

Everyone is told not to re-use passwords, but that doesn't apply to one-time singleton keys such as are typically used in 2FA.

• If I somehow know that last year your regular password was "gzornnplatz", and I'm trying to hack into your account, "gzornnplatz" is a reasonable thing to try; it's considerably more likely to still or again be your password today than some other 11-character string.
• But if I'm trying to guess a 2FA code, and I know that some time ago the generator randomly generated 123456, that tells me nothing about what code it's likely to use today. During any given authentication session, every one of those 1,000,000 codes is equally likely. The security rests on the assumption that there's no way you can possibly try more than a tiny fraction of the codes in that space before the authentication session times out and starts over with a new code (or locks you out for a while).

—scs (talk) 01:20, 18 November 2024 (UTC)[reply]