[math-fun] Mathematicians Catch a Pattern By Figuring Out How To Avoid It
https://science.slashdot.org/story/19/11/28/180213/mathematicians-catch-a-pa... We finally know how big a set of numbers can get before it has to contain a pattern known as a "polynomial progression." <https://www.quantamagazine.org/mathematicians-catch-a-pattern-by-figuring-out-how-to-avoid-it-20191125/> From a report:/A new proof <https://arxiv.org/abs/1909.00309> by Sarah Peluse of the University of Oxford establishes that one particularly important type of numerical sequence is, ultimately, unavoidable: It's guaranteed to show up in every single sufficiently large collection of numbers, regardless of how the numbers are chosen. "There's a sort of indestructibility to these patterns," said Terence Tao of the University of California, Los Angeles. Peluse's proof concerns sequences of numbers called "polynomial progressions." They are easy to generate -- you could create one yourself in short order -- and they touch on the interplay between addition and multiplication among the numbers. For several decades, mathematicians have known that when a collection, or set, of numbers is small (meaning it contains relatively few numbers), the set might not contain any polynomial progressions. They also knew that as a set grows it eventually crosses a threshold, after which it has so many numbers that one of these patterns has to be there, somewhere. It's like a bowl of alphabet soup -- the more letters you have, the more likely it is that the bowl will contain words. But prior to Peluse's work, mathematicians didn't know what that critical threshold was. Her proof provides an answer -- a precise formula for determining how big a set needs to be in order to guarantee that it contains certain polynomial progressions. Previously, mathematicians had only a vague understanding that polynomial progressions are embedded among the whole numbers (1, 2, 3 and so on). Now they know exactly how to find them./ -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/
Google Search: "site:quantamagazine.org terence tao", shows a loud pro Tao bias, with at least two comments from Tao in November alone. The other article [1] follows after [2,3] on arxiv, and Tao is a coauthor on [3]. Presumably he contributed to the proof. Unfortunately, the mathematics of Ref. [3] has recently come under attack as "not so original", thus calling into question the purpose and rigour of the Quanta article, as well as the business of it. As was noticed by Predrag Cvitanović, square eigenvector components in [3] are actually just the central diagonal of the projection operators, discussed in chapter 1 of "Principles of Symmetry, Dynamics, and Spectroscopy": https://modphys.hosted.uark.edu/pdfs/PSDS_Pdfs/PSDS_Ch.1_(4.23.10).pdf The fact that the central diagonal of a particular projector works out in terms of submatrix eigenvalues follows quickly from the fact that determinant and trace are simple functions of eigenvalues. That is all. John P. Raslton, an expert on Neutrino Oscillations and matrix algebra, is mostly done with a short proof that is a lot easier to understand. His write up will also cite PSDS as having the correct theory for calculating "Eigenvectors from Eigenvalues". Here is a quick example in Mathematica: M = {{254, -38, -57}, {-38 , 17, 6}, {-57, 6, 22}} Eigenvalues[M] With[{ev = Eigenvalues@M[[ Complement[Range[3], {#}], Complement[Range[3], {#}]]]}, Expand[Times @@ ev - 7*Total[ev] + 7^2] ] & /@ Range[3] Dot[M - 273*IdentityMatrix[3], M - 13*IdentityMatrix[3]] // MatrixForm Out[] = {273, 13, 7} Out[] = {114, 456, 1026} Out[] = { {114, 228, 342}, {228, 456, 684}, {342, 684, 1026} } I personally feel like we are bikeshedding with Quanta and their cadre of well-liked, world-famous, prize winning, prime-time-tv, whoevers. There is another unanswered question--whether the authors have supported their physics thesis that "eigenvalues are the Rosetta stone for neutrino oscillations in matter". It sounds very rich, but do the authors mean to suggest that their eigenvalues were first stolen by the French, and then stolen again from the French? Any thoughts? Cheers --Brad [1] https://www.quantamagazine.org/neutrinos-lead-to-unexpected-discovery-in-bas... [2] https://arxiv.org/abs/1907.02534 [3] https://arxiv.org/abs/1908.03795 On Fri, Nov 29, 2019 at 2:45 AM Ray Tayek <rtayek@ca.rr.com> wrote:
https://science.slashdot.org/story/19/11/28/180213/mathematicians-catch-a-pa...
We finally know how big a set of numbers can get before it has to contain a pattern known as a "polynomial progression." <https://www.quantamagazine.org/mathematicians-catch-a-pattern-by-figuring-out-how-to-avoid-it-20191125/> From a report:/A new proof <https://arxiv.org/abs/1909.00309> by Sarah Peluse of the University of Oxford establishes that one particularly important type of numerical sequence is, ultimately, unavoidable: It's guaranteed to show up in every single sufficiently large collection of numbers, regardless of how the numbers are chosen. "There's a sort of indestructibility to these patterns," said Terence Tao of the University of California, Los Angeles. Peluse's proof concerns sequences of numbers called "polynomial progressions." They are easy to generate -- you could create one yourself in short order -- and they touch on the interplay between addition and multiplication among the numbers. For several decades, mathematicians have known that when a collection, or set, of numbers is small (meaning it contains relatively few numbers), the set might not contain any polynomial progressions. They also knew that as a set grows it eventually crosses a threshold, after which it has so many numbers that one of these patterns has to be there, somewhere. It's like a bowl of alphabet soup -- the more letters you have, the more likely it is that the bowl will contain words. But prior to Peluse's work, mathematicians didn't know what that critical threshold was. Her proof provides an answer -- a precise formula for determining how big a set needs to be in order to guarantee that it contains certain polynomial progressions. Previously, mathematicians had only a vague understanding that polynomial progressions are embedded among the whole numbers (1, 2, 3 and so on). Now they know exactly how to find them./
-- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/
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There’s gotta be a “Tao neutrino” pun lurking here... Jim Propp On Fri, Nov 29, 2019 at 3:19 PM Brad Klee <bradklee@gmail.com> wrote:
Google Search: "site:quantamagazine.org terence tao", shows a loud pro Tao bias, with at least two comments from Tao in November alone. The other article [1] follows after [2,3] on arxiv, and Tao is a coauthor on [3]. Presumably he contributed to the proof.
Unfortunately, the mathematics of Ref. [3] has recently come under attack as "not so original", thus calling into question the purpose and rigour of the Quanta article, as well as the business of it.
As was noticed by Predrag Cvitanović, square eigenvector components in [3] are actually just the central diagonal of the projection operators, discussed in chapter 1 of "Principles of Symmetry, Dynamics, and Spectroscopy":
https://modphys.hosted.uark.edu/pdfs/PSDS_Pdfs/PSDS_Ch.1_(4.23.10).pdf
The fact that the central diagonal of a particular projector works out in terms of submatrix eigenvalues follows quickly from the fact that determinant and trace are simple functions of eigenvalues. That is all.
John P. Raslton, an expert on Neutrino Oscillations and matrix algebra, is mostly done with a short proof that is a lot easier to understand. His write up will also cite PSDS as having the correct theory for calculating "Eigenvectors from Eigenvalues".
Here is a quick example in Mathematica:
M = {{254, -38, -57}, {-38 , 17, 6}, {-57, 6, 22}} Eigenvalues[M]
With[{ev = Eigenvalues@M[[ Complement[Range[3], {#}], Complement[Range[3], {#}]]]}, Expand[Times @@ ev - 7*Total[ev] + 7^2] ] & /@ Range[3]
Dot[M - 273*IdentityMatrix[3], M - 13*IdentityMatrix[3]] // MatrixForm
Out[] = {273, 13, 7} Out[] = {114, 456, 1026} Out[] = { {114, 228, 342}, {228, 456, 684}, {342, 684, 1026} }
I personally feel like we are bikeshedding with Quanta and their cadre of well-liked, world-famous, prize winning, prime-time-tv, whoevers.
There is another unanswered question--whether the authors have supported their physics thesis that "eigenvalues are the Rosetta stone for neutrino oscillations in matter". It sounds very rich, but do the authors mean to suggest that their eigenvalues were first stolen by the French, and then stolen again from the French? Any thoughts?
Cheers
--Brad
[1] https://www.quantamagazine.org/neutrinos-lead-to-unexpected-discovery-in-bas... [2] https://arxiv.org/abs/1907.02534 [3] https://arxiv.org/abs/1908.03795
On Fri, Nov 29, 2019 at 2:45 AM Ray Tayek <rtayek@ca.rr.com> wrote:
https://science.slashdot.org/story/19/11/28/180213/mathematicians-catch-a-pa...
We finally know how big a set of numbers can get before it has to contain a pattern known as a "polynomial progression." <
https://www.quantamagazine.org/mathematicians-catch-a-pattern-by-figuring-ou...
From a report:/A new proof <https://arxiv.org/abs/1909.00309> by Sarah Peluse of the University of Oxford establishes that one particularly important type of numerical sequence is, ultimately, unavoidable: It's guaranteed to show up in every single sufficiently large collection of numbers, regardless of how the numbers are chosen. "There's a sort of indestructibility to these patterns," said Terence Tao of the University of California, Los Angeles. Peluse's proof concerns sequences of numbers called "polynomial progressions." They are easy to generate -- you could create one yourself in short order -- and they touch on the interplay between addition and multiplication among the numbers. For several decades, mathematicians have known that when a collection, or set, of numbers is small (meaning it contains relatively few numbers), the set might not contain any polynomial progressions. They also knew that as a set grows it eventually crosses a threshold, after which it has so many numbers that one of these patterns has to be there, somewhere. It's like a bowl of alphabet soup -- the more letters you have, the more likely it is that the bowl will contain words. But prior to Peluse's work, mathematicians didn't know what that critical threshold was. Her proof provides an answer -- a precise formula for determining how big a set needs to be in order to guarantee that it contains certain polynomial progressions. Previously, mathematicians had only a vague understanding that polynomial progressions are embedded among the whole numbers (1, 2, 3 and so on). Now they know exactly how to find them./
-- Honesty is a very expensive gift. So, don't expect it from cheap people
- Warren Buffett
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Or how about a closer look at the MACIEJ REBISZ neutrino? https://d2r55xnwy6nx47.cloudfront.net/uploads/2019/11/Neutrinos-Eigenvalues_... Compare with another artist's interpretation, "Farbenphasenraumzeit": https://0x0.st/zlsn.jpg ( GPS WARNING: this pointer is not for the faint of heart! Nor is it an advertisement! ) My taste: the Rebisz picture reminds me too much of his other work in science fiction. I can see that the neutrino is changing it's colour and it's subscript, but I am left to wonder about CP-violation / helicity, and where are the true horses? The circles are too perfect, and what is the point of cluttering up the diagram with a lot of mathematical symbology? It doesn't serve the intended purpose to make the illustration look any more intelligent. If you take away all the text labels, I would estimate ~ zero percent probability that anyone would identify the drawing itself with a neutrino. On the other hand, I have been impressed by some of the graphic design at Quanta, and find the drawing of the young woman doing needle-point by Celia Lownethal as well-drawn, poignant, and even melancholy if not tragic. Happy Holidays or at least try and survive them, --Brad PS. Ralston says that he is "not an expert on anything", but I think he is intentionally understating his position. On Fri, Nov 29, 2019 at 2:26 PM James Propp <jamespropp@gmail.com> wrote:
There’s gotta be a “Tao neutrino” pun lurking here...
Jim Propp
Hello MathFun, none of those integers are prime numbers. Will we find such a "pendulum prime" soon? 1 21 213 4213 42135 642135 6421357 86421357 864213579 10864213579 1086421357911 121086421357911 12108642135791113 1412108642135791113 141210864213579111315 16141210864213579111315 1614121086421357911131517 181614121086421357911131517 18161412108642135791113151719 ... This sequence can be found in the OEIS: https://oeis.org/A053063 Note that this one claims that « a(n) is not prime for any n <= 3000 »: https://oeis.org/A281254 Best, Merry Xmas, É. P.-S. The opening picture (invisible in this mail but visible on my personal blog) is by Kay Rosen: https://bit.ly/2s11v90
Yes. The 121'st of the pendulum numbers (your sequence) is a 255 digit prime, per maple. This was quite a surprise to me. In fact the smallest prime divisors of the pendulum numbers are typically pretty small, and surprisingly erratic. On Sat, Dec 7, 2019 at 6:42 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello MathFun, none of those integers are prime numbers. Will we find such a "pendulum prime" soon?
1 21 213 4213 42135 642135 6421357 86421357 864213579 10864213579 1086421357911 121086421357911 12108642135791113 1412108642135791113 141210864213579111315 16141210864213579111315 1614121086421357911131517 181614121086421357911131517 18161412108642135791113151719 ...
This sequence can be found in the OEIS: https://oeis.org/A053063 Note that this one claims that « a(n) is not prime for any n <= 3000 »: https://oeis.org/A281254 Best, Merry Xmas, É. P.-S. The opening picture (invisible in this mail but visible on my personal blog) is by Kay Rosen: https://bit.ly/2s11v90
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JB: "The 121'st of the pendulum numbers (your sequence) is a 255 digit prime, per maple." There's one more in the first ten thousand: @ 4489 digits, the 1399th.
James, Hans - remarkable work. I updated A053063. Can you do a similar thing for the other sequence that Eric mentioned, A281254? On Sat, Dec 7, 2019 at 3:45 PM Hans Havermann <gladhobo@bell.net> wrote:
JB: "The 121'st of the pendulum numbers (your sequence) is a 255 digit prime, per maple."
There's one more in the first ten thousand: @ 4489 digits, the 1399th. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
And what about the mother of them all, A007908, the "triangle of the gods", which starts 1,12,123,1234,12345,123456,...,1234567891011,... and should contain infinitely many primes. The GIMPS folks checked out to n=344869 but did not find a prime, and their search, see A007908, seems to have been abandoned. On Sat, Dec 7, 2019 at 4:04 PM Neil Sloane <njasloane@gmail.com> wrote:
James, Hans - remarkable work. I updated A053063.
Can you do a similar thing for the other sequence that Eric mentioned, A281254?
On Sat, Dec 7, 2019 at 3:45 PM Hans Havermann <gladhobo@bell.net> wrote:
JB: "The 121'st of the pendulum numbers (your sequence) is a 255 digit prime, per maple."
There's one more in the first ten thousand: @ 4489 digits, the 1399th. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
NJAS: "Can you do a similar thing for the other sequence that Eric mentioned, A281254?" I can tell you that there's no prime hits in the first ten thousand. I might take this deeper tomorrow (unless somebody beats me to it) as all my cores are currently preoccupied.
JB: "The 121st of the pendulum numbers ... is a 255-digit prime..." HH: "There's one more in the first ten thousand: @ 4489 digits, the 1399th." NJAS: "Can you do a similar thing for ... A281254?" The third prime in A053063 is a(40714) with 192464 decimal digits. It wasn't until I determined this that I realized (because I had already calculated the first prime in A281254) that the even-indexed terms of A281254 duplicated the even-indexed terms of A053063. The odd-indexed terms of A281254 are divisible by two so, necessarily, the primes in A281254 are a subset of the primes in A053063, specifically those that appear at even indices.
Hello Math-Fun, I don’t know if the hereunder words write a good and fluid English sentence, but you’ll get the idea: « Randomly draw a letter in this sentence and obtain an E will happen once in seven. » There are indeed 10 E’s and 70 letters altogether. This kind of sentence is easy to build (said my friend Pascal S. from Switzerland), but what would the English word-numbers Y and Z be here: « The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z. » Best, É.
I count 66 letters in the first example. Tom Éric Angelini writes:
Hello Math-Fun, I don’t know if the hereunder words write a good and fluid English sentence, but you’ll get the idea:
« Randomly draw a letter in this sentence and obtain an E will happen once in seven. »
There are indeed 10 E’s and 70 letters altogether. This kind of sentence is easy to build (said my friend Pascal S. from Switzerland), but what would the English word-numbers Y and Z be here:
« The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z. »
Best, É.
Indeed Tom -- I forgot "just": « Randomly draw a letter in this sentence and obtain an E will happen just once in seven. » Thank you for your sharp eye!-)
Le 9 décembre 2019 à 02:00, Tom Karzes <karzes@sonic.net> a écrit :
I count 66 letters in the first example.
Tom
Éric Angelini writes:
Hello Math-Fun, I don’t know if the hereunder words write a good and fluid English sentence, but you’ll get the idea:
« Randomly draw a letter in this sentence and obtain an E will happen once in seven. »
There are indeed 10 E’s and 70 letters altogether. This kind of sentence is easy to build (said my friend Pascal S. from Switzerland), but what would the English word-numbers Y and Z be here:
« The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z. »
Best, É.
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As you guessed, though, Eric, the sentence is not idiomatic English. The problem is the sequence "obtain an E will happen"; it would be more idomatic to say "obtain an E once in seven tries", or "the chances of getting an E are one in seven". On Sun, Dec 8, 2019 at 8:05 PM Éric Angelini <bk263401@skynet.be> wrote:
Indeed Tom -- I forgot "just": « Randomly draw a letter in this sentence and obtain an E will happen just once in seven. » Thank you for your sharp eye!-)
Le 9 décembre 2019 à 02:00, Tom Karzes <karzes@sonic.net> a écrit :
I count 66 letters in the first example.
Tom
Éric Angelini writes:
Hello Math-Fun, I don’t know if the hereunder words write a good and fluid English sentence, but you’ll get the idea:
« Randomly draw a letter in this sentence and obtain an E will happen once in seven. »
There are indeed 10 E’s and 70 letters altogether. This kind of sentence is easy to build (said my friend Pascal S. from Switzerland), but what would the English word-numbers Y and Z be here:
« The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z. »
Best, É.
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Thank you Allan -- here is the correction: « If you randomly draw a letter in this sentence the chances of getting an E are thirteen in eighty-two. » ... The "FOUR" letter challenge is still open! « The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y in Z. » Best, É.
Can I suggest a possible slight improvement to the wording? If you randomly select a letter from this sentence the chances of it being an e are X in Y "select" could also be "choose" or "pick" On Tue, Dec 10, 2019 at 5:06 PM Éric Angelini <bk263401@skynet.be> wrote:
Thank you Allan -- here is the correction:
« If you randomly draw a letter in this sentence the chances of getting an E are thirteen in eighty-two. »
... The "FOUR" letter challenge is still open!
« The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y in Z. »
Best, É.
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Merci Neil! à+ É. Catapulté de mon aPhone
Le 11 déc. 2019 à 09:35, Neil Sloane <njasloane@gmail.com> a écrit :
Can I suggest a possible slight improvement to the wording?
If you randomly select a letter from this sentence the chances of it being an e are X in Y
"select" could also be "choose" or "pick"
On Tue, Dec 10, 2019 at 5:06 PM Éric Angelini <bk263401@skynet.be> wrote:
Thank you Allan -- here is the correction:
« If you randomly draw a letter in this sentence the chances of getting an E are thirteen in eighty-two. »
... The "FOUR" letter challenge is still open!
« The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y in Z. »
Best, É.
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If I've made no mistakes, and using Neil's suggestion: If you randomly select a letter from this sentence, the chances of it being ... ... an 'a' are three in forty-one. ... a 'b' are three in one-hundred and forty-one. ... a 'c' are two in thirty-two. ... a 'd' are nine in one-hundred and seventy-one. ... an 'e' are five in twenty-eight. ... an 'f' are three in sixty-three. ... a 'g' are three in eighty-two. ... an 'h' are five in eighty-two. ... an 'i' are three in forty-one. ... a 'j' are two in one-hundred and eighty. ... a 'k' are two in one-hundred and eighty. ... an 'l' are three in sixty-three. ... an 'm' are two in fifty-four. ... an 'n' are three in thirty. ... an 'o' are six in eighty-one. ... a 'p' are two in one-hundred and eighty. ... a 'q' are two in one-hundred and eighty. ... an 'r' are five in eighty-two. ... an 's' are six in eighty-one. ... a 't' are five in forty-five. ... a 'u' are five in one-hundred and fifty. ... a 'v' are two in one-hundred and eighty. ... a 'w' are seven in two-hundred and twenty-four. ... an 'x' are one in eighty-one. ... a 'y' are four in eighty-one. ... a 'z' are two in one-hundred and eighty. Done by sequential search. Not quite sure how to tackle "If you randomly select... ...an 'a' are Ma/Na, a 'b' are Mb/Nb, a 'c' Mc/Nc, ... On 12/11/19 12:35 AM, Neil Sloane wrote:
Can I suggest a possible slight improvement to the wording?
If you randomly select a letter from this sentence the chances of it being an e are X in Y
"select" could also be "choose" or "pick"
The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z. EA: "The 'FOUR' letter challenge is still open!" If Y = "one" and Z = "one hundred twenty-seven thousand, seven hundred seventy-six", I make the total letter count of the sentence to be 132, with 4 of 'f', 11 of 'o', 6 of 'u', and 9 of 'r'. If the probability is 4*11*6*9/132^4, then we have a hit.
If the probability is 4*11*6*9/132^4, then we have a hit.
... yes, but in my mind once I've picked up a letter, I must erase that letter from the sentence, sorry (how could that figure in the "puzzle-sentence" without using too many words? In French it is a "Tirage sans remise") In that case, aren't the odds 4*11*6*9/132*131*130*129? I've almost a hit in French, by reducing the set of possibilities with the "puzzle-sentence" ending in: "[The probability of...] is one on K" -- with K = 132*131*130*129/4*11*6*9. (I didn't count the digits after the comma, though, which is neither correct, nor very elegant! But my error is below 1/10000... Best, É.
Le 11 décembre 2019 à 13:16, Hans Havermann <gladhobo@bell.net> a écrit :
The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z.
EA: "The 'FOUR' letter challenge is still open!"
If Y = "one" and Z = "one hundred twenty-seven thousand, seven hundred seventy-six", I make the total letter count of the sentence to be 132, with 4 of 'f', 11 of 'o', 6 of 'u', and 9 of 'r'. If the probability is 4*11*6*9/132^4, then we have a hit. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
In English, the technical probability terms are "sampling with replacement" and "sampling without replacement". Perhaps if we say "The odds of picking four letters at random from this sentence, and then finding them to be F, O, U, and R, are Y out Z," the intended protocol will be clearer. On Wed, Dec 11, 2019 at 8:58 AM Éric Angelini <bk263401@skynet.be> wrote:
If the probability is 4*11*6*9/132^4, then we have a hit.
... yes, but in my mind once I've picked up a letter, I must erase that letter from the sentence, sorry (how could that figure in the "puzzle-sentence" without using too many words? In French it is a "Tirage sans remise")
In that case, aren't the odds 4*11*6*9/132*131*130*129?
I've almost a hit in French, by reducing the set of possibilities with the "puzzle-sentence" ending in: "[The probability of...] is one on K" -- with K = 132*131*130*129/4*11*6*9. (I didn't count the digits after the comma, though, which is neither correct, nor very elegant! But my error is below 1/10000... Best, É.
Le 11 décembre 2019 à 13:16, Hans Havermann <gladhobo@bell.net> a écrit :
The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z.
EA: "The 'FOUR' letter challenge is still open!"
If Y = "one" and Z = "one hundred twenty-seven thousand, seven hundred seventy-six", I make the total letter count of the sentence to be 132, with 4 of 'f', 11 of 'o', 6 of 'u', and 9 of 'r'. If the probability is 4*11*6*9/132^4, then we have a hit. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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"The odds of picking four letters at random from this sentence, and then finding them to be F, O, U, and R, are Y out Z".
... this is it, I guess, thanks Allan -- wait: "X out Z" or "X out of Z"? I prefer "pick" to "select" (suggested by Neil) as "select" seems to imply that the sampler carefully chooses -- instead of being blind, no?
Le 11 décembre 2019 à 15:41, Allan Wechsler <acwacw@gmail.com> a écrit :
In English, the technical probability terms are "sampling with replacement" and "sampling without replacement". Perhaps if we say "The odds of picking four letters at random from this sentence, and then finding them to be F, O, U, and R, are Y out Z," the intended protocol will be clearer.
On Wed, Dec 11, 2019 at 8:58 AM Éric Angelini <bk263401@skynet.be> wrote:
If the probability is 4*11*6*9/132^4, then we have a hit.
... yes, but in my mind once I've picked up a letter, I must erase that letter from the sentence, sorry (how could that figure in the "puzzle-sentence" without using too many words? In French it is a "Tirage sans remise")
In that case, aren't the odds 4*11*6*9/132*131*130*129?
I've almost a hit in French, by reducing the set of possibilities with the "puzzle-sentence" ending in: "[The probability of...] is one on K" -- with K = 132*131*130*129/4*11*6*9. (I didn't count the digits after the comma, though, which is neither correct, nor very elegant! But my error is below 1/10000... Best, É.
Le 11 décembre 2019 à 13:16, Hans Havermann <gladhobo@bell.net> a écrit :
The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z.
EA: "The 'FOUR' letter challenge is still open!"
If Y = "one" and Z = "one hundred twenty-seven thousand, seven hundred seventy-six", I make the total letter count of the sentence to be 132, with 4 of 'f', 11 of 'o', 6 of 'u', and 9 of 'r'. If the probability is 4*11*6*9/132^4, then we have a hit. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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"Out of" is correct. On Wed, Dec 11, 2019, 9:53 AM Éric Angelini <bk263401@skynet.be> wrote:
"The odds of picking four letters at random from this sentence, and then finding them to be F, O, U, and R, are Y out Z".
... this is it, I guess, thanks Allan -- wait: "X out Z" or "X out of Z"?
I prefer "pick" to "select" (suggested by Neil) as "select" seems to imply that the sampler carefully chooses -- instead of being blind, no?
Le 11 décembre 2019 à 15:41, Allan Wechsler <acwacw@gmail.com> a écrit :
In English, the technical probability terms are "sampling with replacement" and "sampling without replacement". Perhaps if we say "The odds of picking four letters at random from this sentence, and then finding them to be F, O, U, and R, are Y out Z," the intended protocol will be clearer.
On Wed, Dec 11, 2019 at 8:58 AM Éric Angelini <bk263401@skynet.be> wrote:
If the probability is 4*11*6*9/132^4, then we have a hit.
... yes, but in my mind once I've picked up a letter, I must erase that letter from the sentence, sorry (how could that figure in the "puzzle-sentence" without using too many words? In French it is a "Tirage sans remise")
In that case, aren't the odds 4*11*6*9/132*131*130*129?
I've almost a hit in French, by reducing the set of possibilities with the "puzzle-sentence" ending in: "[The probability of...] is one on K" -- with K = 132*131*130*129/4*11*6*9. (I didn't count the digits after the comma, though, which is neither correct, nor very elegant! But my error is below 1/10000... Best, É.
Le 11 décembre 2019 à 13:16, Hans Havermann <gladhobo@bell.net> a écrit :
The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z.
EA: "The 'FOUR' letter challenge is still open!"
If Y = "one" and Z = "one hundred twenty-seven thousand, seven hundred seventy-six", I make the total letter count of the sentence to be 132, with 4 of 'f', 11 of 'o', 6 of 'u', and 9 of 'r'. If the probability is 4*11*6*9/132^4, then we have a hit. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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HH: "The odds of picking at random the four letters F, O, U, R in this sentence and in that order are one out of one hundred twenty-seven thousand seven hundred seventy-six." If the probability is 4*11*6*9/132^4 ... EA: ... aren't the odds 4*11*6*9/(132*131*130*129) ? [brackets added] OK. Second attempt: "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are two out of two hundred nineteen thousand six hundred eighty-seven." Letter count = 132. F count = 5. O count = 11. U count = 6. R count = 8. 5*11*6*8/(132*131*130*129) = 2/219687
Waow, yes, that's it, wonderful work, Hans ! à+ É. Catapulté de mon aPhone
Le 13 déc. 2019 à 16:30, Hans Havermann <gladhobo@bell.net> a écrit :
HH: "The odds of picking at random the four letters F, O, U, R in this sentence and in that order are one out of one hundred twenty-seven thousand seven hundred seventy-six." If the probability is 4*11*6*9/132^4 ...
EA: ... aren't the odds 4*11*6*9/(132*131*130*129) ? [brackets added]
OK. Second attempt: "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are two out of two hundred nineteen thousand six hundred eighty-seven."
Letter count = 132. F count = 5. O count = 11. U count = 6. R count = 8.
5*11*6*8/(132*131*130*129) = 2/219687
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HH: "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are two out of two hundred nineteen thousand six hundred eighty-seven." I spent a considerable amount of time looking for additional solutions using the exact same template and blogged the results yesterday. Today, Greg Ross of Futility Closet decided to give it a shout-out: https://www.futilitycloset.com/2019/12/20/truth-in-advertising-7/
The odds of randomly picking two words from this statement and having them be BRAVO HANS, are two out of twenty two! Best, É. Catapulté de mon aPhone
Le 20 déc. 2019 à 18:36, Hans Havermann <gladhobo@bell.net> a écrit :
HH: "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are two out of two hundred nineteen thousand six hundred eighty-seven."
I spent a considerable amount of time looking for additional solutions using the exact same template and blogged the results yesterday. Today, Greg Ross of Futility Closet decided to give it a shout-out:
https://www.futilitycloset.com/2019/12/20/truth-in-advertising-7/ _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (11)
-
Allan Wechsler -
Brad Klee -
Hans Havermann -
James Buddenhagen -
James Propp -
Neil Sloane -
Ray Tayek -
Tom Karzes -
William R Somsky -
Éric Angelini -
Éric Angelini