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Thread: Cycles, Revisited

02182016, 06:18 AM #16
For the Canada Lotto 6/49 draw for 2/17 only one of our unhit numbers hit: 28.
This leaves 12 numbers still to hit in this cycle:1,3,6,10,11,14,16,17,22,30,46,49.
For the 11th draw of the cycle we can expect:
0 hits = .18
1 hit =.27
2 hits =.32
3 hits = .18
4 hits = .03
One or two hits from this group are our best bets, I'll pick 2 for this round.
Good Luck!

02182016, 06:28 AM #17Originally Posted by cdrake
Originally Posted by cdrake
Good Luck!

02202016, 03:48 AM #18
Last night was the 21 draw of the current cycle on the UK Thunderball, only 13,24 & 33 hadn't hit in this cycle, well 2 of those numbers hit last night (24 & 33), leaving just 13 remaining in the current cycle.
I had one hit 3 but no thunderball so just £10

02202016, 05:40 AM #19Registered User
 Join Date
 Feb 2011
 Posts
 867
Hello ice, you can do the cycle, but only the last digit endings (09)
In each position of 49/6 and 60/6
example =
01 05 12 26 38 56 last digit = 1,5,2,6,8,6
Then it would be only the cycle of endings in each position in the vertical
BJECTIVE = see the cycle by position by position but only last digit

02202016, 08:33 AM #20
Thanks for the reply Icewynd.

02202016, 10:34 AM #21Registered User
 Join Date
 Jun 2015
 Posts
 47
Originally Posted by Frank
The number of balls left to come out can be worked out by
49x(42/49)^n), so for n=1 (after one draw), there are 42 balls statistically (and in reality!) left to come out. After two draws (n=2), it is 36, etc.
This is an exponential function, so the value will never be zero. In the past, I've tried working out the number of draws for the number of balls being 0.5 (the lowest number that rounds to 1) as an attempted approximation. This gives a value of 29.7 draws.
For those with Excel, use the calculation =LOG(0.5/49,42/49)
The above calculation can be used to work out the number of draws before there are b balls left to come out
=LOG(b/49,42/49)
If anyone wants to adapt the above formula, the second fraction is the probability of a ball not being drawn for the lottery, so e.g. for the National Lottery 59 balls, 6 balls drawn, it would be =LOG(b/59,53/59).
Ion Saliu mentions something with e and the degree of certainty on his site here:
[url]http://saliu.com/Saliu2.htm[/url]
I haven't worked out yet how to work out the median number of draws for all balls to have appeared at least once for a particular lottery. Homework tonight?

02212016, 03:46 AM #22Registered User
 Join Date
 Feb 2011
 Posts
 867
Hello ice, in the case of number, digit initial and final
Position by position, to cycle in each position example of 49/6
To cycle the 1st position of the 01 limit to 28
2nd place 0235 until the 6th position
So instead of making the whole of the six positions together do for each position
Separately by bias limit.
Ie the 49/6 we .because 6 cycles is each position has its cycle can do?

02212016, 06:18 AM #23
For last night's Lotto 6/49 draw, we hit 2 of our unhit numbers: 10 and 30. That leaves us with 10 numbers still unhit: 1,3,6,11,14,16,17,22,46,49.
For the 12th draw in the cycle the probabilities are:
0 hits =.32
1 hit =.26
2 hits =.34
3 hits=.05
4 hits=.00
5 hits=.02
So our choices would seem to be eliminating all 10 digits from play for the next game or picking 1 or 2 to hit. I'm going with zero hits from this group.
Good Luck!

02212016, 06:20 AM #24Originally Posted by TheConcept

02212016, 02:14 PM #25Originally Posted by TheConcept
Hi The Concept, thanks for putting me out of my misery and saving me some time. Actually its quite logical once you see it, but it would have taken me a while to get as far as the Excel log function to plot it. I have actually succeeded in working out the law of the line based on the natural logarithm e, not from mathematics but from trials in Excel using random lottery draws obtained from Random.org.
For a 7 from 49 lottery, the UK results were too little in number so I used 7000 randomly created lottery results.
Firstly I carried out tests just to measure the cycle completion lengths within the sample and came up with 228 cycles from which the average cycle length was 30.57 draws. The Median was 29, Mode was 25, and the MIN was 14, MAX 63. Your formula evaluates to 29.74 but I notice that you don't have to change the 0.5 in that formula by much to have a large variation in the final value. I discovered that of you use 0.44 instead of 0.5 then you get the same average as I did.
The full Frequency distribution is show in chart 4. All charts are in a zip file ... https://www.mediafire.com/?wk1t5d3dv893byb.. and the plot of the cycle lengths for 749 is in chart 3. You could get a trendline to do much anything depending on whether it was linear, polynomial, moving average or power, the one shown is power which looks just below 30. Despite the chart titles the bonus ball WAS included.
The next trial on the data did what I was discussing with Icewynd, it looked at the remaining balls at each draw within each cycle through all of the cycles and took an average value for balls left for each position in a cycle. The result enabled me to plot a very smooth exponential decay curve (chart 2) and then I took natural logs of each point to determine the slope of the straight line graph that resulted. The slope was 0.1528. I concluded that the law of the line was n= 49 x e^(0.1528 x d) where n = number of balls left in a cycle at d draws onto the cycle. I added the theoretical curve that this equation creates to the original plot but it lay exactly on top blotting it out. I have shown that curve separately though (chart 1). However (42/49)^n is simpler once you know it.
I did a similar excercise for the UK 759 lottery including bonus using again random data to generate 6000 draws. This produced 162 cycles and :
the average cycle length was 36.99.
max= 68
min 21
median= 35.5
mode 33
Your Log formula came up with 37.77 for the calculated average cycle length.
There are plots of the Frequency distribution and the cycle lengths available for those who are iinterested. Charts 5 and 6.

02222016, 08:22 AM #26Registered User
 Join Date
 Jun 2015
 Posts
 47
Originally Posted by Frank
I'll have a look at your charts that you've done. I'm actually surprised at how much the observed results tend to fit with statistical calculations. It leads me to think that there has to be some way of exploiting this.

02252016, 06:11 AM #27
For the Lotto 6/49 draw of 2/24/16 we hit one of our remaining unhit numbers, "14". We now have 9 numbers that have not hit in this cycle: 1,3,6,11,16,17,22,46,49.
The 13th draw of the cycle gives us another excellent chance to eliminate all these numbers from play. For the 13th draw, the probabilities are:
0 hits = 31%
1 hit = 42%
2 hits = 19%
3 hits = 6%.
I'm going with zero hits again and will be eliminating all 9 numbers from my plays.
Good Luck!

02282016, 05:57 AM #28
Lotto 6/49 for 2/27 hit 2 of our unhit numbers: 11 and 22. This leaves us with 7 numbers still to hit in this cycle: 1,3,6,16,16,46,49.
For the 14th draw of the cycle we have another excellent chance to eliminate all these from play:
0 hits = 35%
1 hit = 45%
2 hits = 18%
Good Luck!

03032016, 05:42 AM #29
And it worked! For the March 02, 2016 draw, none of our unhit numbers hit, so I correctly eliminated them all.
For the next draw we still have 7 unhit numbers: 1,3,6,16,17,46,49
On the 15th draw of the cycle the probabilities are:
0 Hits = 44%
1 hit = 35%
2 hits=16%
3 hits = 3%
So I would select at most one of these 7 numbers for play.
Good Luck!

03062016, 06:01 AM #30
No hits again for our group of 7 unhit numbers: 1,3,6,16,17,46,49
For the 16th draw of the cycle, the probabilities are:
0 hits = .45
1 hit = .45
2 hits = .08
3 hits = .02
So, basically 50/50 chance of zero or 1 from this group. I'll go with 1 hit this time.
Good Luck!
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