Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

August 8

Closed plane curve questions

Got some questions about plane curves. In particular, I'm pretty sure that the answer to all of these questions is yes, but proofs elude me for the moment.

1. Does every closed plane curve admit some line that touches the curve at exactly one point? Similarly, do they all admit some line that touches the curve at exactly two points?

2. Does every concave closed plane curve admit some line, lying entirely outside of the open region enclosed by the curve, that touches the curve at exactly two points? Essentially, an exactly two-point tangent line?

3. In the isometric problem, there's a trick where a "dent" in the boundary of a concave closed plane curve can be reflected across a two-point tangent line to increase the enclosed area while maintaining boundary length. Can every concave closed $C^{1}$ plane curve be turned convex by a finite number of such reflections? GalacticShoe (talk) 03:27, 8 August 2023 (UTC)

For the purpose of this discussion, let a tangent of a closed plane curve be a line that intersects the curve, but not the interior of the region enclosed by the curve. So the line given by the equation $x+y=0$ in the Cartesian plane is a tangent of an axes-oriented square in the first quadrant with the origin as one of its corners.

1. While I do not immediately see a proof, I think that indeed every closed plane curve has some tangent that intersects the curve solely at one point. No tangent of a circle touches it at more than one point, and each tangent of a square either touches it at a single point or at an infinte collection of points, never exactly two, so the second part of 1 should be answered with no.

2. By "concave plane curve", do you mean one enclosing a non-convex region? Take a square and make a dent halfway one of its sides. This is a counterexample.

3. You can put infinitely many smaller and smaller dents in the side of a square, like each next one half the size of the previous one. Then you'll need infinitely many reflections. You can also make a smaller out-dent on the first in-dent, which becomes an in-dent after reflection. In that smaller out-dent you can make an even smaller in-dent and repeat the process, alternating in and out, ad infinitum.

 --Lambiam 09:04, 8 August 2023 (UTC)

I think there may some confusion about what is meant by a "closed curve". If you insist that it's the image of a one-one differentiable map from the circle to the plane, and that the inverse map is differentiable, then a square would not be allowed. On the other hand if you only insist that the map be continuous then you allow space filling curves and come of the questions become pretty meaningless. Even if you add the conditions that the map is one-one and it's inverse is continuous, you'd still allow curves which have no tangent line at all in the usual sense. --RDBury (talk) 14:00, 8 August 2023 (UTC)

That's an excellent point; for the purposes of this question, I take a "closed curve" by the continuity definition. Since this allows for space-filling curves, I think this now means that the second half of the first question is proven false, since there appear to be closed space-filling curves, like the Moore curve. This does leave the question open for Jordan curves, however, since no space-filling curves are simple. GalacticShoe (talk) 14:19, 8 August 2023 (UTC)

Hey Lambiam, thanks for the response!

1. I should clarify for this question that in terms of lines touching the curve at exactly two points, I also mean lines that can intersect the shape rather than lying tangent to it; I imagine that every closed plane curve has some one-point tangent line that can be translated to intersect the curve at exactly two points, but I don't know how one would prove this.

2. That's exactly right, thanks for the counterexample. I was thinking too much of smooth curves when I was coming up with this question, and upon reflection even with smoothness there would be counterexamples (e.g. by smoothing out the corners and only the corners of the shape you described.) For bookkeeping purposes I'm just going to write a bold note here saying that this is resolved in the negative.

3. I was wondering, when you mention the denting process, are the dents made "smoothly"? For the purposes of the question I'm looking for a $C^{1}$ curve, and I'm curious whether an infinite amount of "smooth" denting can create a shape that is still $C^{1}$. GalacticShoe (talk) 14:10, 8 August 2023 (UTC)

Re 3, I think this is even possible while resulting in a $C^{\infty }$ curve. Define function $\rho$ on the interval $(-1,1]$ by:

$\rho (x)=0$ for $x=0,1$;

$\rho (x)=\sin({\frac {1}{x}})\exp \left({\frac {-1}{x^{2}(1-x^{2})}}\right)$ otherwise.

Now take the curve with polar representation $r(\varphi )=1+\rho \left({\frac {\varphi }{\pi }}\right),-\pi <\varphi \leq \pi .$  --Lambiam 16:00, 8 August 2023 (UTC)

This is a great example but I was wondering how you can determine whether the resultant curve is indeed concave? It would seem that the nature of polar curves allows for some 'leeway' in whether perturbations to the curve make concave sections. For example, $r(\theta )=1+{\frac {1}{6}}\sin(2\theta )$ is a convex curve despite the fluctuations in ${\frac {1}{6}}\sin(2\theta )$. Similarly, I'm having a hard time determining whether $1+\rho ({\frac {\theta }{\pi }})$ becomes concave in infinitely many places. Thanks again! GalacticShoe (talk) 23:08, 8 August 2023 (UTC)

I had some doubts myself and should not have posted this without checking. I am fairly convinced, though, a provably correct counterexample can be constructed.  --Lambiam 10:32, 9 August 2023 (UTC)

After thinking more about it, I'm even fairlier convinced. The basic step would be to have $r(\varphi )$ behave in a neighbourhood of $0$ like $\left(1+\rho (\varphi /\pi )\right)/\cos \varphi .$ In the mapping $r(\varphi )\mapsto (x,y)$ we then have $x-1=\rho (\varphi /\pi ),$ which switches from positive to negative and back infinitely often in any neighbourhood of $0.$ Function $r$ can always be glued, with preservation of $C^{\infty }$-hood, to a function making it periodic with period $2\pi$.  --Lambiam 18:57, 9 August 2023 (UTC)

Re 1. The Latin word tangens from which we have our word tangent literally means "touching", so the use of the word "touch" for lines suggests that tangent lines are meant. So is it correct to say that you are interested in constraints on curves such that for each curve $C$ meeting certain conditions there exists some line $L$ such that the cardinality of $C\cap L$ is $1$ (or $2$)?  --Lambiam 20:27, 8 August 2023 (UTC)

That is indeed what I was interested in. In particular, whether all closed plane curves $C$ admit some line $L$ such that $\#(C\cap L)=1$, and whether all Jordan curves $C$ admit some line $L$ such that $\#(C\cap L)=2$. GalacticShoe (talk) 23:09, 8 August 2023 (UTC)

The dragon curve consists of an infinite sequence of similar space-filling segments joined at each junction by one point. Its boundary consists of an infinite sequence of Jordan curves, joined likewise. I suspect this dragon-segment boundary is a counterexample to the existence of a two-point intersection; in fact, if there is more than one point of intersection, I think there are countably infinitely many.  --Lambiam 10:45, 9 August 2023 (UTC)

It does seem likely that all lines intersecting the dragon curve do so in either one or more than two points, but I think that, as a space-filling curve, it can't be Jordan. At least, that's what I'm reading from this StackExchange post; it seems that although space-filling curves are indeed continuous, and can very well be closed, they can't be injective. That being said, I get the feeling that some non-space-filling fractal might work as a counterexample, although I struggle to imagine what such a fractal would look like. GalacticShoe (talk) 13:12, 9 August 2023 (UTC)

I was talking about the boundary, which is a chain of similar Jordan curves. An approximation of this Jordan curve, only a few iterations but hopefully conveying the idea, can be seen here.  --Lambiam 19:42, 9 August 2023 (UTC)

Ah, I see what you mean; the boundary of the dragon curve certainly does seem as good a counterexample as any, and I can't think of any way a line could intersect the curve in exactly/only two points. I will have to look into whether there are explicit ways of representing the boundary so as to formalize this, but as it stands it seems likely that this works as a counterexample. GalacticShoe (talk) 21:07, 9 August 2023 (UTC)

Just for posterity's sake, I should note here that for no. 2, I think it's possible for the counterexample rounded square to be $C^{\infty }$ as well. I haven't constructed any explicit examples, but I imagine one could use flat functions to keep most of the sides perfectly straight while still rounding off the corners. GalacticShoe (talk) 13:30, 9 August 2023 (UTC)

Define function $\gamma$ by

$\gamma (x)=0~$ if $~x\leq -{\tfrac {1}{2}},$

$\gamma (x)=1~$ if $~x\geq +{\tfrac {1}{2}},$

$\gamma (x)={\tfrac {1}{2}}\left(1+\tanh \,\tan(\pi x)\right)~$ otherwise.

Then, if $f$ is $C^{\infty }$ on $[a,+{\tfrac {1}{2}}],$ where $a<-{\tfrac {1}{2}},$ and $g$ is $C^{\infty }$ on $[-{\tfrac {1}{2}},b],$ where $+{\tfrac {1}{2}}<b,$ so is function $h$ defined by

$h(x)=(1{-}\gamma (x))f(x)+\gamma (x)g(x).$

on the interval $[a,b].$ So such gluing to round the corners is possible.  --Lambiam 20:15, 9 August 2023 (UTC)

On reflection, I think the first half of #1 may be simpler than I expected. I suspect that if one can prove that every convex curve admits such a line (which is almost certainly the case), then for any concave curve $C$, one can just take the line corresponding to the convex hull of $C$. GalacticShoe (talk) 21:40, 9 August 2023 (UTC)

Daniel Mathias on Math StackExchange makes an even simpler argument: rather than the convex hull of $C$, consider an arbitrary circle circumscribed around $C$. At a point of intersection $P$, the tangent line of the circle suffices. GalacticShoe (talk) 22:39, 9 August 2023 (UTC)

August 10

Finding rational coefficients of a cubic polynomial that fits 4 data points that have been floored to an integer

I have 4 data points:

• (204, 5422892)
• (205, 5722486)
• (207, 6343357)
• (213, 8386502)

I have information that these data points were generated with a cubic polynomial

y = ax ^ 3 + bx ^ 2 + cx + d

with a, b, c, and d being positive rational (but not necessarily integral) coefficients, and with the resulting y values being floats that were floored to an integer (the x values are known to be true integers, not floored floats). I also have information that the coefficients are relatively "nice", "simple", or "short" numbers with only a few digits like 50 or 1/2 rather than anything that is a long string of many digits.

I need to determine what those coefficients are. I tried submitting the query

"find cubic polynomial that fits (204, 5422892), (205, 5722486), (207, 6343357), (213, 8386502)"

to Wolfram Alpha, but Wolfram Alpha gives me the answer

(589/36) x ^ 3 - (116363/18) x ^ 2 + (10691263/12) x - 46197659

which has coefficients that are rational but are not positive and not "nice" (in the sense used above). I think the difficulty is that Wolfram Alpha is using Lagrange interpolation to try to find a cubic polynomial that is an absolute exact fit to the given data points, but there is actually flooring to an integer being applied to the y values.

So can anyone help me...

(1) find the coefficients for this specific case of the 4 data points given above?

(2) show me how to, in general, given any 4 data points that are known to be generated with a cubic polynomial with positive rational coefficients with the y values then floored to an integer, find those coefficients? (I am aware of Lagrange interpolation, but I think that is what Wolfram Alpha is doing above and is what is giving wrong results due to not being able to take into account flooring.)

—SeekingAnswers (reply) 11:49, 10 August 2023 (UTC)

(Update to original post: I am virtually certain that the y values are being floored rather than having any other type of rounding.) —SeekingAnswers (reply) 17:46, 10 August 2023 (UTC)

I've actually solved this kind of thing, but the method I used isn't simple and I don't guarantee it's the most practical. What you've got is a system of linear inequalities starting with 5422892-1/2 ≤ 204³a + 204² + 204c + d ≤ 5422892+1/2 (assuming simple rounding is used). This is a total of 8 inequalities in 4 unknown, and it produces a 4-dimensional polytope in (a, b, c, d)-space. If there happens to be an integer point in it, and if you can find it, or them if there are more than one, then you're done. Trying to find an integer solution changes the problem from an example of linear programming to an example of integer programming. If there is no integer solution (and it's not hard to see that there isn't in this case), then you can try different "nice" denominators, for example solve with a=A/2, b=B/2, c=C/2, d=D/2 where A, B, C, D are integers. If you get to a=A/36, b=B/36, c=C/36, d=D/36 you can stop because that's just the Lagrange solution. I'd suggest as a first step, round the Lagrange coeffs to the nearest integer, then subtract the values of that polynomial from the points to get smaller, more tractable numbers. You can add the rounded Lagrange coeffs back in when you're done to get a solution to the original problem. Also offset the x values by 204 so you x values of 0, 1, 3, 9, also to make the numbers smaller and more tractable. To put the answer in the original range change x to (x-204). I don't know what's publicly available in terms of IP solvers, but if you state the problem as a system of inequalities you might find something to do most of the actual work for you. --RDBury (talk) 16:22, 10 August 2023 (UTC)

PS. Less elegant but perhaps easier to implement, for a given denominator N there are about N⁴ possible combinations of values you can get for the values of your polynomial at the four points. For N≤36 that's low enough to use a brute force search. For each set of values find the Legrange polynomial that fits those values, if the denominators you get are factors of N then you have a solution. With N=32 that's about 1,000,000 combinations to try, and it probably wouldn't take a computer too long to check them all. A lot of the work, for example finding the Lagrange basis, only has to be done once. --RDBury (talk) 16:42, 10 August 2023 (UTC)

A cubic polynomial that can give rise – whether with rounding to the nearest integer or with rounding down – to the given data set cannot have only non-negative coefficients. The constant coefficient of any such polynomial is necessarily less than −45 600 000, while that of x² is less than −6 400.  --Lambiam 20:20, 10 August 2023 (UTC)

Yes, the problem would have to be reworked a bit to fit the usual paradigm, but there are standard techniques for doing this. Convert each each inequality to an equality with a slack variable. For example -130 ≤ x ≤ 50 becomes -130+s = x, x+t = 50, s, t ≥ 0. Since x has unrestricted sign it can be eliminated to give s+t=180, s, t ≥ 0. Once you have a solution you'd have to use back substitution to get the original variables. I would think a good LP/IP solver would do this kind of thing for you. Btw, it seems modern spreadsheet programs have solvers built in, but I've never tried using them. There are (of course) YouTube tutorials for using them. --RDBury (talk) 23:45, 10 August 2023 (UTC)

Hmm, okay, I see the problem. The context is that I am trying to reverse-engineer the behavior of some software. Related parts of the software that I have been able to reverse-engineer all use cubic polynomials with positive rational "nice" (in the sense given above) coefficients and then apply flooring, so I assumed it was true of this part as well, but it looks like at least the "positive" part of that pattern is being broken. —SeekingAnswers (reply) 01:56, 11 August 2023 (UTC)

August 11

Maximum number of days

If we want to separate the numbers 1, 2, 3, …, n^2 to n parts, each part has n numbers, every day, and any two numbers can be in the same part at most one day, what is the maximum number of days such that this is possible? For example, for n = 3, 4 days is possible:

• Day 1: (1,2,3) (4,5,6) (7,8,9)
• Day 2: (1,4,7) (2,5,8) (3,6,9)
• Day 3: (1,5,9) (2,6,7) (3,4,8)
• Day 4: (1,6,8) (2,4,9) (3,5,7)

But 5 days is not possible. 218.187.65.9 (talk) 17:34, 11 August 2023 (UTC)

This is actually a problem in finite geometry, in fact the example you've given is known as the Hesse configuration. It's not hard to see that n+1 is an upper limit, there are n²-1 numbers ("points") other than 1, and each day you use up n-1 of them, so you can have at most (n²-1)/(n-1)=n+1 days. Each group would correspond to a line, and each day a set of parallel lines in this geometry. If you count each day as a "point at infinity" you get a finite projective plane of order n. (Sorry to throw a lot of jargon at you, but it's the easiest way to think of the problem.) If n is prime then you can create examples relatively easily as follows: Relabel the numbers 1 ... n² as ordered pairs (0, 0), (0, 1), ... (0, n-1), (1, 0), (1, 1), ... (1, n-1), ... (n-1, 0), (n-1, 1), ... (n-1, n-1). For day k starting with k=0, group the pairs (a, b) according to the value of a+kb (mod n). You get the last day by grouping on b. So with n=3 you get

• Day 0: {(0, 0), (0, 1), (0, 2)}, {(1, 0), (1, 1), (1, 2)}, {(2, 0), (2, 1), (2, 2)}
• Day 1: {(0, 0), (1, 2), (2, 1)}, {(0, 1), (1, 0), (2, 2)}, {(0, 2), (1, 1), (2, 0)}
• Day 2: {(0, 0), (1, 1), (2, 2)}, {(0, 1), (1, 2), (2, 0)}, {(0, 2), (1, 0), (2, 1)}
• Last day: {(0, 0), (1, 0), (2, 0)}, {(0, 1), (1, 1), (2, 1)}, {(0, 2), (1, 2), (2, 2)}

This method fails when n is not a prime. But when n is a prime power it can be fixed using finite fields; I won't go into details but they should be in the articles mentioned above. See also Galois geometry. It's an open problem whether n+1 is possible for n not a prime power. What I know of what is known is that it's been proven in the negative for n=6 and 10, with 10 being fairly recent and involving a computer search. That leaves 12 as the next candidate. --RDBury (talk) 06:32, 12 August 2023 (UTC)

Interesting stuff, RD! Thanks! --Trovatore (talk) 18:51, 12 August 2023 (UTC)

Hyperoperation: Does the n-th hyper operator relate to n or to n-1 or to n-2?

Does the relation a[n]b (the n-th hyperoperation, i.e. n=1 for addition, n=2 for multiplication, n=3 for exponentiation, n=4 for tetration, n=5 for pentation, n=6 for hexation, etc. relate to n or n-1 or n-2? The Graham’s number has 64 towers of Knuth's up-arrow notation, but Knuth's up-arrow notation has n-2 instead of n up-arrows for the n-th hyperoperation, a[n]b. 218.187.65.9 (talk) 17:40, 11 August 2023 (UTC)

I think the $a[n]b$ notation was developed by enthusiasts on Usenet; it's not clear that it's really become standard in more formal (not mathematically formal but "register"-wise) venues. The hyperoperation article frankly has a problematic title and should be moved to something like generalizations of exponentiation, which I suggested some time ago but haven't followed up on.

That said, at a quick glance, the $a[n]b$ notation seems to be shifted by two from the Knuth notation — natural-number exponentiation $a^{b}$ is $a[3]b$ but $a\uparrow ^{1}b$. --Trovatore (talk) 19:41, 11 August 2023 (UTC)

August 14

Trying to figure out the probability that my players pass this dice roll

Hi all. I have a party of four Call of Cthulhu RPG players who are going to find a damaged radio in their next session. None of them put skillpoints into the skill needed to repair it, meaning all four of them have the default 10% chance of success. I'm trying to figure out what their odds are of success using the following assumptions:

• In CoC, players roll a d100 (actually two dice, a d10 and a d-percentile which represents the 'tens' place of their roll result) for skillchecks, and their goal is to roll low. In this case, when a player attempts the check, they succeed if they roll a 01, 02, 03, 04, 05, 06, 07, 08, 09, or 10 out of the 100 sides.
• Knowing my players, they're going to pair off, two of them will use the "help" action to give the other two a 'bonus die' on their attempt to repair the radio. (Update: This means the two helping do not get to make a check to fix the radio themselves.) What this means is that both attempters get to roll their d-percentile a second time and use the lower of the two values (remember, low values are better in CoC). Only one of the two attempters needs to pass the check to repair the radio.

So, assuming the party does what I expect, what are the odds they successfully roll to repair the radio? --Aabicus (talk) 20:30, 14 August 2023 (UTC)

First, isn't 00 a valid dice role? An any case, the actual mechanics of generating a 10% chance are irrelevant. Also, I don't think pairing really matters at all since the upshot is that if anyone gets a win in four tries then it's a success. It's easier to find the probability of failure. All four players would have to fail, there's a 90% chance of each player failing, so there is a .9⁴=65.61% chance that all of them will fail. That means a 34.39% chance of success. --RDBury (talk) 22:13, 14 August 2023 (UTC)

00 is considered "100" on a d100, ergo a failure in CoC rules. There's a consideration I didn't mention that explains why the four don't all do the checks individually (rolling 95-100 is a fumble, and if any of them fumble the radio would break irreparably. They've been burnt by fumbles in the past so they prefer to do the bonus dice thing which makes fumbles far less likely to roll). Which is why I'm hoping to know the odds if two of them use the bonus dice style of rolling, instead of straight 4 checks at 10% each. --Aabicus (talk) 22:22, 14 August 2023 (UTC)

Ignoring the fumble for a second, it doesn't matter. Think of it like this: If the helpers help, they get to reroll one die for their teammate: the d-percentile die. If they don't help, they get to reroll both of their teammate's dice (they actually get to roll their own dice, but there's no difference between rolling dice that belong to you and rolling dice that belong to someone else). If anything, not helping might be better. But it turns out in this specific case, the probabilities work out the same.

With fumbling, there needs to be a clarification I think. Does a teammate have to announce ahead of the roll whether they're helping? Or can they look at the result and say "I'll spend my help action to reroll the d-percentile die"? Closhund/talk/ 23:23, 14 August 2023 (UTC)

They've got to declare beforehand that they're helping a specific ally's attempt (and sacrificing their own opportunity to try by doing so), instead of taking their own attempt to fix the radio. That's a fair point, how in the end the party is still rolling 4 times, and they still only need a single one of those checks to pass. Like you mentioned, the only difference seems to be that the "4 straight checks" has four opportunities to fumble, whereas the "2 checks with bonus dice" has significantly lower odds of fumble (either attempter would need to fumble on both their regular and bonus roll.)

My brain's having trouble committing to the thought that "reroll both dice" and "reroll only the tens die" are essentially equivalent when it comes to a player's odds of passing the check, but I can kinda see why that'd be the case. Indeed, I can't think of a concrete reason why it wouldn't be, it still boils down to whether that tens die comes up 00 vs any other number.--Aabicus (talk) 23:37, 14 August 2023 (UTC)

There should be 100 possible outcomes of the two-dice roll. Are the possible outcomes 00 through 99, or 01 through 100? A two-player team fails if the first player's roll fails and then the second player's roll also fails. We may assume the rolls are independent. This means that the probability of a double failure is the product of the probabilities of the single failures. If 10 out of the 100 equally likely outcomes mean success and 90 mean failure, the probability of a single failure is 90/100 = 0.9. That of a double failure is then 0.9 × 0.9 = 0.81.. The probability of the complementary event, team success, is its complement with respect to 1, 1 − 0.81 = 0.19, or 19%.  --Lambiam 22:21, 14 August 2023 (UTC)

Rolling a 00 is considered 100 on a d100, so the outcomes are 00-99 but 00 is still a fail from the perspective of the players. Does that 19% include the 'bonus dice' mechanic whereby both attempters get to reroll their d-percentile and hope for a lower digit in the tens-column of their result? --Aabicus (talk) 22:27, 14 August 2023 (UTC)

So the "fumble" mechanic makes things a bit more interesting. With taking the best of two I gather that the chance of a fumble is .05² = .0025. For each pair of rolls, the chance of a fumble or immediate loss is .05, the chance to not win otherwise is .9²-.0025=.8075 and the chance of a win is 1-.9²=.19. You only roll a second time if there is a "not win", so multiply those odd by .8075, you get .00201875 chance of fumble on the second pair of rolls, .65205625 of a not win, and 0.153425 of a win. The totals are .343425 to win, and .656575 to lose. Note that the fumble rule does not significantly change the outcome, and that makes sense considering it's so unlikely.

In the context of a game, small differences in probability aren't really noticeable. To distinguish between 34% and 35% to any statistical significance would take hundreds of repetitions, and most people aren't going to do that, and be willing to carefully record the results, unless they're being paid for the effort. Popular games tend to have complicated mechanisms built in to make find exact odds difficult to compute; people seem to prefer having some kind of suspense in the process. --RDBury (talk) 01:55, 15 August 2023 (UTC)

Thank you for breaking it down! This is pretty much exactly what I was hoping to know, they have roughly a 34% chance of fixing the radio. I'm also able to follow your train of calculation so I can apply it to situations in the future to determine roughly how likely or unlikely some certain check will be for this party. Thanks again! --Aabicus (talk) 08:09, 15 August 2023 (UTC)

It's still not clear to me. Let's call the teams team A and team B, and the players A1, A2, B1 and B2. A1 rolls both the d100 and the d%ile and gets 23, say. Does player A1 now re-roll the d%ile, and if they get 73 player A2 can now also roll both dice and so on? Or does A1, after getting 23, let A2 re-roll the d%ile in the hope of getting 03? If both A1 and A2 can roll both dice and re-roll the d%ile, the probability of one-player failure is 0.81 and of team failure is 0.3439. Also, do you want to know the probability of "at least one of teams A and B gets through" or of "A gets through"?  --Lambiam 06:52, 15 August 2023 (UTC)

After rereading I now think that the "attempters" from the question are A1 and B1, and that A2 and B2 never roll the dice. The probability that an individual attempter fails is then 0.81. The probability that both attempters fail is 0.3439. This ignores fumbling. Apparently, there is one radio, and if one team breaks it, the other team does not get a chance. Can A1 roll the dice, and if getting 95, still reroll the d%ile die? Or is it then game over (as far as the radio is concerned) for everyone?  --Lambiam 07:11, 15 August 2023 (UTC)

In that scenario, A1 would get to use their bonus die and reroll their d-percentile (assuming a bonus die is in play, granted from A2 in your example) and the skillcheck wouldn't be a fumble unless that bonus die also came up a fumble. If there isn't a bonus die in play (e.g. all four players chose to attempt to fix the radio, no helping), any one of them rolling a fumble breaks the radio. (And each attempter would roll one-at-a-time, not all at once, so there's no chance of "A1 fumbled, but B1 fixed the radio at the same time" sort of trouble.) --Aabicus (talk) 08:04, 15 August 2023 (UTC)

The possible outcomes for a single attempter are success, (simple) failure and fumble. The probabilities of these three outcomes are 0.190 for success, 0.756 for failure, and 0.054 for fumble. The radio gets repaired if team A succeeds, or if team A fails but does not fumble and team B succeeds. This gives a probability of repair of 0.190 + 0.756 × 0.190 = 0.33364.  --Lambiam 09:06, 15 August 2023 (UTC)

Awesome, thank you so much! --Aabicus (talk) 18:47, 15 August 2023 (UTC)

In future you can note that for games like CoC, the probability distribution for the minimum of two uniform dice rolls looks like a one-sided triangular distribution, which has a S-shaped cumulative distribution (i.e., probability that an assisted check succeeds). So if you can follow along a bit on the calculations here, you could also try plotting it out on a graph with different choices for the success threshold. (For a very basic introduction to simple dice probabilities, that includes the PDF and CDF, see a 2018-05-09 blog post by Mazur.

Responding to RDBury: FYI most RPGs I've examined don't actually have complex dice mechanisms from a probability perspective -- at most they have a triangular PDF (the most curvature you can get out of two fair or Fudge dice rolls), but most use uniform distributions on top of which they add bells and whistles that are just equivalent linear shifts (see e.g. most editions of D&D for the most egregious, most popular example, that somehow makes it also the most confusing and mentally difficult). The most sophisticated popular RPG mechanic in probabilities, though still relatively easy to calculate analytically, as well as to process mentally during gameplay, are the White Wolf games. A consideration with dice mechanisms is of course always the speed and complexity of mental calculation. But from my dive into the subject of tabletop gaming, most of the change in usability comes from design and testing over probability theory. Some may even purposely design against usability -- it seems D&D designers and players embrace their illusion of mathematical complexity and the sense of exclusivity that their frustrating early mechanics brings -- and I can see the appeal from watching especially retrogamer groups. (And if actually do you know someone willing to pay for my effort on this, then please let me know.) SamuelRiv (talk) 11:49, 15 August 2023 (UTC)

• Just as an aside on the gameplay issue; as a DM, your role is to keep the game going and make it fun and challenging. You can always bend the rules a bit for the sake of gameplay. Dice rolling is a useful way to add tension and randomness to a game session, and present unforeseen challenges that force you and the players to think and adapt to the situation quickly and in novel ways. Having an "out" to keep the plot moving along, even if you run into a situation where basically no one fixes the radio, would be a good thing to plan for... Just my 2 cents. --Jayron32 13:24, 15 August 2023 (UTC)

  Always good to keep in mind! I'm fully expecting them to fail on repairing the radio, the stakes are higher if they don't manage to call for help and need to find their own escape off the island. But if they beat the odds on the check (or somebody comes up with an ingenious alternate method for fixing the radio using a skill other than Mechanical Repair) I'm more than happy to let them succeed and introduce a new plot element through it. I just wanted to know how likely it was that happens or if I should implement some gimme like a Radio Operator's Manual they could find in a drawer. But at 1-in-3 odds I think I'm just gonna let them court lady luck. --Aabicus (talk) 19:05, 15 August 2023 (UTC)