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• lottoarchitect

  

  Greece
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  November 18, 2003
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  Apr 19, 2004, 11:54 am

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  Hi everyone, my questi

  If you have something to do, at least do it well...
• RJOh

  

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  mid-Ohio
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  Apr 19, 2004, 2:16 pm

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  I recently added a feature to my program that allows me to compare a drawing with the previous drawings (any amount) to see if the hit frequency(hot or cold) can be an indicator of which ones to wheel.  For example when looking at the previous 20 draws for a 5/37 game(Buckeye5), numbers that have hit 2-3 times repeat most. It also shows that numbers that  have hit 2-3 times is usually a list of 18-25 numbers and contain 4 of the winning numbers about 5% of the time.  With PowerBall (5/53) the choice is even more confusing.

  RJOh

   * you don't need to buy every combination, just the winning ones * 

         
  
• BobP

  

  Dump Water Florida
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  Apr 19, 2004, 2:21 pm

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  Just apply probability you'll get the same answer or the programs did even worse.  BobP
• lottoarchitect

  

  Greece
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  November 18, 2003
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  Apr 19, 2004, 4:46 pm

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  So, RJOh you say that you pick 18-25 numbers and in 20 draws, your chance to have 4 correct numbers is 5% in a 5 of 37 game?

  BobP, all those programs do worse than probability?

  For example, if i pick X numbers in a game 5 of 37, then the probability is:

  Picked>      10            15            20

  0 match : 18.5%        6%          1.4%

  1 match : 40.3%      25.2%      10.9%

  2 match : 30.2%      37.1%      29.6%

  3 match :  9.7%      24.1%      35.6%

  4 match :  1.3%        6.9%      18.9%

  5 match : 0.05%      0.7%        3.5%

  So, if we pick 20 numbers for a game 5 of 37, our chances are 18.9% to have 4 numbers in our selection. RJOh, do you perform worse than luck? It looks strange you get only 5% for 4 match even if you use 18 numbers (chance 13.3% to match 4).

  If you have something to do, at least do it well...
• RJOh

  

  100

  mid-Ohio
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  March 24, 2001
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  Apr 19, 2004, 6:19 pm

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  Apagogeas,

  I was surprised at those results also because I was looking for a way to pick a few numbers to wheel and willing to accept losing if there was a good chance of a match5 in the near future.  I was figuring that any strategy that would work in the future, had probably worked in the past.

  According to your figures, if I pick 20 numbers the group is more likely to have 3 numbers(35.6%) than 2 numbers (29.6%) than 1 number (10.9%).  I would have thought it would be just the opposite.  I know that figures don't lie, but I have to question those figures. The figures I posted was what I observed, I would have thought that calculated figures would have been closer to them..

  RJOh

   * you don't need to buy every combination, just the winning ones * 

         
  
• lottoarchitect

  

  Greece
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  Apr 19, 2004, 7:02 pm

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  There is a simple equation to determine the chance to match x numbers. I'll describe it to you. It works only for fields that don't allow same numbers to be drawn (no Pick-3/4 games).

  Suppose you have a game a of b (eg 5 of 45, a=5 b=45). You want to pick n numbers (eg 10 numbers, n=10) and you want to find the chance to have m matches (eg 3 matches, m=3) in your selection.

  Then, chance %= [ nCm * (b-m)C(a-m) / bCa ] * 100%

  where nCk =n!/ (k!*(n-k)!)
  

  If you have something to do, at least do it well...
• RJOh

  

  100

  mid-Ohio
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  Apr 19, 2004, 8:06 pm

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  I was going to use the equation to calculate the chance% or a match4 in a 5/37 game with 20 numbers,
  a=5
  b=37
  n=20
  m=4
  but you didn't define n,C or k.  nCk looks like the combination formula.

  RJOh

   * you don't need to buy every combination, just the winning ones * 

         
  
• lottoarchitect

  

  Greece
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  Apr 19, 2004, 8:26 pm

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  nCk is the combination formula

  If you have something to do, at least do it well...
• lottoarchitect

  

  Greece
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  Apr 19, 2004, 8:35 pm

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  There is a mistake in the formula

  game 5 of 45 -> a=5, b=45

  picked numbers -> n

  match -> m

  Then, chance %= [ nCm * (b-n)C(a-m) / bCa ] * 100%

  If you have something to do, at least do it well...
• RJOh

  

  100

  mid-Ohio
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  Apr 20, 2004, 2:13 am

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  5/20 = 15504 possible combinations of 5's - 4845 combinations of 4's
  5/37 = 435897 possible combinations of 5's - 66045 combinations of 4's
  15504/435897 = 3.6%  and  4845/66045 = 7.3%
  My guess is the correct solution is between those two figures.

   * you don't need to buy every combination, just the winning ones * 

         
  
• lottoarchitect

  

  Greece
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  November 18, 2003
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  Apr 20, 2004, 10:55 am

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  The division you apply is not valid because you apply it in different sets of numbers. This is why you get wrong results. I'll explain the above equation just to make it clear why it works.

  The parameters are as above a,b,n,m

  We have the equation %=[ nCm * (b-n)C(a-m) / bCa ] * 100%
  

  Now, bCa = the total tickets available for our game.

  nCm = In how many different ways we can have m numbers in a set of n numbers.

  Now, for each of the above nCm ways, we can arange the remaining a-m numbers (to form a complete ticket) in (b-n)C(a-m) possible ways. Thus, all the tickets that have m matches in n numbers, is the multiplication of the above 2 sets, nCm * (b-n)C(a-m). This result shows all tickets that can have m matches in a set of n picketd numbers.

  Finally we divide by bCa to get a rate from 0 to 1 and multiply with 100% to get the percentage.

  If you have something to do, at least do it well...
• lottoarchitect

  

  Greece
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  Apr 22, 2004, 2:20 pm

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  Ouaou! Ok, no more ads! I boug

  If you have something to do, at least do it well...
• lottoarchitect

  

  Greece
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  November 18, 2003
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  Apr 26, 2004, 7:40 pm

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  It seems there is a problem

  If you have something to do, at least do it well...
• BobP

  

  Dump Water Florida
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  June 5, 2002
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  Apr 27, 2004, 3:06 am

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  The odds of any set of twenty numbers containing the prize winning numbers in a 5/37 game are . . .
  
  5 in 20: 1 in 28.12
  4 in 20: 1 in 5.29
  3 in 20: 1 in 2.81
  2 in 20: 1 in 3.37 (positive)
  1 in 20: 1 in 9.16 (positive)
  
  I don't do math, but I have good software and I know how to use it. ;-) BobP
• lottoarchitect

  

  Greece
  Member #2,815
  November 18, 2003
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  Apr 27, 2004, 4:51 am

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  These rates are correct BobP,

  Actually, you say it in other words, because

  1 in 20 : 1 in 9.16=1/9.16=10.9%

  2 in 20 : 1 in 3.37=1/3.37=29.7%

  etc

  If you have something to do, at least do it well...