Wikipedia:Reference desk/Archives/Science/2020 November 21

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November 21

Why are there some cases where one's blood type, semen type, and saliva type don't match?

Apparently in Soviet serial killer Andrei Chikatilo's case, his blood and saliva were of type A whereas his semen was of type AB. What exactly causes such mismatches to occur? Also, were there any other famous cases of such mismatches? Futurist110 (talk) 05:03, 21 November 2020 (UTC)

I'm not sure this is well studied. Various sources like [1] [2] claim it's because he was non-secretor. But this explanation seems to need major work. As I understand it, for a normal non-secretor, this would mean you can't reliably detect the A or B (or other) antigens in his non-blood bodily fluids (and I mean as antigens, not via DNA testing). I could perhaps imagine a partial ofor weak secretor may for some reason have reliable detection of either A or B in his other bodily fluids, while not the other. In other words, if you were to say someone is reliably A or B in their semen and saliva, and AB in blood type, this could probably be explained by secretor status. Although if you are only testing a few times it seems also possibly you would just detect A or B by random chance but not the other, which would again explain A or B in their semen and saliva and AB in blood type. I guess on that note, you could perhaps by chance detect one or both in saliva or semen but none or one in semen or saliva. So perhaps you would get A or B in semen or saliva and AB in blood type + saliva or semen. (I find it harder to imagine that you would get this reliably i.e. if you test enough over time it's a persistent result unless there is a condition resulting in red blood cells being persistently present in the semen or saliva, which would seem to be something you would detect anyway.) Anyway the problem is how being a non-secretor explains someone having A (or B) blood and AB saliva or semen. That specific case seems far harder to explain from secretory status from my understanding. Nil Einne (talk) 06:30, 21 November 2020 (UTC) 12:43, 21 November 2020 (UTC)

I wonder whether something like Cis AB, especially when combined with some FUT2/secretory variants may explain the phenotype. Nil Einne (talk) 06:54, 21 November

This may be due to somatic mosaicism, caused by a mutation occurring during mitosis in the developing embryo, particularly in the early blastula stage. Moreover, though very rare, some people are natural chimeras, having developed from the merger, in their mother's womb, of two (or possibly more) fertilized eggs or blastulae. In either case, the cells producing the blood cells and those producing the semen may then have different genotypes.  --Lambiam 09:02, 21 November 2020 (UTC)

Chimerism could also explain genetic differences between tissues in the same person. 167.95.98.207 (talk) 14:28, 23 November 2020 (UTC)

Isn't that what I said in my second sentence?  --Lambiam 12:11, 26 November 2020 (UTC)

What animal has the toughest/most resilient bones, and what animal has the toughest/most resilient flesh? (not shells or coverings, actual meat.

For a Reddit debate. Thanks in advance. --Squeeps10 ^{Talk to me}_{please ping me} 06:05, 21 November 2020 (UTC)

Strongest bones/skeleton would be found in the large burrowing animals. Wombats, badgers, etc have an interlocking skeletal structure (I guess that's to protect them in case of a cave-in) Their skeletons can easily take the weight of a human standing on their back. 49.182.12.60 (talk) 20:52, 22 November 2020 (UTC)

I was hoping more for most durable individual bones (IE if I removed a bone, how resistant that bone would be to breaking), not skeletal structures. But thanks for the response. {{u|Squeeps10}} {Talk} Please ping when replying. 23:23, 22 November 2020 (UTC)

Finite vs. Infinite Universe

First: my highest level of education is high school, so there are almost certainly things I've never thought of with regard to this question that are nevertheless so obvious to most that I'll end up seeming really dumb just for asking. But I had this sort of thought experiment that I wanted to run by some real people

I was thinking about whether or not the universe can physically be infinite in size. So, let's say the universe were a very large, but finite, two dimensional grid of one-foot by one-foot squares. (Let's say it's 100 total squares ... which, come to think of it, actually isn't that large, but whatever.) Now suppose we have some object (an apple, let's say) sitting on one of the squares. We can definitively say that the apple is on the square that it's on, but we can also say that the probability of the apple being on any given square is 1/100, or 1%.

Ok ... now expand the grid so it has infinitely many squares. (Not just a huge number, but literally an infinite number.) We're still looking at the apple, right? But now, isn't the probability of the apple actually being on that square exactly 0? It's just a limit, right? If x is the number of squares, 1/x is the probability of the apple being on a given square. So, as x approaches infinity, the probability approaches 0. Like I said, I'm certain there's a flaw in my thinking. I'm just wondering what it is. Thanks! TheRiseOfSkittlez (talk) 16:44, 21 November 2020 (UTC)

The natural density of prime numbers per counting number is exactly 0 but infinitely many primes exist. Sagittarian Milky Way (talk) 17:25, 21 November 2020 (UTC)

Counting on my fingers I find that I have 40% density of prime fingers. Is this not natural? 84.209.119.241 (talk) 01:19, 22 November 2020 (UTC)

The math(s) with infinity is a bit complex. 1/∞ can be zero or it can be infinitesimal. If it were exactly 0, the universe would be devoid of apples. If it were infinitesimal, there would be an infinite number of apples, spread - infinitly sparsely - on an infinitly larger infinity of squares.

I recommend poking your nose into Rudy Ruckers thin treatise on this sort of math(s). Reading a page may be quick, considering the concepts thereof takes a while. --Cookatoo.ergo.ZooM (talk) 17:37, 21 November 2020 (UTC)

In math, there is no such thing as 1/∞. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:52, 21 November 2020 (UTC)

@BB: As you are a person who refuses to grasp the decimal system, your pontification may be fallible. If memory serves me right, you have also argued on this desk that infinity does not exist. --Cookatoo.ergo.ZooM (talk) 19:28, 21 November 2020 (UTC)

No, I've argued what my old math teachers told me: Infinity is not a number, it's a concept of boundless expansion. Hence, 1 divided by infinity is mathematic nonsense. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:37, 21 November 2020 (UTC)

Yeah, so this is the sort of thing that high-school math teachers are indeed wont to say. As with lots of things, depending on what you mean, it can be true, false, or a lie to children.

High-school math teachers, I think, mostly want to keep their students from plopping ∞ in as a value for variables, and then applying rules that don't work in that case. For example, it's true (in almost any context where it's meaningful) that 1+∞ = 2+∞, but you can't "cancel the infinities" to get 1=2.

It's also possible that they've internalized a philosophical view that held sway for millennia, from Archimedes through Aquinas to Gauss, that only potential infinity is meaningful. This view began to crumble in the mid-19th century, and arguably was dealt its death blow by Georg Cantor, though it has modern proponents.

In any case, there are important mathematical contexts in which it is meaningful and true to write 1/∞ = 0. Probably the most salient one is the Riemann sphere, which admittedly has little to do with the question posed by the OP. --Trovatore (talk) 23:02, 21 November 2020 (UTC)

The limit statement below is the right way to express it. There is no such thing as 1 divided by infinity, because infinity is not a number. But it is valid to say that the "limit" of 1 / n, as n "approaches infinity" (i.e. gets larger and larger) is 0. It just can't ever get to 0. But it can get as close as you want it to, by using ever larger values of n. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:19, 21 November 2020 (UTC)

Bugs, did you follow the link I posted? While argument from authority is a fallacy, it is also true that I am a mathematician and you are not, so while that certainly doesn't prove I'm right, it might at least be enough cause to follow the link and see what it says. --Trovatore (talk) 23:31, 21 November 2020 (UTC)

Estimated values: 0.9 probability that BB comprehends math including the decimal system, with residual uncertainty about the superiority of the metric system. This minor personal defect reflects a lack of revolutionary zeal. 1.0 probability that Trovatore is a mathematician. The OP may or may not thank him for citing the Riemann sphere as a way to extend a flat universe to infinity in every direction, always handy to know. 0.9 probability that BB followed the link to potential infinity and if so,1.0 probability that BB saw that it says ".. potential infinity is often formalized using the concept of limit." which is prezactly what he has done. 84.209.119.241 (talk) 01:12, 22 November 2020 (UTC)

To swear by what old maths teachers told one certainly indicates a willingness to accept dicta by authority, possibly bolstered by senescence. But ordinal arithmetic, older than the oldest maths teachers alive, calculates with infinities; rather than infinity being just a number, it reckons with an infinitude of infinite ordinal numbers.  --Lambiam 09:04, 22 November 2020 (UTC)

The OP actually calculates the mathematical probability of guessing correctly the location of an apple when that is not already known. If a universe is divided into 100 locations then that probability is indeed 1/100 or 1% provided the italicized condition applies. One also assumes preknowledge that there exists exactly one apple. The example universe might be more realistically 3-dimensional if it is sliced into a grid of 100 cubes rather than squares, without changing the calculation. Moving to the second example of a universe with an infinite number of locations (squares or cubes, as you please), the probability calculation is indeed 1/∞ = 0. There is no contradiction or paradox here and the math holds

Lim x -> ∞ ( 1/x ) = 0.

Consternation arises if one tries to apply probability calculation to something that is already known. If the apple is reliably observed at a location then the probability of it existing there is exactly 1 = certain. The size of the universe has no effect on that calculated certainty unless there is some unspecified uncertainty in making the observation. See articles Probability for a general discussion and Probability theory for a rigorous mathematical treatment. 84.209.119.241 (talk) 19:34, 21 November 2020 (UTC)

The original calculation is missing an assumption, namely that the apple is equally likely to be on any square. That's apparently an application of the principle of indifference, which is not really a universally valid thing.
But I think the more fundamental confusion here is that people (not just the OP) seem to believe uncritically that if something has probability zero, then it can't happen. That's a fallacy unfortunately propagated in most elementary expositions of probability theory, because it's true in a lot of simple situations. But in general, it's just flat false. See almost surely for more details. --Trovatore (talk) 20:01, 21 November 2020 (UTC)

Almost surely a random point of an infinite universe is not in me. Sagittarian Milky Way (talk) 20:17, 21 November 2020 (UTC)

Well, there's a nuance there. It's not entirely clear that "a random point of an infinite universe" is meaningful. You can't, for example, put a translation-invariant probability measure on R³, and actually it's for reasons related to the OP's question.

R³ (usual Euclidean 3-space) can be partitioned into a countably infinite collection of cubical blocks, and the probability of being in any of them should be the same. But if that probability is 0, then because probability measures are countably additive, the probability of being anywhere in the universe would also be 0, which is a contradiction. However, if the probability is anything greater than 0, then the probability of being anywhere in the universe would be ∞, which doesn't make sense either.

However, you can put a finitely additive probability measure on R³,hmm — see below and for some purposes that may be enough. --Trovatore (talk) 21:20, 21 November 2020 (UTC)

I should be careful here. You can't put a translation-invariant finitely additive probability measure that measures all subsets of R³, as shown by the Banach–Tarski paradox. What I had in mind was something like the limiting density of the intersection of the subset with a ball, as the size of the ball goes to infinity (the analogue of what Saggitarian Milky Way was calling "natural density"). That's a fapm, but there are even some pretty simple sets on which it's not well-defined. --Trovatore (talk) 19:59, 23 November 2020 (UTC)

So the OP is correct in having the intuition that, using maths speak, you can't have a uniform probability distribution over an infinite universe such as Rⁿ (or, for that matter, Zⁿ) for n > 0. However, it does not follow that the universe is finite, even if it is uniform. But an exercise in which one must put a single apple in an absolutely random spot in a universe of infinite extent is unsolvable. Whatever procedure you follow, some spots will be more likely than others. In contrast, there is no mathematical argument against the possibility of putting an infinite number of apples on the squares (large enough to contain an apple) of an infinite checkers board, such that the probability of any square being occupied is a given positive value p (say 1 in 10¹⁰⁰, in any case not exceeding 1), but that revealing the occupancy status (yes or no) of any set of squares provides no new information about the status likelihoods of any other square.  --Lambiam 23:52, 21 November 2020 (UTC)

@TheRiseOfSkittlez: Relevant here is Olbers' paradox. It is a calculation that explains why, with our current observations, it is very unlikely for the universe to be infinitely large because that would mean that, if stars are distributed evenly throughout it, the sky would be infinitely bright (contrary to how it actually looks, being mostly dark). RedPanda25 02:54, 22 November 2020 (UTC)

That paradox assumes that the universe is infinitely large and infinitely old. It really doesn't apply anymore, now that the Big Bang is considered well-established. --Trovatore (talk) 02:58, 22 November 2020 (UTC)

Olbers' paradox also assumes that the universe is not expanding.  --Lambiam 08:43, 22 November 2020 (UTC)

A well motivated model of an infinite universe is the eternal inflation model. There are then an infinite number of copies of the exact physical state of the entire observable part of the universe. Quantum mechanics makes the number of possible physical states each finite region of the universe can be in, to be a finite number. Count Iblis (talk) 09:02, 22 November 2020 (UTC)

Given the existence of one apple and an infinite universe for placement, as well as a uniform probability distribution, then the nonzero infinitesimal probability of it does not imply the existence of any additional apples or an exact zero probability for any position in the universe either, it simply implies, as stated, an infinitesimal probability for each and every position that when taken together always sum to one over the entire universe such that the apple is placed somewhere. Infinitesimals are not real numbers or zeros however, thus, alternatively, these probabilities can be set to zero or almost never. But to express the summation of infinitesimal probabilities: n(1/n)=1 and n=∞ and, importantly, the infinities must have a one-to-one correspondence without which the limit would be indeterminate instead of one as it needs to be. I think that's right, but expect to be persuaded otherwise given prior objections. -Modocc (talk) 18:48, 22 November 2020 (UTC)

We can entertain a mental construction for picking a random number in the unit interval, given a truly random and fair coin with sides marked "0" and "1", by starting with "0." and creating a binary expansion by appending a (countably) infinite number of flipping outcomes. This will result in some infinite expansion such as "0.1111000010111111110001..." (truncated here to avoid the need of infinite disk space on the Wikimedia servers), or, in decimal notation, "0.940426..." . The probability of hitting exactly this number in this mental model of throwing a dart at the unit interval is, of course, infinitesimally small. For the one-apple infinite universe, however, we do not just need an nonzero infinitesimal probability. We would need a distribution with an everywhere infinitesimal probability density. No mental construction for, to keep it simple, picking a random natural number will give you that.  --Lambiam 12:49, 23 November 2020 (UTC)

No specific random natural number other than 0 can be given, sure, for the others are not large or small enough. Darts, apples or binaries are instances of the exact same mental abstraction since binaries can represent the reals which we use to determine objects' positions in both space and time. For nonzero reals x(1/x) = 1 and I have assumed this particular PDF requires an infinitesimal numbering system for which this works too, but I'm too ill-equipped to either critique or support my assumptions. Individually the limits regarding probability and the sample space size are 0 and ∞, yet when these are taken together their limit has to be 1 when integrating my supposed infinitesimal PDF curve, yet I am not in a position to discern if this is a valid application. -Modocc (talk) 17:35, 23 November 2020 (UTC)

I am not aware of any successful effort to formulate probability theory in such a way that you can get a uniform probability distribution on a countably infinite set, even by allowing probabilities to be infinitesimal. The first thing you might think of trying would be to use the infinitesimals from nonstandard analysis, but they don't work at all for this purpose. Let me know if you want details. --Trovatore (talk) 19:40, 23 November 2020 (UTC)

Thanks. I'm not planning on counting infinitesimals and using them, typically because I tend to prefer the more parsimonious solutions. By no "uniform probability distribution on a countably infinite set" I take that to mean one cannot be integrated in the usual fashion if there exists ranges that have zero probability of being selected. However, maybe we just haven't modeled the problem properly yet, for I don't think that's quite true here. One can take a fair coin or die flipped an infinite number of times to select a random permutation from such a set and partition its members (such as odd vs even numbers) in various ways to ensure fairness. And the OP's squares can also be systematically partitioned to test the fairness of the distribution, i.e. given twenty apples how many fall on white squares vs black on an infinite checkered pattern. That an apple is present ex post facto if one is looking at it doesn't diminish the improbability, but in my experience most things are more probable than imagined. -Modocc (talk) 08:06, 24 November 2020 (UTC)

By flipping a coin say 4 times, giving consecutively 1-0-1-1 (representing oddity vs. evenness) and selecting one half of the remaining numbers, you can restrict N to the subset of numbers whose binary expansion ends on 1101, such as 13, 1975308637, and 1580246913758024691373. So you are down to only a 16th part of the naturals, that is ... precisely as many as you started with. (See also Hilbert's hotel.) It is just not possible to bring the infinite flip sequences into a one-to-one correspondence with the naturals; see Cantor's diagonal argument. After only a finite number of flips, you have not made a dent in whittling down the size of the choice space; at time Aleph-naught you have outflipped it. If a "proper" way of modelling the problem is one that allows some solution of picking (or defining, while remaining logically consistent, a notion of) a "truly random" natural number where no number has better chances than any other, then the reason we just haven't modeled the problem properly yet is that there is just no way of doing that.  --Lambiam 15:53, 24 November 2020 (UTC)

The binary tree's powers of two increase the tallies that the clones register.

The sample space size never changes, so if a coin flips only zero, it's unclear to me what is being outflipped. Moreover, if we take the limit of the OP's PDF and we find zero, with an apple no less, as I've noted, the limit of x(1/x) is 1 as it should be. To make my conceptual case clearer, consider an infinite binary tree that begins with the radix point and its edges contain either 0 or a power of 2 that increases with tree depth (see illustration). A self-replicating robot duplicates itself at every node (it initially starts out at the root node with the radix point) with one robot then visiting the node's left child and the other visiting the node's right child, as they each tally the powers of 2 they encounter. Some tallies increase in size others don't, and with an infinite depth every finite nonnegative integer is tallied once by a clone. None have been excluded. Now let the robot start with a cart of apples. Each node is visited by one clone and a fair coin is used to decide which duplicate gets to carry each apple. That way all the clones (thus tallies) have a fair shot of retaining any of the apples whose own total number remains constant throughout, from start to finish. At each level of the tree, the apples' distribution should be uniform and remain so even after the clones complete their infinite summations. Thus we would expect half the apples to land on odd numbers. Now I've done it, I'm at odds with Cantor because I've created a uniform distribution of apples upon an infinite set of clones. Yikes. Maybe we could call it an Apple machine (see Bean machine). Plus we will have to append infinite sets of fraction tallying clones (that traverse trees with negative powers of 2) and dedicate a few bonus apples to the Pi descendant.. -Modocc (talk) 22:49, 24 November 2020 (UTC)

Given a random finite binary sequence s, however long, the set of natural numbers whose binary expansion ends on s has cardinality $\aleph _{0}$. Given a random infinite binary sequence s, with probability 1 it contains infinitely many 1s, and then there exists no natural number whose binary expansion ends on s. That is what I meant by outflipping the space: you go from cardinality $\aleph _{0}$ to cardinality 0. The tree construction can, in the limit, select with uniform probability a real number in the unit interval – no problem there. It cannot be modified to do the same for selecting a natural number.  --Lambiam 12:07, 26 November 2020 (UTC)

So you are saying we cannot choose from an infinite number of repeated finite numbers embedded in a string. Correct? I agree with that and I should have stated that the army of clones has to march and duplicate in sync to be effective, otherwise we end up talking about n + infinity (which is why I [almost] abandoned this years ago but now I don't see this as a problem either upon seeing how useful the in sync algorithm is here). Note that with the tree their number doubles with their children (see the illustration): there are never duplicates of a number as the tree grows in depth. It is from the tree's infinite leaf nodes, all at the same infinite depth, that has all the unique natural numbers as required. The imaginary bot hoard is just a way to facilitate the distribution of the apples to all of them. I'll add that we can ignore the tree and just exponentially duplicate the bot an infinite number of times with instructions as to what to do with the apples. Of note, if the reals are countable so is Cantor's set T since there is a bijection between them and he set up a strawman. -Modocc (talk) 18:33, 26 November 2020 (UTC)

I don't want this strange exchange to pass into the archives unanswered, for someone to find later and think that Modocc had some way of doing this impossible thing and no one had a response to it. I didn't respond earlier because I had no idea what he/she meant by "in lockstep", and I still don't really, but it doesn't seem to matter, because if you look at Modocc's post from 22:49, 24 November 2020, it's well enough specified: Down any branch of the tree, we're to add 0 to a running tally whenever we go left, and 2^d whenever we go right, where d is the depth of the node.

That does indeed result in all natural numbers having a branch assigned to them. The problem is, for almost all branches, we take the right child infinitely many times, so the final tally is infinite, not a natural number. If we assign probabilities to natural numbers in this way, every natural number will have probability 0. So the method does not give a uniform probability distribution on the natural numbers.

I note that Modocc has followed up on this point on the math reference desk (WP:RD/Math#Reals are countable: an informal proof, I think..; will eventually be at Wikipedia:Reference desk/Archives/Mathematics/2020 November 27#Reals are countable: an informal proof, I think..). There he/she has claimed to have refuted Cantor's theorem on the uncountability of the reals. It is not going well. --Trovatore (talk) 19:39, 28 November 2020 (UTC)

@Trovatore: I've just now read your reply, hence my apology for taking so long to answer. Distributing resources between parallel processes is a bit old-fashion these days, but the swarm branches to the left child infinitely just as many times as branching to the right. Regarding numbers in general, pi requires a complete summation of its expansion as is the case with all the numeric binary expansions. Each number is unique, thus, assign a unique color to every real number, alias them to a line, then spread them out like a rainbow unto to the squares, reflect them back, their number remains constant throughout. I claim too that the distribution of an apple by the swarm to these numbers can be made uniform.. In any case, to me, the argument that .111... = 1 is false because it's nonterminating with an arbitrary differential is on an equal footing with Cantor's argument. An infinitesimal(or an arbitrary diagonal compliment nonterminating branch) is uncounted. With completed infinities it's untenable. We can certainly demonstrate/show/prove completed infinities exist, and parallel algorithms such as this one exist, as well as the math and statistics that are entailed by it. -Modocc (talk) 01:46, 23 December 2020 (UTC)