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Re: [math-fun] playing card color guessing game
by Dan Asimov 26 Dec '18

26 Dec '18

Hmm, strategy seems clear: guess the majority color remaining, if any. So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum): E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc. —Dan Allan Wechsler wrote: ----- I know how to analyze this, but I haven't tried to actually do the work. But I'm going to make a wild-ass stab based on an intuition. Is it $32.87? On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck(a)gmail.com> wrote: > the usual 52-card playing deck is shuffled and then revealed one-by-one to > a player who guesses the color (red or black) of each card prior to its > being revealed. the player earns one dollar for each card whose color he > guesses correctly; there is no penalty for being wrong. what is a fair > entry price to play? ----=-

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[math-fun] playing card color guessing game
by Thane Plambeck 26 Dec '18

26 Dec '18

the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play?

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Re: [math-fun] playing card color guessing game
by Dan Asimov 26 Dec '18

26 Dec '18

I don't know the answer, but it's a beautiful question. And what does the entry price approach as 52 —> oo ??? —Dan Thane Plambeck wrote: ----- the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? -----

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Re: [math-fun] Klein bottle puzzle (SPOILER)
by Dan Asimov 25 Dec '18

25 Dec '18

Elements of the fundamental group of a space X can be any oriented closed curve that starts and ends at the basepoint, which is just some point chosen for the purpose. As long as X is path-connected, the resulting group π_1(X) will be independent of the choice. The fundamental group of the Klein bottle can be given the presentation π_1(K) = < a, b | a b a^(-1) b = 1 > which when made abelian must become the group H_1(K) = Z + Z_2. Since K is a surface instead of being higher-dimensional, lots of these elements of π_1(K) cannot be represented by *simple* closed curves. But Yes — each of the 5 classes of simple closed curves on K can be given an orientation ... and then as long as each curve contains the basepoint, they can then be thought of as elements of π_1(K) that have been identified to their inverses. *However*, there is a subtlety: The continuous deformation that can make two closed curves in a space X equivalent in π_1(X) must be through curves that all start and end at the same basepoint. This means that *more curves might be equivalent* in the present situation: Any two simple closed curves on K are equivalent (in this question) if there is a continuous map H : S^1 x [0, 1] —> K with one curve being H(( , 0)) : S^1 —> K and the other H(( , 1)) S^1 —> K. —Dan Cris Moore wrote: ----- On Dec 25, 2018, at 2:16 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: PS For more fun with the Klein bottle, identify the exactly 5 classes of (unoriented) simple closed curves on it that cannot be deformed one to the other. I take it these are the elements of the fundamental group, with inverses identified? -----

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Re: [math-fun] Klein bottle puzzle (SPOILER)
by Dan Asimov 25 Dec '18

25 Dec '18

Defeated by the Klein bottle! Its mapping class group is not Z_2, as I had erroneously claimed. As Allan and Andy pointed out, sources state that the mapping class group is Z_2 + Z_2. So, what are these four classes of self-homeomorphisms of K ??? Any homeomorphism h : X —> X of a space X to itself induces an automorphism on each homology group H_n(X). For the Klein bottle, the group H_1(K) = Z + Z_2. There are just 4 automorphisms of Z + Z_2. (The generator u of Z_2 must go to itself, and the generator 1 of Z can go to 1, -1, 1+u, or -1+u.) Can these 4 automorphisms all be induced by homeomorphisms K —> K ??? As it turns out, the answer is Yes. And by basic facts, homeomorphisms that induce unequal automorphisms on any H_n cannot be continuously deformable to each other, so (assuming that Z_2 + Z_2 is the correct group), these must be all the cases. * The 1 |—> 1 case is induced by the identity K —> K. * The 1 |—> -1 case is induced by reversing the cylinder before identification. But the other two cases? These come from homeomorphisms of K that in my naïveté I had not suspected: Follow either of the above maps on the cylinder with the operation of *twisting one end of the cylinder 360º while holding the other end fixed* . (Because of the orneriness of the Klein bottle, it doesn't matter which way you twist it.) This does not appear to be the same homeomorphism as just flipping the orientation of the cylinder. In fact, if flipping the orientation of the cylinder C means reflecting it about a plane P of symmetry that cuts C into two rectangles (and then identifying the ends of C to get K), this self-homeomorphism of K can be deformed to the identity by sliding the Klein bottle around itself once. —Dan PS For more fun with the Klein bottle, identify the exactly 5 classes of (unoriented) simple closed curves on it that cannot be deformed one to the other. Andy Latto wrote: ----- On Mon, Dec 24, 2018 at 10:35 PM Allan Wechsler <acwacw(a)gmail.com> wrote: > > Wikipedia gives a different answer. See the article "Mapping Class Group", > in the section "Examples", subsection "Nonorientable surfaces", where they > say it's Z2 x Z2. > > > On Mon, Dec 24, 2018, 9:52 PM Dan Asimov <dasimov(a)earthlink.net wrote: > > > Think of the Klein bottle K as a cylinder S^1 x [0,1] after its two > > boundary circle S^1 x 0 and S^1 x 1 have been identified by a reflection > > of S^1 via (x, y) |—> (x, -y). > > > > Then you cannot rotate the cylinder after identification (which has become > > K) > > around its axis, because its end circles must rotate in opposite > > directions. > > The two things you *can* do are combinations of > > > > a) slide the cylinder along its length, and > > > > b) reverse the cylinder's direction. You can also flip the orientation of the circle. That gives the other Z/2Z Andy > > > > All of the slides are continuously deformable to each other, so there > > are two equivalence classes, so the group is Z/2. (Not a proof, of course.) > > > > —Dan > > > > ----- > > Here's a puzzle I think I know the answer to, but I don't have a proof: > > > > Let K denote the Klein bottle, the ideal surface you get when you identify > > the top and bottom edges of the unit square [0,1] x [0,1] normally, by > > > > (x,0) ~ (x,1), > > > > but identify the left and right edges by a flip: > > > > (0,y) ~ (1,1-y). > > > > Puzzle: > > ----------------------------------------------------------------------- > > Consider the space Homeo(K) of self-homeomorphisms of the Klein bottle. > > I.e., Homeo(K) consists of all continuous bijections > > > > h : K —> K > > > > having a continuous inverse. > > > > > > Two self-homeomorphisms h_0, h_1 of K are *in the same path component* of > > Homeo(K) if there > > is a continuous *family* > > > > {h(t) | 0 <= t <= 1} > > > > of homeomorphisms h(t) in Homeo(K) such that h(e) = h_e for e = 0, 1. > > > > The continuity of this family just amounts to there being a continuous map > > > > H : K x [0,1] —> K > > > > such that the restriction of H to any time-slice K x {t}: > > > > H | K x {t} —> K > > > > is the homeomorphism h(t) : K —> K. > > > > QUESTION: > > ————————— > > How many path components does Homeo(K) have? > > ----------------------------------------------------------------------- -----

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Re: [math-fun] Klein bottle puzzle (SPOILER)
by Dan Asimov 24 Dec '18

24 Dec '18

Think of the Klein bottle K as a cylinder S^1 x [0,1] after its two boundary circle S^1 x 0 and S^1 x 1 have been identified by a reflection of S^1 via (x, y) |—> (x, -y). Then you cannot rotate the cylinder after identification (which has become K) around its axis, because its end circles must rotate in opposite directions. The two things you *can* do are combinations of a) slide the cylinder along its length, and b) reverse the cylinder's direction. All of the slides are continuously deformable to each other, so there are two equivalence classes, so the group is Z/2. (Not a proof, of course.) —Dan ----- Here's a puzzle I think I know the answer to, but I don't have a proof: Let K denote the Klein bottle, the ideal surface you get when you identify the top and bottom edges of the unit square [0,1] x [0,1] normally, by (x,0) ~ (x,1), but identify the left and right edges by a flip: (0,y) ~ (1,1-y). Puzzle: ----------------------------------------------------------------------- Consider the space Homeo(K) of self-homeomorphisms of the Klein bottle. I.e., Homeo(K) consists of all continuous bijections h : K —> K having a continuous inverse. Two self-homeomorphisms h_0, h_1 of K are *in the same path component* of Homeo(K) if there is a continuous *family* {h(t) | 0 <= t <= 1} of homeomorphisms h(t) in Homeo(K) such that h(e) = h_e for e = 0, 1. The continuity of this family just amounts to there being a continuous map H : K x [0,1] —> K such that the restriction of H to any time-slice K x {t}: H | K x {t} —> K is the homeomorphism h(t) : K —> K. QUESTION: ————————— How many path components does Homeo(K) have? ----------------------------------------------------------------------- -----

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Re: [math-fun] New Year's oddity
by Keith F. Lynch 24 Dec '18

24 Dec '18

Dave Dyer <ddyer(a)real-me.net> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> I can't find any numerology in my post. > I'm sure that's not your intent, but just as astronomical observations > and the ability to track the position of astronomical bodies was > joined with a system of divination to make astrology, the ability to > perform arcane searches for mathematical curiosities opens the door > for lunatics or charlatans to make non-mathematical use. I'm not responsible for any use that other might make of my posts. And even if I *were* responsible, I'd be in good mathematical company: "I remember once going to see him [Ramanujan] when he was lying ill at Putney. I had ridden in taxi-cab number 1729, and remarked that the number seemed to be rather a dull one, and that I hoped it was not an unfavourable omen. 'No', he replied, 'it is a very interesting number; it is the smallest number expressible as the sum of two [positive] cubes in two different ways.'" -- Hardy Watch out for those unfavorable omens. Remember that it's bad luck to be susperstitious. (What would have happened if Hardy had ridden a different taxi? Would Ramanujan have found something just as interesting about *its* number? I like to think so.) > It's easy to stray from abstract math into that territory - witness > the recent discussion about messages from god in the expansion of PI, I blame Carl Sagan for that one. > and for that matter, what's the mathematical significance of "2018"? In about a week, none at all. :-) We're confronted with the number of each year for a whole year, so we might as well have fun exploring that number. If we get bored with it, we'll get a replacement in a year. It's also a fun programming challenge. Or rather lots of programming challenges. For instance how many different numbers can you get by interpolating various mathematical symbols between the digits of the current year? Some years ago I thought about looping, not over numbers, but over operations. I noticed that * + , - . / were consecutive ASCII characters. I used the comma for concatenation (e.g. 4,2 = 42) and . is of course a decimal point. The other four are of course the standard four arithmetic operations. Then I thought about looping over the ways to parenthesize these operations, and "discovered" the Catalan sequence. Real math, real programming. Today I've been thinking about how I would go about determining which is the last Fermat number which does *not* contain 2019 as a substring. I'm convinced that there is one, and that there isn't a last power of two with that property. I've since answered my own question as to which are the first powers of two that *begin* with 2018 and 2019: 4665 and 2044 respectively. Proving that there must be powers of two (or of any other positive integer except one or a power of ten) that begin with any desired integer is serious math. Someone who doesn't know that might be tempted to ascribe mystical significance to the fact that a number that's meaningful to them has that property.

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Re: [math-fun] New Year's oddity
by Keith F. Lynch 24 Dec '18

24 Dec '18

Dave Dyer <ddyer(a)real-me.net> wrote: > The line between math and numerology is pretty thin. Wikipedia says, "Numerology is any belief in the divine or mystical relationship between a number and one or more coinciding events. It is also the study of the numerical value of the letters in words, names and ideas. It is often associated with the paranormal, alongside astrology and similar divinatory arts." I can't find any numerology in my post. The distribution of substrings is pretty much pure math, even if it is contaminated by base-ten cooties. For instance, is there likely to be a *highest* power of two that does *not* contain 2019? How about a highest factorial that doesn't contain it? I find these to be interesting questions. Neil Sloane <njasloane(a)gmail.com> wrote: > Keith, we do have the sequence 10, 0, 1, 5, 2, 8, 4, 15, 3, 12, 10, > 40, 7, ..., which of course you will immediately recognize. It also > has an easily remembered A-number, A030000 Thanks. I notice that its value for the 9634th term is wrong. I also calculated a second 10,000 terms. I'll upload them when I get a tuit with fewer corners.

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Re: [math-fun] 1,409, 481, 24961, 28721, ...?
by Dan Asimov 24 Dec '18

24 Dec '18

This is very interesting! It might also be fun to also generalize the question to higher powers, like the squarefree part of (a^6 + b^6 + c^6)^(1/3). But also it might be fun to look at the same question in other number systems, like the fields Z/p. —Dan Bill Gosper wrote: ----- the squarefree part of integers of the form sqrt(a^4+b^4+c^4). -----

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[math-fun] A remark by Neil Bickford just helped me (= PSLQ) find
by Bill Gosper 23 Dec '18

23 Dec '18

DedekindEta[I Sqrt[5/2]] == 2 2^(7/8) 5^(3/8) (1 + Sqrt[10])^(1/4) (7/40)! (3/8)! Sqrt[(√2 + √5) (1/20)! (1/5)! (1/4)! (2/5)! (9/20)!/(3 (1/10)!)]/(3 π^(5/4) (3/40)! (7/20)!) I had not imagined needing (7/40)! —rwg But I still can't find DedekindEta[I Sqrt[7/2]]

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