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  1. #1
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    Cycles, Revisited

    Some of you who have been around for a while may remember the discussion of completion cycles for the 6/49 game -- the length of time that it takes to hit all 49 numbers in the game. The rest of you can read about it here:
    http://www.lottoforums.com/lottery/lotto-tips-strategies/12945

    I was just looking at this data again and I thought I would share what is currently happening in the Canandian Lotto 6/49 game.

    The current cycle started on 1/16/2016 and 7 games have been drawn. There are 19 numbers that remain to be hit, a little higher than the average of 16 at this point in the cycle.

    We can expect some of the unhit numbers to be drawn tonight. Historically,
    5% of games have drawn have hit 0 of the unhit numbers
    21% have hit 1
    18% have hit 2
    34% have hit 3
    18% have hit 4, and
    3% have hit 5.
    So, 2, 3, or 4 of the unhit numbers should hit tonight. Four has been hitting above its average lately, so I would go with 3 or possibly 2 of these numbers:
    1.2.3.7.8.9.10.11.14.16.17.18.22.27,28,30,33,46,49 .

    Pick your favorites, or just make some random selections.

    Good Luck!

  2. #2
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    Icewynd, I just tried your link and it goes nowhere, except back to the home page ?

  3. #3
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    Thanks for the heads up Frank. Here's a working link:
    http://www.lottoforums.com/lottery/lotto-tips-strategies/12945-eliminate-up-18-numbers-6-a.html

  4. #4
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    And we hit 3 of the unhit numbers: 8,12 and 33, leaving 16 unhit at the 9th game of the cycle vs. an average of 14.

    For the 9th game of the cycle we can expect:
    0 unhit =6%
    1 =21%
    2 =35%
    3 =24%
    4 =11%

    So, most likely 1,2 or 3 of the unhit numbers will come out in Saturday's draw. I'm going with 2. The remaining numbers are:

    1,2,3,6,10,11,14,16,17,18,22,27,28,30,47,49.

    Good Luck!

  5. #5
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    For anyone following LT's set of 20 numbers (which get excellent results) the numbers marked in red are the ones that match the set of numbers that have not yet hit in the current cycle:

    01,04,05,06,09,10,12,17,20,22
    23,31,35,36,37,39,43,44,45,46

    These would make good picks if you are planning on incorporating 2 or 3 of these numbers into your sets for play.

    Good Luck!

  6. #6
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    3 more numbers from the unhit group last night: 2, 18, 27.

    Wednesday's draw will be the 10th in the cycle. There are now 13 numbers remaining that have not hit in this cycle:
    1,3,6,10,11,14,16,17,22,28,30,46,49.

    For the 10th draw we can expect:
    0 hits 13%
    1 hit 27%
    2 hits 34%
    3 hits 15%
    4 hits 8%
    5 hits 2%

    As you can see, 1 or 2 hits from this group are most likely, but we can see that the probability of zero hits is growing as we go on. If I were playing 3 tickets I would probably include 0,1 and 2 of these picks. I will choose 1 hit from this group (and 5 from the already-hit group) as this cycle has been hitting higher numbers for the last few games.

    Good Luck!

  7. #7
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    Ice,

    I've done a bit of investigation into the UK Thunderball since the draw changed from 34 to 39 balls, a total of 903 draws, analysis is without the bonus ball as it is drawn from a different pool.

    There have been 31 full cycles at an average of 28.5 draws per cycle. The current cycle is 19 draws old and has only 3 balls yet to come out, they are 13,24 & 33. Also for the current cycle the number of balls drawn per draw are so:

    5,3,4 4,3,4,3,2,1,2,0,1,1,0,1,1,0,1,0

    Cartref

  8. #8
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    Interesting, Cartref! I would have thought that the average cycle length would have been quite a bit shorter for 39 numbers than the average of 31 games for 49 numbers.

    I see you have spotted the exciting possibility of this method -- that around mid-cycle you start to get draws where none of the unhit balls are drawn (and thus, can be successfully eliminated).

    Good Luck!

  9. #9
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    Yes this does happen in most cycles, the magic is when does this happen? .

    Following is the number of draws for each of the 31 cycles:

    18,33,20,27,26,30,32,24,29,35,34,31,30,38,25,28,38 ,20,27,25,25,16,34,33,26,33,29,22,27,45,23

    There are a couple of exceptionally long cycles which does push the average up.

    Also for the 3 largest cycles; 38, 38 & 45.

    The last ball drawn took an extra 13, 18 & 22 more draws before coming out.

  10. #10
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    I'm pretty sure there is a mathematical law that determines what the average cycle span is for a particular lottery matrix. It's definitely an exponential decay curve if you plot balls still to come out in the cycle against draws into the cycle. As Icewynd has already demonstrated, the probability of NOT matching the balls already drawn in the cycle ( hence adding to the count of unique balls drawn) reduces the further into the cycle you go. This is a classic property of exponential decay. Somewhere the exponent of the natural constant e contains an expression involving the size of the lottery ( how many balls) , and how many are drawn. Working out what this expression is, may be beyond me, but I'd love to know what it is.

  11. #11
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    Having said that, The new 'improved' UK 659 lottery is on its 37th draw and is 37 draws into its first ever cycle. Including bonus balls, with three balls still to appear (5,33,45), the rate of reduction of undrawn balls is currently:- 7, 7,6,4,7,4,1,2,2,2,2,0,1,0,0,2,2,0,1,1,0,3,1,0,0,0, 0,0,0,0,0,0,1,0,0,0,0.

    Plot that as a graph and it looks horrible. One would need to get data for multiple cycles and take the average for each point before a plot would look like a decay curve.

  12. #12
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    Well, Frank, even after more than 100 cycles on the Canadian 6/49 the rate of decay is only steady to about 19 draws and then starts to jump around -- so the "tail of the beast" is the most likely to vary from the established decay rate. It is unlikely that there is a lotto in existence with enough data to provide a smooth curve out past this point.

  13. #13
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    Yes, I would tend to agree, but the maybe first 19 steady (averaged) points would be enough to establish the law. The method is to take logarithms of the points on the vertical axis and plot those and see if it approximates to a straight line.The slope of the line is the exponent of e. As you cannot take logs of zero (doesn't exist) and the log of 1 is 0 and these values are on "the tail of the beast " as you put it they are of no value anyway.

  14. #14
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    Good work Icewynd. I was a little lost for a minute as my sheet wasn't adding up to what you where saying until I figured out that you where including the bonus number. So just to back up a little bit. A new cycle will start when all of the numbers finally have hit. Right now the #14 for the Canadian 649 last occurrence was Oct31/15. Last question Icewynd. When use your formula (draws/frequency)-skip. When the number in question is in the most recent draw do you give it a skip of "0" or "1"? Thanks for all of your input.

  15. #15
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    Quote Originally Posted by Frank
    Having said that, The new 'improved' UK 659 lottery is on its 37th draw and is 37 draws into its first ever cycle. Including bonus balls, with three balls still to appear (5,33,45)........
    I notice that TWO of the Undrawn balls were drawn last night, 5 and 33 leaving just number 45 to complete the cycle. I didn't bet on them so don't get my free ticket prize.

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