< Earlier Kibitzing · PAGE 287 OF 287 ·
Later Kibitzing> 
Jul0316
  john barleycorn: Still, my question: "When does the player lose?" There is no misreading possible as it is not mentioned in the game description. the question:
<If the casino is to break even playing this game, how much should they charge — what should be the value of X?> is absolutely meaningless in that case. 

Jul0316   beatgiant: <john barleycorn>
The player loses money if the charge X he paid is higher than the amount he won. 

Jul0316   beatgiant: The gambling game in question is the wellknown "St. Petersburg Paradox." Wikipedia has a good article on it. 

Jul0316
  WannaBe: Here's the link, I'll save y'all from typing it in google. =) https://en.wikipedia.org/wiki/St._P... 

Jul0416
  john barleycorn: <beatgiant: <john barleycorn>
The player loses money if the charge X he paid is higher than the amount he won.> Ah, thank you. so the player pays an amount X at the beginning and it is automatically collected by the casino irrelevant of the outcome of the coin flips. 

Jul1416
  al wazir: It's convention time!
As is traditional, a banquet is held at the conclusion of the Republicratic convention. Two hundred senior partisans, selected by invitation from the party's most principled princes and dignified dignitaries, stream into the ballroom where tables have been set up. When they arrive they find that seats has been preassigned. Every setting has a name tag, and the invitees are required to sit at the place with their own tag. Chaos!
Two hundred mostly elderly, increasingly hungry, slightly tipsy (from prebanquet cocktails) guests mill around, bumping into chairs and tables, stepping on each other's toes, as they try to locate their places. Some give up the search and sit at random, only to be displaced when the rightful occupant finds his name and claims the place. A few headstrong mavericks  most of them longserving legislators accustomed to their own safe electoral seats  redistribute the name tags. As a result, some tables have more tags than chairs. Is there a better way to do it? (Cf. https://www.amazon.com/ArtComputer...) 

Jul1416   Schwartz: Just a small amount should search at the same time, unless there's a lot of room. 

Jul1416
  john barleycorn: get them seated by trained stuff. 

Jul1416
  Tiggler: More cocktails! Then bring them in on wheelbarrows. 

Jul1416
  Tiggler: Open seating is the reason Southwest Airlines boards twice as fast and saves billions. 

Jul1516
  al wazir: <Tiggler: Open seating is the reason Southwest Airlines boards twice as fast and saves billions.> Good point. But in this scenario the seating isn't open. Since you mention airline seating, I should point out that some airlines *platoon* boarding passengers. Their tickets are labeled "Group 1," "Group 2," and so on, and the groups are assigned seats in such a way that Group 1 goes to the back of the plane where it won't interfere with Group 2. Then Group 2 takes the seats forward of them, and so on. So one solution to the problem would be to give the banqueters tickets with table numbers. Alternatively, they could simply be told ahead of time what table they are assigned to, and at appropriate intervals the Maître d' could announce over the PA system "Table 1 now seating ... Table 2 now seating ...," etc. But I'm not convinced that there isn't a better way. I recently attended an outdoor luncheon where they must have been around 200 guests, if not more. Owing to a conflict (it was a chess conflict, as a matter of fact), I arrived an hour after everyone else had been seated, and even knowing the table I was supposed to sit at I had a hell of a time finding my seat. So if everyone arrived at the same time it must have been a mess. 

Aug1016
  WannaBe: Collatz Conjecture:
http://www.popularmechanics.com/tec... 

Aug1016
  john barleycorn: I know Collatz from his work on linear optimization but never knew he has his own problem . ha the easier the problem is to formulate the harder to solve. Thanks for pointing it out 

Aug1316
  al wazir: Collatz conjecture: See Louis Stumpers (kibitz #4488) et seq. 

Aug1616
  WannaBe: Since we like to discuss math here, thought I'd post this: Saw "The Man Who Knew Infinity" last night, solid movie. Very enjoyable. I give it an "A". 

Aug1616
  alexmagnus: The Collatz conjecture mention here, huh. But as R.K. Guy said: <"Don't try to solve these problems!"> (so was the title of a paper including several unsolved "elementary" math problems, among them the Collatz) Another problem from the same paper is similar, but involving division by 3 instead of 2: Start with a natural number n. If it is divisible by 3, multiply it by 2/3. If the remainder from division by 3 is 1, take 4n/31/3. And if the remainder is 2, take 4n/3+1/3. The "unsolvable" problem: does the sequence beginning with n=8 ever go into a cycle or does it diverge? Also, what if we trace the same sequence from n=8 <backwards>?! It appears to go from large numbers, through n=8 and then back to large numbers. Why n=8? Because for n<8 the sequence does go into a cycle (1 just repeats endlessly, 2 forms a (2,3)cycle, 4 forms a (4,5,7,9,6) cycle). There is only one other cycle known  (44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66). Whether any more exist, is another open question. 

Aug1716
  al wazir: Here's a puzzle inspired by the Collatz Conjecture. Take any positive integer, double it and add 1. If the result is a prime, stop. If not, double it again and add 1. If this number is a prime, stop. If not, keep repeating the process. Will the sequence of numbers resulting from this procedure always end in a prime? Note that the sequences created this way increase as powers of 2, i.e., exponentially. The distribution of primes is known to be logarithmic. (This is the content of the Prime Number Theorem, https://en.wikipedia.org/wiki/Prime....) Primes are rather abundant for small integers; they become less abundant the higher you go. But the '2x+1' procedure yields numbers that are even farther apart. So what we have here is a situation analogous to Arctic foxes hunting lemmings: the farther north you go, the rarer members of both species become, and by the time you reach the North Pole it is doubtful whether any kills take place. I conjecture that the answer is no. Future mathematicians will refer to this as the Al Wazir Conjecture. Here are some examples.
1, 3.
2, 5.
3, 7.
4, 9, 19.
5, 11.
6, 13.
7, 15, 31.
8, 17.
9, 19.
10, 21, 43.
11, 23, 47.
12, 25, 51, 103.
13, 27, 55, 111, 223.
14, 29.
15, 31.
16, 33, 67.
17, 35, 71.
18, 37.
19, 39, 79.
20, 41. 

Aug1716
  al wazir: Oops. The eleventh line of that list should read 11, 23. Sorry. 

Aug1716
  alexmagnus: <Al wazir> The answer is no. Your numbers have a form 2n+1, 4n+3, 8n+7 and so on  that is n*2^k+2^k1 = (n+1)2^k1. So, we need a number x for which x*2^k1 is always composite  then x1 is your counterexample. The numbers x from above are known as Riesel numbers. It is proven that there are infinitely many of them. The smalles number <proven> to be Riesel is 509203 (for it, x*2^k1 will always be divisible by one of 3, 5, 7, 13, 17, 241). There are some smaller numbers for which it is unknown whether they are Riesel or not (smallest of them being 2293). So, if you start your sequence at 509202, you'll never end up with a prime. 

Aug1716
  alexmagnus: <for it [509203], x*2^k1 will always be divisible by one of 3, 5, 7, 13, 17, 241> In case you need the proof, here the remainders of 2x+1 recursion as a cycle for 509202: 3: (1, 0)
5: (0, 1, 3, 2)
7: (3, 0, 1)
13: (11, 10, 8, 4, 9, 6, 0, 1, 3, 7, 2, 5)
17: (3, 7, 15, 14, 12, 8, 0, 1)
241: (180, 120, 0, 1, 3, 7, 15, 31, 63, 127, 14, 29, 59, 119, 239, 238, 236, 232, 224, 208, 176, 112, 225, 210) So, the numbers divisible by those have the k index which satisfies one of: 0 mod 2
1 mod 4
2 mod 3
7 mod 12
7 mod 8
3 mod 24
Now search for numbers that fit neither congruence.
First line eliminates all even numbers. Second line leaves only number 3 mod 4 behind. Third line: only 3 mod 4 AND 0 or 1 mod 3  that is (by the Chinese remainder theorem), either 3 mod 12 or 7 mod 12. The next line eliminates the numbers 7 mod 12  leaving only with 3 mod 12. The next line tells that our 3 mod 12 numbers may not be 7 mod 8  that is, they are 0, 1, 2, 3, 4, 5 or 6 mod 8. 0, 2, 4, and 6 fall out because they are not 1 mod 2. 1 and 5 fall out because they are not 3 mod 4. So, our number is 3 mod both 8 and 12  that is, 3 mod 24. But the last line excludes numbers 3 mod 24.
That means, no counterexamples exist. 

Aug1716
  Domdaniel: <alexmagnus> That's beautiful, thank you. I knew that recursive loops existed, but not where to find them. 

Aug1716
  al wazir: <alexmagnus>: Outstanding! That shoots downs my bid to become an eponym. But it looks as if Riesel was ahead of me in line. 

Aug1716
  Domdaniel: <al wazir>  < it looks as if Riesel was ahead of me >
That's nothing  I've 'discovered' large chunks of 18th century maths. I once even 'invented' Bernoulli numbers, about 300 years after Bernoulli. And don't get me started on Euler, who repeatedly pipped me to the post... with 250 years to spare. 

Aug1716
  john barleycorn: <Domdaniel: ... I once even 'invented' Bernoulli numbers, about 300 years after Bernoulli. ...> We all know how hard it is to invent something that is already invented. 

Aug1716
  alexmagnus: When I was 10, I "invented" logarithms.
Actually I often come up with some mathematical ideas  but Googling quickly reveals those ideas are well known. This is actually why I knew about the Riesel numbers  I once had ideas along similar lines as <al wazir> (slightly differnt  but eventually also arriving at the question of existence of what is known as Riesel numbers). The proof I posted if from me though  I'm pretty sure if you google you'll find some quicker proof (after all, my proof comes from <knowing> that 509203 is a Riesel number, while whoever made the original proof arrived at it first somehow!) 


< Earlier Kibitzing · PAGE 287 OF 287 ·
Later Kibitzing> 


