Pick 3/4 Meter Made Math MovesPrev TopicNext Topic

• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 2:16 pm

  ▲▼

  folks, here's an example.

  we'll use a small set to show the changing 'a' value.

  if a small data set is {1, 6, 8, 0}, then a and b are the following:

  when 'a' is	then 'b' is
  1	6
  6	8
  8	0

  'a' is never constant.

  if 'a' is constant, the data is constant and therefore no randomness.

  huh, holding 'a' constant proves it's not random.

  The One Over None
  I Know...
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 4:02 pm

  ▲▼
  In response to dddwww

  Quote: Originally posted by dddwww on Jan 17, 2016

  The first step in analyzing should be asking "what is the probability of this occurring"

  While you seeing something in 6,7,8 that is just because your mapping of gaps into 1-10 makes those obvious.  Any other set of 3 gaps showing the same pattern would be just as likely.  There are 128 different combination of 3 digits so there are 128 different sets of 3 gaps that would be just as interesting as this particular set of three gaps.

  You got interested when the gap was 27 drawings long.  What are the odds that any of three specific clockwise moves won't show up in a set of two drawings?  70%

  So what are the odds that they won't show up 27 consecutive times?  .7 to the 27th power or about 1 in 15,000.

   

  Since there are 128 different sets of three gaps the odds that you'll see a gap of 27 drawings for any set of 3 clockwise moves is 1 in 120

  You have a sample size of 10,000 so there are almost certainly MANY gaps of 27 drawings for other sets of 3 clockwise moves.  You just can't see them because you only have placed 7 such sets adjacent to each other in your graph.

   

  My conclusion, just another example of how you can manipulate data to make patterns appear in large data sets that might seem unusual but really aren't

  well, it's clear you haven't read the entire topic.

  had you done so, you would have seen the full blow out of our graph, shown here:

  

  what you're seeing is the entire set of pick 3 numbers for both ball drawn and computer drawn numbers.

  the lottery post can only show a portion at a time; you'll have to scroll right.

  everyone can see the relatively smooth graph scrolling right then change suddenly at the end.

  we too questioned the unusual gap, so, we ran 10 x 10,000 samples or 100,000 samples to see if this is expected in 100,000 samples.

  we don't see a happenstance of 3 clustered wide gaps.

  this suggest the happenstance of 3 wide gaps is greater than 100,000 samples.

  once again we posted the graph of that sample run, seen here:

  

  The One Over None
  I Know...
• dddwww

  

  United States
  Member #171,725
  January 11, 2016
  127 Posts
  Offline
  Jan 17, 2016, 7:41 pm

  ▲▼
  In response to JADELottery

  Quote: Originally posted by JADELottery on Jan 17, 2016

  Actually, 'a' can't always be constant in a stream of data, so, your point is invalid.

  If a stream of data is d_n = {d₀, d₁, d₂, d₃, d₄, ...}, then a =  d_n and b = d_n+1.

  'a' can be any value at any point in the data.

  My point certainly is valid.

  You stated ""However, if one or both of a and b are non-random, then the difference, (b - a), must be non-random""

  An example of a non-random stream of numbers is a stream of constants.  Adding a stream of constants to a stream of random numbers does not produce a non-random output stream.

  As a starting point, consider two lists of length 1.  a and b.  a is a constant and b is a random number.  Adding them produces a random number.  The same is true regardless of the lengths of the lists.

  This refutes your claim and that is my point.

  You new claim is that "a can be any value at any point in the data".  That is not the definition of a non-random number.  A non-random number has a predictable value and can't have "any value at any point".

  If a() is a discrete function it would not be random.   a(n) + b(n) would generate a sum that is random with a distribution around a() instead of the normal set of random numbers people generate that have a distribution around a fixed number.  But the result is also random.  The position of an electron is represented by a probability distribution function that does not have a distribution around a fixed number but instead it is a distribution around the equation describing the orbit for a particular electron shell.
• dddwww

  

  United States
  Member #171,725
  January 11, 2016
  127 Posts
  Offline
  Jan 17, 2016, 7:43 pm

  ▲▼
  In response to lakerben

  Quote: Originally posted by lakerben on Jan 17, 2016

  

  I guess if you can't refute something you wave a flag?

  I'm always open to someone correcting some specific flaw in my analysis.
• dddwww

  

  United States
  Member #171,725
  January 11, 2016
  127 Posts
  Offline
  Jan 17, 2016, 7:54 pm

  ▲▼
  In response to JADELottery

  Quote: Originally posted by JADELottery on Jan 17, 2016

  folks, here's an example.

  we'll use a small set to show the changing 'a' value.

  if a small data set is {1, 6, 8, 0}, then a and b are the following:

  when 'a' is	then 'b' is
  1	6
  6	8
  8	0

  'a' is never constant.

  if 'a' is constant, the data is constant and therefore no randomness.

  huh, holding 'a' constant proves it's not random.

  "However, if one or both of a and b are non-random, then the difference, (b - a), must be non-random"

  That's what you original said.  Now you say "If 'a' is constant, the data is constant and therefore no randomness, holding 'a' constant proves it's not random"

   

  Make up your mind, you say "if 'a' is non-random" and then complain that if I make 'a' non-random then it's not random and somehow that's a problem for you?

  If you take a SINGLE stream of data and make every point both 'a' and 'b' (in your case a(n) = b(n-1) then both 'a' and 'b' are random or both 'a' and 'b' and not random.  There is no case where one is random and the other isn't as in fact they are identical streams of data, merely shifted. 

   

  Your original point is incorrect (random + non-random != non-random) and your last statement that a(n) = b(n-1) eliminates the possibility of a() and b() not either being both random or both non-random.
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 8:01 pm

  ▲▼
  In response to dddwww

  Quote: Originally posted by dddwww on Jan 17, 2016

  My point certainly is valid.

  You stated ""However, if one or both of a and b are non-random, then the difference, (b - a), must be non-random""

  An example of a non-random stream of numbers is a stream of constants.  Adding a stream of constants to a stream of random numbers does not produce a non-random output stream.

  As a starting point, consider two lists of length 1.  a and b.  a is a constant and b is a random number.  Adding them produces a random number.  The same is true regardless of the lengths of the lists.

  This refutes your claim and that is my point.

  You new claim is that "a can be any value at any point in the data".  That is not the definition of a non-random number.  A non-random number has a predictable value and can't have "any value at any point".

  If a() is a discrete function it would not be random.   a(n) + b(n) would generate a sum that is random with a distribution around a() instead of the normal set of random numbers people generate that have a distribution around a fixed number.  But the result is also random.  The position of an electron is represented by a probability distribution function that does not have a distribution around a fixed number but instead it is a distribution around the equation describing the orbit for a particular electron shell.

  "An example of a non-random stream of numbers is a stream of constants.  Adding a stream of constants to a stream of random numbers does not produce a non-random output stream."

   

  no, they can be anything.

   

  "As a starting point, consider two lists of length 1.  a and b.  a is a constant and b is a random number.  Adding them produces a random number.  The same is true regardless of the lengths of the lists."

   

  no,  there's only one list to work {d₀, d₁, d₂, d₃, d₄, ...}.

   

  "You new claim is that "a can be any value at any point in the data".  That is not the definition of a non-random number.  A non-random number has a predictable value and can't have "any value at any point"."

   

  no, nothing new here, the 'd' values in {d₀, d₁, d₂, d₃, d₄, ...} can be anything.

   

  "If a() is a discrete function it would not be random.   a(n) + b(n) would generate a sum that is random with a distribution around a() instead of the normal set of random numbers people generate that have a distribution around a fixed number.  But the result is also random.  The position of an electron is represented by a probability distribution function that does not have a distribution around a fixed number but instead it is a distribution around the equation describing the orbit for a particular electron shell."

   

  no, we never used 'a' as a function, only a value in a function, (b - a).

   

  no + no + no + no = invalid.

  The One Over None
  I Know...
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 8:06 pm

  ▲▼
  In response to dddwww

  Quote: Originally posted by dddwww on Jan 17, 2016

  "However, if one or both of a and b are non-random, then the difference, (b - a), must be non-random"

  That's what you original said.  Now you say "If 'a' is constant, the data is constant and therefore no randomness, holding 'a' constant proves it's not random"

   

  Make up your mind, you say "if 'a' is non-random" and then complain that if I make 'a' non-random then it's not random and somehow that's a problem for you?

  If you take a SINGLE stream of data and make every point both 'a' and 'b' (in your case a(n) = b(n-1) then both 'a' and 'b' are random or both 'a' and 'b' and not random.  There is no case where one is random and the other isn't as in fact they are identical streams of data, merely shifted. 

   

  Your original point is incorrect (random + non-random != non-random) and your last statement that a(n) = b(n-1) eliminates the possibility of a() and b() not either being both random or both non-random.

  An example of 'a' being constant would be {0, 0, 0, 0}.

  That wound make the following table:

  when 'a' is	then 'b' is
  0	0
  0	0
  0	0

  'a' becomes all 0's, no change, no randomness.

  The One Over None
  I Know...
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 8:11 pm

  ▲▼
  In response to dddwww

  Quote: Originally posted by dddwww on Jan 17, 2016

  I guess if you can't refute something you wave a flag?

  I'm always open to someone correcting some specific flaw in my analysis.

  Really, is that the best you come up with...  pfff.

  The One Over None
  I Know...
• dddwww

  

  United States
  Member #171,725
  January 11, 2016
  127 Posts
  Offline
  Jan 17, 2016, 9:04 pm

  ▲▼

  I wrote some quick and dirty C code that JUST looks at the 6,7 and 8 simultaneous gaps

  It confirms my math which said the odds are roughly 1 gap of 27 every 15000 draws

   

  My code runs on a Linux based virtual machine and I set it to 15000 draws and ran it a few times.  The smallest 6,7,8 gap I saw was 19 and most were in the mid to high 20's
  
  
  As I stated before, the 6/7/8 is only one of 128 different combination of 3 clockwise moves that would also exhibit exactly the same probability of having long simultaneous gaps.
  
  
  
  
   That makes such gaps quite common and the graphic method used by the OP is simply not up to the job of spotting most of those 128 combinations.
  
  
  
  
  You'll have to reformat the following, I have no idea how to format text pasted to a message.
  
  
  
  
   Change trials = 15000 if you want more or fewer, change the if ((move != 6) && (move != 7) && (move != 8)) to look for longest simultaneous gaps with other combinations
  
  
  
  
  I'm too lazy to pass trials and a command line parameter or write the code to look for maxgaps for all 128 combinations of clockwise moves. 
  

  
  
  
  
  
  
  #include <stdio.h>
  
  
  #include <stdlib.h>
  
  
  #include <time.h>
  
  
  
  
  main()
  
  
  {
  
  
   int i, j, a, b, move, s[10], maxgaps[10], currgaps[10];
  
  
   int trials = 15000;
  
  
   int specialcurrgaps = 0;
  
  
   int specialmaxgaps = 0;
  
  
  
  
   srand(time(NULL));
  
  
   /* Some stats of clockwise moves */
  
  
  
  
    for (i=0; i<10; i++) s[i] = 0, maxgaps[i] = 0, currgaps[i] = 0;
  
  
  
  
   b = rand()%10;
  
  
   /* Generate a stream of random digits between 0 and 9 */
  
  
   for (i=0; i<trials; i++) {
  
  
   a = rand()%10;
  
  
   if (b>a)
  
  
   move = 10+a-b;
  
  
   else
  
  
   move = a-b;
  
  
  
  
    s[move]++;
  
  
  
  
   /* Update gaps */
  
  
   /* The gap for move has ended, check to see if new longest gap */
  
  
   if (currgaps[move] > maxgaps[move]) {
  
  
   maxgaps[move] = currgaps[move];
  
  
   }
  
  
  
  
   /* Update all gaps */
  
  
   for (j=0; j<10; j++) {
  
  
   if (j==move) currgaps[j] = 0;
  
  
   else currgaps[j]++;
  
  
   }
  
  
  
  
   /* Now check for gaps 6, 7 and 8 all active at once */
  
  
   if ((move != 6) && (move != 7) && (move != 8))
  
  
    specialcurrgaps++;
  
  
  
  
   else {
  
  
   if (specialcurrgaps > specialmaxgaps)
  
  
    specialmaxgaps = specialcurrgaps;
  
  
   specialcurrgaps = 0;
  
  
   }
  
  
  
  
   b = a;
  
  
   }
  
  
  
  
   /* Print results of trials */
  
  
   printf(" %i trials\n", trials );
  
  
  
  
   printf ("\nSpecial Gap Max Size: %i\n", specialmaxgaps );
  
  
  }
  
  
  
  
  
  
  
• dddwww

  

  United States
  Member #171,725
  January 11, 2016
  127 Posts
  Offline
  Jan 17, 2016, 9:13 pm

  ▲▼
  In response to JADELottery

  Quote: Originally posted by JADELottery on Jan 17, 2016

  Really, is that the best you come up with...  pfff.

  That's certainly the best I can come up for in reply to a message which contained nothing but a waving flag.

   

  I did post some actual code so other people can confirm that in fact my analysis was correct and such gap lengths are common in a 15000 draw sequence.  I'd be happy to look at your code where  you claim it doesn't find long gaps to be common as it is clear you are doing something wrong.

   

  My conclusion is that there is no flaw in the drawings, what you noticed is not unusual.  My proof consists of a mathematical analysis of the probability followed by a simulation using C code that I posted here for review that confirms my mathematical analysis.  I suggested that your use of a graphical view that only puts 7 of the 128 combinations adjacent to each other effectively hides other combinations that also have similar gap lengths and as the probability of finding a gap of 27 in any combination of 3 clockwise move values requires far less than 15000 draws.  I haven't verified that with a simulation.
• dddwww

  

  United States
  Member #171,725
  January 11, 2016
  127 Posts
  Offline
  Jan 17, 2016, 9:14 pm

  ▲▼
  In response to JADELottery

  Quote: Originally posted by JADELottery on Jan 17, 2016

  An example of 'a' being constant would be {0, 0, 0, 0}.

  That wound make the following table:

  when 'a' is	then 'b' is
  0	0
  0	0
  0	0

  'a' becomes all 0's, no change, no randomness.

  That would be an example of both a and b being non-random

   

  Show me an example where a is non-random and b is random
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 9:19 pm

  ▲▼
  In response to dddwww

  Quote: Originally posted by dddwww on Jan 17, 2016

  That would be an example of both a and b being non-random

   

  Show me an example where a is non-random and b is random

  actually, that is your expression of 'a' being constant.

  The One Over None
  I Know...
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 9:42 pm

  ▲▼

  an example would be using the small set of {d₀, d₁, d₂, d₃}; where d₂ is non-random.

  when 'a' is	and 'b' is	(b - a) becomes
  d0 'random'	d1 'random'	(d1 - d0) 'random'
  d1 'random'	d2 'non-random'	(d2 - d1) 'non-random'
  d2 'non-random'	d3 'random'	(d3 - d2) 'non-random'

  The One Over None
  I Know...
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 9:44 pm

  ▲▼
  In response to dddwww

  Quote: Originally posted by dddwww on Jan 17, 2016

  I wrote some quick and dirty C code that JUST looks at the 6,7 and 8 simultaneous gaps

  It confirms my math which said the odds are roughly 1 gap of 27 every 15000 draws

   

  My code runs on a Linux based virtual machine and I set it to 15000 draws and ran it a few times.  The smallest 6,7,8 gap I saw was 19 and most were in the mid to high 20's
  
  
  As I stated before, the 6/7/8 is only one of 128 different combination of 3 clockwise moves that would also exhibit exactly the same probability of having long simultaneous gaps.
  
  
  
  
   That makes such gaps quite common and the graphic method used by the OP is simply not up to the job of spotting most of those 128 combinations.
  
  
  
  
  You'll have to reformat the following, I have no idea how to format text pasted to a message.
  
  
  
  
   Change trials = 15000 if you want more or fewer, change the if ((move != 6) && (move != 7) && (move != 8)) to look for longest simultaneous gaps with other combinations
  
  
  
  
  I'm too lazy to pass trials and a command line parameter or write the code to look for maxgaps for all 128 combinations of clockwise moves. 
  

  
  
  
  
  
  
  #include <stdio.h>
  
  
  #include <stdlib.h>
  
  
  #include <time.h>
  
  
  
  
  main()
  
  
  {
  
  
   int i, j, a, b, move, s[10], maxgaps[10], currgaps[10];
  
  
   int trials = 15000;
  
  
   int specialcurrgaps = 0;
  
  
   int specialmaxgaps = 0;
  
  
  
  
   srand(time(NULL));
  
  
   /* Some stats of clockwise moves */
  
  
  
  
    for (i=0; i<10; i++) s[i] = 0, maxgaps[i] = 0, currgaps[i] = 0;
  
  
  
  
   b = rand()%10;
  
  
   /* Generate a stream of random digits between 0 and 9 */
  
  
   for (i=0; i<trials; i++) {
  
  
   a = rand()%10;
  
  
   if (b>a)
  
  
   move = 10+a-b;
  
  
   else
  
  
   move = a-b;
  
  
  
  
    s[move]++;
  
  
  
  
   /* Update gaps */
  
  
   /* The gap for move has ended, check to see if new longest gap */
  
  
   if (currgaps[move] > maxgaps[move]) {
  
  
   maxgaps[move] = currgaps[move];
  
  
   }
  
  
  
  
   /* Update all gaps */
  
  
   for (j=0; j<10; j++) {
  
  
   if (j==move) currgaps[j] = 0;
  
  
   else currgaps[j]++;
  
  
   }
  
  
  
  
   /* Now check for gaps 6, 7 and 8 all active at once */
  
  
   if ((move != 6) && (move != 7) && (move != 8))
  
  
    specialcurrgaps++;
  
  
  
  
   else {
  
  
   if (specialcurrgaps > specialmaxgaps)
  
  
    specialmaxgaps = specialcurrgaps;
  
  
   specialcurrgaps = 0;
  
  
   }
  
  
  
  
   b = a;
  
  
   }
  
  
  
  
   /* Print results of trials */
  
  
   printf(" %i trials\n", trials );
  
  
  
  
   printf ("\nSpecial Gap Max Size: %i\n", specialmaxgaps );
  
  
  }
  
  
  
  
  
  
  

  We'll examine the code.

  The One Over None
  I Know...
• JADELottery

  

  Eagle Sun

  IA, MN, WI
  United States
  Member #21
  December 7, 2001
  5,177 Posts
  Offline
  Jan 17, 2016, 9:47 pm

  ▲▼
  In response to dddwww

  Quote: Originally posted by dddwww on Jan 17, 2016

  That's certainly the best I can come up for in reply to a message which contained nothing but a waving flag.

   

  I did post some actual code so other people can confirm that in fact my analysis was correct and such gap lengths are common in a 15000 draw sequence.  I'd be happy to look at your code where  you claim it doesn't find long gaps to be common as it is clear you are doing something wrong.

   

  My conclusion is that there is no flaw in the drawings, what you noticed is not unusual.  My proof consists of a mathematical analysis of the probability followed by a simulation using C code that I posted here for review that confirms my mathematical analysis.  I suggested that your use of a graphical view that only puts 7 of the 128 combinations adjacent to each other effectively hides other combinations that also have similar gap lengths and as the probability of finding a gap of 27 in any combination of 3 clockwise move values requires far less than 15000 draws.  I haven't verified that with a simulation.

  Excellent.

  Ignore the others' replies.

  The One Over None
  I Know...