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December 4

Subring of prime ring

Is a subring of a non-commutative prime ring always itself a prime ring? If yes, give a proof; otherwise, give a counterexample. GeoffreyT2000 (talk) 05:51, 4 December 2017 (UTC)

This looks like homework so I'll let you work out the details, but it looks like you might want to try taking R to be the ring of 2x2 matrices over a field and D to be the subring of diagonal matrices. --RDBury (talk) 17:50, 4 December 2017 (UTC)

No LaTeX \phantom in WP? Work around?

As the title says ... I did not succeed to find something in Help:Displaying a formula and also not in other places, so I ask for cheap tricks, possibly saving me to dig into arrays with elaborate aligning. Purgy (talk) 18:08, 4 December 2017 (UTC)

No, \phantom isn't supported. What are you trying to do? For the future, this would probably be better at Help Talk:Displaying a formula or WT:WPM. –Deacon Vorbis (carbon • videos) 18:40, 4 December 2017 (UTC)

Thanks for the confirmation, and I'll tell my woes there. Sorry, for having been wrongly here. Purgy (talk) 08:53, 5 December 2017 (UTC)

Um, DV, Help talk:Displaying a formula is for discussing ways to improve the page Help:Displaying a formula; it's not a support forum. @Purgy Purgatorio: a better place to try would probably be Wikipedia:Help desk, since that's specifically set up to help users "use or edit Wikipedia". (time passes, as I start to type up a reply at Help:Displaying_a_formula…) I see that users have begun to ask you followup questions at Help talk:Displaying a formula#Work around for LaTeX \phantom?, so maybe you will end up getting some help there. Alternatively, you could just start editing whichever article you want to contribute to, and if someone reverts your changes because of problems with the TeX coding, pull the math content out of the page history, post it to the talk page, and ask for help from users who know how to edit math. Presumably, if the article calls for heavy math coding, there will be users watching the article who can do that.) - dcljr (talk) 10:28, 5 December 2017 (UTC)

@Dcljr, thank you, for your hints to the bureaucratically correct use of "Help talk:Displaying a formula", for pointing me to the universal "Wikipedia:Help desk", and for mentioning the possibility of discussing my problem on some specifically cognizant talk page. — It's just that I got simply no reply to my question referring to this at that talk page, and that I see the Help desk crowded with people not really interested and involved in math layout (as seemingly is the whole upper squad of WP). Obviously, my usurpational attack on this here Reference desk, which supplied me both with further contacts and with a confirmation of my conjecture, is at least, albeit not fully de rigeur, quite successful. Thanks for caring, anyhow. :) Purgy (talk) 19:42, 5 December 2017 (UTC)

December 5

Matching birthdays

I know that in a room full of people you need very few before you have a pair with a common birthday (something like 23 - I haven't looked it up).

If you have two rooms of people, each with 'N' people, how big does N have to be before you (on average) get a match? How big for 'Y' matches (in particular 'Y=20'? -- SGBailey (talk) 07:30, 5 December 2017 (UTC)

Am I wrong? Sureley if you had a room with (say) 23 people in it, the probability if a match is ‘x’. If you then built a dividing wall in the room, making, in effect, two rooms, the probability of a match would still be 'x’, irrespective of whether the division was 11/12, or 10/13, or 22/1 etc? — Preceding unsigned comment added by 82.38.221.49 (talk) 09:40, 5 December 2017 (UTC)

I am interpreting the question differently. I think the question is how many people need to be in each of 2 rooms (same number in each room) so that there is a greater than 50% probability that a pair in one room or the other (or both) have the same birthday. If there are n people in each room and the probability of a birthday match in one room is P(n) then we want to find n such that (1-P(n))² < 0.5 i.e. probability of no match in either room is less than 0.5. So P(n) > 1 - sqrt(0.5) which is approximately 0.29. To make P(n) greater than 0.29, I get n = 17, which makes P(n) approximately 0.315 and (1-P(n))² approximately 0.469. Gandalf61 (talk) 10:09, 5 December 2017 (UTC)

I obviously phrased the question poorly. N people in each room. How many are needed so someone in room A matches a person in room B? (EG mixed sex couples arrive at a party - how long before a man has the same birthday as a woman?) How many people in each room before 20 matches are likely? It is possible for there to be 1 million in each room and not have a match (if room A is filled with Jan to Jun and Room B with Jul to Dec). -- SGBailey (talk)

Just a quick note, you need to be a little more careful what you're looking for. Something like "how many people have to be in the room before there's a match?" doesn't make sense. You need to ask something like "how many people have to be in the room until there's at least a 50% chance of a match?". –Deacon Vorbis (carbon • videos) 14:06, 5 December 2017 (UTC)

Have you seen Birthday_problem? AlfonsoAnonymous (talk) 19:01, 5 December 2017 (UTC)

Yes, I even refer to it (though not by name) in the first line of my original question.

Also I did mention "on average" and "matches are likely" as an (obviously failed) attempt to indicate that certainty isn't possible and probability is the name of the game. So does anyone know the answer or perhaps how to evaluate it. -- SGBailey (talk) 19:21, 5 December 2017 (UTC)

The probability of no collision can be written as an explicit double sum involving Stirling numbers of the second kind, which is good enough for computation: here are the probabilities of no collision for up to 20 people. {{0, 1.}, {1, 0.99726}, {2, 0.989086}, {3, 0.975611}, {4, 0.957054}, {5, 0.933714}, {6, 0.905959}, {7, 0.874221}, {8, 0.838981}, {9, 0.800759}, {10, 0.760099}, {11, 0.71756}, {12, 0.6737}, {13, 0.629065}, {14, 0.584179}, {15, 0.539533}, {16, 0.49558}, {17, 0.452724}, {18, 0.411317}, {19, 0.371661}, {20, 0.333997}}.

On the other hand, the expected number of matches is extremely straightforward to compute by linearity of expectation: it is $n^{2}/365$ for n people in each group. (Both answers disregard leap years.) --JBL (talk) 21:40, 5 December 2017 (UTC)

Thank you - that was what I was seeking. -- SGBailey (talk) 07:15, 6 December 2017 (UTC)

I know you referred to the one point about needed 23 people before the probability is above 50%, but because you did not refer to it by name I wasn't sure if you had found the relevant Wikipedia article. It has more information than just that one number, and I thought some of that information may be helpful. In fact, the page I linked mentions the double sum involving Stirling numbers of the second kind that I believe JBL refers to. (Though JBL may be referring to a different formula, I am not totally sure. The point is that the article has plenty of relevant information.) AlfonsoAnonymous (talk) 11:30, 7 December 2017 (UTC)

Yes, that's right. (Your first answer was a good one, and would have been even if a section addressing this specific variation hadn't existed.) --JBL (talk) 12:37, 7 December 2017 (UTC)

Successor function paradox loop?

If the successor function is defined for $n\in \mathbb {N}$

${\begin{aligned}(1)\quad &n+0=n\\(2)\quad &S(n)=n+1\\(3)\quad &n+S(m)=S(n)+m=S(m+n)\end{aligned}}$

then how are we trying to prove associativity for addition, while definition (3) requires the proofs of associativity and commutativity below?

$n+(m+1)=(n+1)+m=(n+m)+1$

יהודה שמחה ולדמן (talk) 15:30, 5 December 2017 (UTC)

I don't think I understand your question. How does the definition require the proofs below? What proofs below? What are you actually asking? –Deacon Vorbis (carbon • videos) 15:55, 5 December 2017 (UTC)

What the proof of associativity for addition actually wants to do, is to prove the general associativity n+(m+c) = (n+m)+c, by the special case of associativity - wherein c=1, so where is the loop? Is the loop the (legitimate) desire to prove "general associativity" - by "special associativity" - i.e. by (3)? I don't see here any loop, because (3) - being a part of the definition of the successor function - does not need any proof, because - like all other definitions - (3) is like an axiom. HOTmag (talk) 16:21, 5 December 2017 (UTC)

Also, it sounds like you've got things backwards. Addition is defined in terms of the successor function, not the other way around. Maybe that's where your confusion arises. On a side note, the article you linked with the proofs is kind of a train wreck. I'm not sure whether it should be improved or just submitted to AfD. Does anyone else have any thoughts? –Deacon Vorbis (carbon • videos) 17:20, 5 December 2017 (UTC)

I believe the proof of associativity in the article is correct. However I don't think we should be giving proofs in Wikipedia unless they're fairly trivial or of particular note - we should point people at citations for the proofs. Dmcq (talk) 18:30, 5 December 2017 (UTC)

Let me ask: You are saying all 3 properties are actually defined to already have both associativity and commutativity? יהודה שמחה ולדמן (talk) 19:26, 5 December 2017 (UTC)

Your statement seems to have a category error. What does it mean for one of the listed properties to be "defined to already have associativity and commutativity" ? The properties are as you have listed them. Nothing more.--129.74.112.165 (talk) 21:14, 5 December 2017 (UTC)

According to the linked article, the successor function isn't actually defined but is subject to the Peano axioms. You can then prove addition is associative based on those axioms and the definition of addition in terms of the successor function. You might want to read 'Foundations of Analysis' by Edmund Landau for a good, if slightly dated, exposition of this; it's available for free on the internet. --RDBury (talk) 21:27, 5 December 2017 (UTC)

PS. I see that Landau is already used as a reference in the second linked article. To respond to Dmcq, a few years back there was a long discussion on Wikipedia talk:WikiProject Mathematics about the policy for proof on WP. The proofs in the article do seem unencyclopedic and imo should be moved to WikiBooks. A major problem with proofs in an encyclopedia is there needs to be some context for a proof to be valid; what is being assumed for the proof and what is the theorem to be used for. In this case there is a flaw in that the so-called inductive definition (A1 and A2) in the article does not define anything; it's just a pair of functional equations. In order for these to become a definition there has to be a theorem which states that this pair of functional equations has a unique solution, and this theorem isn't in the article. Since the + operation hasn't been properly defined it makes no sense to even ask whether it's associative. Landau goes to some lengths to address this issue and to do the job properly you really need to view the book as a whole, not just pull out bits and pieces. --RDBury (talk) 22:19, 5 December 2017 (UTC)

PPS. Wikipedia:WikiProject Mathematics/Proofs has some material on when and how to include proofs in WP. Not saying it's binding or even that I agree with everything in there though. --RDBury (talk) 22:32, 5 December 2017 (UTC)

December 6

A question about a countable bijection

Does there exist any function (not as recurrence relation) for monomial order? Alireza Badali (talk) 12:16, 6 December 2017 (UTC)

I presume that you are interested in the infinite-dimensional case, and don't want to appeal to the axiom of choice? — Charles Stewart (talk) 13:36, 6 December 2017 (UTC)

I'm not interested in logic I want just make a group on ${\mathbb {N}}$ isomorph to the $({\mathbb {Q}},+)$ Alireza Badali (talk) 16:33, 6 December 2017 (UTC)

IIUC, you can do what you are asking by showing that the rationals are countable and then throwing out duplicates. You don't need monomials, if that is really what you want. — Charles Stewart (talk) 16:38, 6 December 2017 (UTC)

Yes this sequence:

$(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),(2,2k-3),...$ $,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),...,(k,k),...$ $,(2k-2,2),(2k-1,1),...$

of course Professor Daniel Lazard gave me its formula:

${\begin{cases}(a_{1},b_{1})=(1,1)\\(a_{n+1},b_{n+1})={\begin{cases}(1,a_{n}+1)&b_{n}=1\\(a_{n}+1,b_{n}-1)&{\text{else}}\end{cases}}\end{cases}}$

but I need a formula not as recurrence relation. Alireza Badali (talk) 17:00, 6 December 2017 (UTC)

I understand what you want, but I don't see what this has to do with monomials. Also, your recurrence function does not throw out duplicates. — Charles Stewart (talk) 22:21, 6 December 2017 (UTC)

Please ignore monomial and only that sequence is important! Alireza Badali (talk) 13:11, 7 December 2017 (UTC)

I've changed the title of this question. — Charles Stewart (talk) 13:23, 7 December 2017 (UTC)

Given the recurrence you were given and thec trivial injection from the naturals into the positive rationals, you get a bijection, but you need the axiom of choice to get it directly. — Charles Stewart (talk) 13:26, 7 December 2017 (UTC)

Take a look at https://math.stackexchange.com/questions/193263/countability-of-mathbbq — Charles Stewart (talk) 13:29, 7 December 2017 (UTC)

Good title, thank you, but really what is for doing with axiom of choice here$\color {red}{?}\color {green}{!}\color {blue}{?}\color {fuchsia}{!}$ Alireza Badali (talk) 13:53, 7 December 2017 (UTC)

Maybe Cantor pairing function is what you want. --JBL (talk) 00:46, 9 December 2017 (UTC)

Thank you. Alireza Badali (talk) 10:58, 9 December 2017 (UTC)

Two questions about density

Question $1$: Let $r:{\mathbb {N}}\to (0,1)$ is a function given by $r(n)$ is obtained as put a point at the beginning of $n$ like $r(34880)=0.34880$ and similarly consider $\forall k\in {\mathbb {N}}\cup \{0\}$ $r_{k}:{\mathbb {N}}\to (0,1)$ by $r_{k}(n)=10^{-k}\cdot r(n)$ and ${\mathbb {P}}$ is the set prime numbers now is $\{t^{i}\,|\,t\in r({\mathbb {P}})\}$ dense in the $\{z\in {\mathbb {C}}\,|\,|z|=1\}$ and is $\{s\cdot t^{i}\,|\,s,t\in \bigcup _{k\in {\mathbb {N}}\cup \{0\}}r_{k}({\mathbb {P}})\}$ dense in the $\{z\in {\mathbb {C}}\,|\,0.1\leq |z|\leq 1\}$.

Question $2$: Is $\{(p^{2}+q^{2})^{-0.5}\cdot (p+qi)\,|\,p,q\in {\mathbb {P} }\}$ dense in the $\{z\in {\mathbb {C} }\,|\,|z|=1,Re(z)\geq 0,Im(z)\geq 0\}$. Alireza Badali (talk) 19:02, 6 December 2017 (UTC)

r(1)=0.1, and r(01)=0.01, but 1=01, so r is not a function. Bo Jacoby (talk) 22:41, 6 December 2017 (UTC).

That seems like a cheap shot. You could equally argue that 5 = 101₂ but r(5) = 0.5 ≠ r(101₂) = 0.101₂ = 0.625. Each natural number has a unique base-10 representation without leading zeros, and a charitable reading of this question has n represented thusly. -- ToE 02:41, 7 December 2017 (UTC)

What is $01\color {red}{?!}$ Alireza Badali (talk) 13:39, 7 December 2017 (UTC)

Is that stated quite how you wish? Forget for a moment r(ℙ). Can you think of any t ∈ (0,1) such that tⁱ is close to -1? -- ToE 03:02, 7 December 2017 (UTC)

Or, keeping r(ℙ) in mind, is there likely to be any t ∈ r(ℙ) such that tⁱ is close to 1? -- ToE 03:18, 7 December 2017 (UTC)

And why are you complicating things by projecting this into the complex plane. Isn't what you are really getting at asking if r(ℙ) is dense in (0.1,1) and if {p/q | p, q ∈ ℙ} is dense in ℝ⁺? -- ToE 03:24, 7 December 2017 (UTC)

Complex plane wasn't my idea and a Professor (that likes be unknown) offered it to me but I don't know how should I operate with complex plane. Alireza Badali (talk) 13:39, 7 December 2017 (UTC)

How would you address the real number, "less complex", related problems I offered? How would you then draw a connection from there to the complex plane? (See Dense set#Properties.) For extra credit, determine what is wrong with your professor's statement of the first problem, correct it, then solve it. -- ToE 14:16, 7 December 2017 (UTC)

The proved theorem for reals is: Let $\mathbb {P}$ is the set prime numbers and $S$ is a set that has been made as below: put a point at the beginning of each member of ${\mathbb {P}}$ like $0.2$ or $0.19$ then $S=\{0.2,0.3,0.5,0.7,...\}$ is dense in the interval $[0.1,1]$ of real numbers. that its proved extension is: For each subinterval of $[0.1,1)$ like $(a,b),\,\exists m\in {\mathbb {N}}$ that $\forall k\in {\mathbb {N}}$ with $k\geq m$ then $\exists t\in (a,b)$ that $t\cdot 10^{k}\in {\mathbb {P}}$.

and this: Assume $S_{1}=\{a/10^{n}\,|\,a\in S$ for $n=0,1,2,3,...\}$ then $S_{1}$ is dense in the interval $(0,1)$ and $S_{1}\times S_{1}$ is dense in the $(0,1)\times (0,1)$.

But I want make a form for sets $S$ & $S_{1}$ & $L$ in the complex plane and below conjecture that is an equivalent to Goldbach conjectureas: Let $L=\{(a,b)\,|\,a,b\in S_{1}$ & $0.01\leq a+b<0.1$ & $\exists m\in {\mathbb {N} },\,a\cdot 10^{m},\,b\cdot 10^{m}$ are prime numbers & $a\cdot 10^{m}\neq 2\neq b\cdot 10^{m}\}$ but conjecture: For each even natural number like $t=t_{1}t_{2}t_{3}...t_{k}$, then $\exists (a,b),(b,a)\in L\cap$ $\{(x,y)\,|$ $x+y=0.0t_{1}t_{2}t_{3}...t_{k}\}$ such that $0.0t_{1}t_{2}t_{3}...t_{k}=a+b$ & $10^{k+1}\cdot a,10^{k+1}\cdot b$ are prime numbers. Alireza Badali (talk) 16:44, 7 December 2017 (UTC)

My questions were rhetorical, intended as guiding hints. (We really don't need or want to see your proof.) If you have the solution in ℝ, then look and see how the problem is mapped over to ℂ and then apply Dense set#Properties. Do you understand what tⁱ does? -- ToE 17:01, 7 December 2017 (UTC)

Indeed I have no plan for complex plane but I can solve it but main difficulty for me is this question. Alireza Badali (talk) 18:24, 7 December 2017 (UTC)

• On the top of this page, it says We don't do your homework for you, though we’ll help you past the stuck point. This looks plainly like a homework question (though an advanced one). What have you tried so far? Tigraan^{Click here to contact me} 10:45, 7 December 2017 (UTC)

My notes is on real numbers but a Professor (that likes be unknown) offered it to me but I don't know how should I operate with complex plane. Alireza Badali (talk) 13:39, 7 December 2017 (UTC)

Thank you so much guys. Alireza Badali (talk) 18:24, 7 December 2017 (UTC)

December 8

Three-way vertex query

As far as I can see by drawing some particular cases, an undirected graph with all vertices being the meet of three edges, and only one edge joining any particular pair of vertices, must have an even number of vertices. Is this the case? Further, how many different such graphs have exactly six vertices? - I can find just two, one with three vertices inside an outside loop of three and the other with two inside an outer loop of four. → 31.50.229.60 (talk) 19:30, 8 December 2017 (UTC)

For the first statement, see handshaking lemma (and yes, you are correct). For the enumeration of cubic graphs, see [1] (and again you are correct). --JBL (talk) 19:58, 8 December 2017 (UTC)

Thanks. On checking, my two six-vertex graphs are topologically the same, and the second actual one is invalid for my purposes as it requires edge-crossing. →31.50.229.60 (talk) 13:37, 9 December 2017 (UTC)

December 9

Sines in a triangle

Let ABC be a triangle and O a point in the interior, then sin ∠BAO sin ∠CBO sin ∠ACO = sin ∠OAC sin ∠OBA sin ∠OCB. This isn't hard to prove being an application of the Law of Sines, and I assume it's well known, so does anyone know a reference for it? --RDBury (talk) 17:33, 9 December 2017 (UTC)

Have you looked at Law of sines? Dolphin (t) 20:46, 9 December 2017 (UTC)

That article seems to only cover a simple triangle, but what I'm looking for is more a relation between the angles in a complete quadrangle. Besides, the Law of Sines is certainly not the only way to prove the relation. It seems to be similar to Ceva's theorem but using sines, I don't see the exact connection though. --RDBury (talk) 23:59, 9 December 2017 (UTC)

Ok, now I see the connection to Ceva's theorem. In fact, the relations seems to be known as the 'trigonometric form of Ceva's theorem'. I found a reference on p66 (equation 10) of Proceedings of the Edinburgh Mathematical Society vol XX (1901-1902) but it's not given a name there. I'd still be interested to know if there is an earlier or more definitive reference though. --RDBury (talk) 00:56, 10 December 2017 (UTC)

@RDBury: The article indeed covers a single triangle, but we have just three single triangles here. Consider the ABO triangle: ${\tfrac {|BO|}{\sin \angle BAO}}={\tfrac {|AO|}{\sin \angle OBA}}$, which implies $\sin \angle BAO=\sin \angle OBA{\tfrac {|BO|}{|AO|}}$. Similary from triangles BCO and CAO we have $\sin \angle CBO=\sin \angle OCB{\tfrac {|CO|}{|BO|}}$ and $\sin \angle ACO=\sin \angle OAC{\tfrac {|AO|}{|CO|}}$. Multiply all three equalities side-wise ad we get:
$\sin \angle BAO\cdot \sin \angle CBO\cdot \sin \angle ACO=\sin \angle OBA{\tfrac {|BO|}{|AO|}}\cdot \sin \angle OCB{\tfrac {|CO|}{|BO|}}\cdot \sin \angle OAC{\tfrac {|AO|}{|CO|}}$
All line segments' lengths will get reduced and the claimed equality gets proven. --CiaPan (talk) 21:19, 10 December 2017 (UTC)

That's pretty much the proof I had in mind, but it's not covered in the article other than you're using the theorem that the article is about. Like I said, it's not hard to prove but you need to say more than just "By the law of sines." The problem I was having was finding a reference since it was too easy not to be already known. --RDBury (talk) 23:42, 10 December 2017 (UTC)

December 10

Roulette

The numbers on a roulette wheel are not sequential. They run (on the French wheel):

0-32-15-19-4-21-2-25-17-34-6-27-13-36-11-30-8-23-10-5-24-16-33-1-20-14-31-9-22-18-29-7-28-12-35-3-26

and (on the American wheel):

0-28-9-26-30-11-7-20-32-17-5-22-34-15-3-24-36-13-1-00-27-10-25-29-12-8-19-31-18-6-21-33-16-4-23-35-14-2.

These are not random sequences - black and red numbers alternate according to the rule that even numbers between 2 and 10 and 20 and 28 (inclusive) are black. The French sequence was not standardised originally, so how did these sequences originate? Is there some underlying pattern and did the arrangement of the dominant manufacturer prevail as happened with the typewriter? How many different sequences can be generated within the constraint of the red/black rule? 92.27.49.50 (talk) 13:29, 10 December 2017 (UTC)

One of your questions is easy to answer. If you already have a rule that assigns red to half of the numbers and black to the other half, and have already constructed and painted your wheel, then there are 18! ways of numbering the black pockets and 18! ways of numbering the red pockets. See factorial and permutation. Thus the total number of different sequences which can be generated within the constraint of the red/black rule is the product of those two numbers, or (18!)² ≈ 4.1×10³¹ (a very big number). Without following a predetermined red/black rule (but still keeping the zeros where they are), then the total number of different sequences is 36! ≈ 3.7×10⁴¹ (an even bigger big number). -- ToE 16:26, 10 December 2017 (UTC)

The relevant section of our article, Roulette#Roulette wheel number sequence, gives the sequences but not their history. -- ToE 16:33, 10 December 2017 (UTC)

There is a pattern in the placement of high (19-36) and low (1-18). The French wheel alternates high, low, high, low, ... in either direction from the zero, meeting half-way across the wheel at two adjacent lows, 10 & 5. This, combined with the alternating red and black pockets, ensures that half of either color is high and half is low. The American wheel alternates once, then alternates in pairs, that is, H,L,H,H,L,L,H,H,L,L,..., clockwise around the wheel, starting from either zero. Since the pockets adjacent to a zero (either zero) are the same color on an American wheel, this pattern also ensures that half of either color is high and half is low. Since 18 is not divisible by 4, each half of the American wheel could not alternate in pairs while keeping the same number of high and lows on each half, hence the initial singleton alternation. -- ToE 17:03, 10 December 2017 (UTC)

Additionally, for the half of the American wheel clockwise from 0 to 00, the highs are all even and the lows are all odd, with the opposite the case for the other half of the wheel, and with the consecutive numbers 1&2, 3&4, etc. diametrically opposed on the wheel. -- ToE 17:47, 10 December 2017 (UTC)

Looking at the sequences, there appears to be an effort to make it so that the bets listed at Roulette#Outside bets don't cluster on one half of the wheel. I presume that this is to avoid the possibility of a skilled player being able to hit one side more often than the other. I wonder if either of the traditional sequences are actually optimized for this. Sounds like an interesting math puzzle. --Guy Macon (talk) 19:58, 10 December 2017 (UTC)

Detail correction: it's not the people betting who launch the ball, it's the croupier. But it doesn't matter, since no honest casino wants their croupiers to be able to influence who wins. --69.159.60.147 (talk) 05:34, 11 December 2017 (UTC)

I see what is almost one more pattern in the American wheel, but I don't quite know what to make of it. Since one side of the wheel determines the other, it is sufficient to look for patterns within and between the high/even and the low/odd sequences on the first half of the wheel, separating them as 28-26-30-20-32-22-34-24-36 and 9-11-7-17-5-15-3-13-1. For relative ordering of the first sequence, subtract 18 and divide by two, and for the second sequence, add one and divide by two, giving 5-4-6-1-7-2-8-3-9 and 5-6-4-9-3-8-2-7-1. If you cast out the 5 (which is from the singleton high/low alternation before they start alternating in pairs) then you get two quite similar sequences: 4-6-1-7-2-8-3-9 and 6-4-9-3-8-2-7-1. Rotate the first sequence left by two (i.e., take the 4-6 from the start and put it on the end), and you have the the reverse of the second sequence. This seems too significant to be mere coincidence, but not significant enough to have much meaning. The individual sequence 1-7-2-8-3-9-4-6 isn't quite broken at the end, as it comes in pairs of n, (n mod 4) + 6. This reflects the pattern in the alternating high, high and low, low pairs where within each pair there is one number from the upper half of the range and one from the lower, with the difference between them 12 except for when that take the higher number beyond its range in which case the difference is 4. For example, break the even highs into the lower highs, 20, 22, 24, & 26, the middle high 28, and the upper highs, 30, 32, 34, & 36. 28, the middle high comes first, then the pair 26 & 30, namely the fourth of the lower highs and the first of the upper highs, then 20 & 32, namely the first of the lower highs and the second of the upper highs, then 22 & 34, namely the second of the lower highs and the third of the upper highs, and so on. Using the symbols Even/Odd, High/Low, Upper/Middle/Lower, & numbers 1..4, we have EH:(M--L4-U1--L1-U2--L2-U3--L3-U4). The odd lows run OL:(M--U1-L4--U4-L3--U3-L2--U2-L1). (And the two are appropriately interleaved.) So there is a pattern there, with the two running in the opposite direction, but jumbled up just a bit at the start. -- ToE 20:03, 10 December 2017 (UTC)