Cycles, Revisited

[TheConcept]

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.

Hi Frank. If I use the example of a 49 ball lottery with the bonus ball, the probability of a number not being drawn is 42/49. The probability of a number not being drawn after n draws is (42/49)^n. (^n = to the power of n)

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:
http://saliu.com/Saliu2.htm

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?

 

[jack]

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 02-35 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?

 

[Icewynd]

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!

 

[Icewynd]

Hi Frank. If I use the example of a 49 ball lottery with the bonus ball, the probability of a number not being drawn is 42/49. The probability of a number not being drawn after n draws is (42/49)^n. (^n = to the power of n)

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.

Thanks, TheConcept! This accords well with the empirical data.

 

[Frank]

Hi Frank. If I use the example of a 49 ball lottery with the bonus ball, the probability of a number not being drawn is 42/49. The probability of a number not being drawn after n draws is (42/49)^n. (^n = to the power of n)

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:
http://saliu.com/Saliu2.htm

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?

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.

 

[TheConcept]

Hi The Concept, thanks for putting me out of my misery and saving me some time.

You're welcome.

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.

 

[Icewynd]

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!

 

[Icewynd]

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!

 

[Icewynd]

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!

 

[Icewynd]

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!

 

[Icewynd]

2 hits from our unhit group this draw: 3 and 16. That leaves us with 5 unhit numbers: 1,6,17,46,49

For the 17th draw of the cycle the probabilities are:
0 hits =52%
1 hit =34%
2 hits= 5%
3 hits= 5%

So zero or 1 are the most likely scenarios.

Good Luck!

 

[Icewynd]

BTW, you might be interested to know that 39 is the hottest number during this cycle with 7 hits and 34 is second with 6 hits.

34 hit in the first draw of the cycle, while 39 hit in the second draw. They both got their second hit within 3 draws. This is a good way to identify upcoming hot numbers.

Good Luck!

 

[Icewynd]

Just an update on the cycle. There are still 2 stragglers to hit:46 and 49.

At this stage of the cycle it is the norm to get no hits, so it is likely to take a few draws to hit these. There are 10 more draws until the cycle gets to average length so likely sometime in that span.

Good Luck!

 

[Bestbrain]

New cycle for Canada Lotto Max

Hi Icewynd, Is there any new cycle for Lotto Max? If yes, what date was it started?

 

[Icewynd]

Hi Bestbrain,

I have never tracked the cycles for LottoMax, so I don't know. What I would suggest is that you just pick an arbitrary date and start your tracking. You might want to pick a draw 6 or 7 draws back and see what has hit and what hasn't since that date. The formulas listed on this thread should help you calculate how many numbers to expect from each group.

Good Luck!

 

[Icewynd]

I've sort of let this thread slide because there is not much action at this stage of the cycle. However, those of you who are paying attention would have noticed that 46 hit at the end of March (and repeated in the next draw!). That leaves us with 1 unhit number: 49.

Good Luck!

 

[blitzed]

hiya Icewynd, cool...hopefully it can be exploited!

goodluck!
blitzed

I've sort of let this thread slide because there is not much action at this stage of the cycle. However, those of you who are paying attention would have noticed that 46 hit at the end of March (and repeated in the next draw!). That leaves us with 1 unhit number: 49.

Good Luck!

 

[Icewynd]

So the 49 finally hit on April 9th, which completes the cycle. The most recent draw was the first in the new cycle with 7 unhit numbers being drawn (since all 49 numbers went back into the unhit pool).

So, on the next draw we can have one of the 7 repeat (or, rarely 2 or more) or all 7 from the unhit pool of 42 numbers.

And that's how it goes...

Good Luck!

 

[Bestbrain]

What are the 42 unhits numbers?

 

[Icewynd]

Every cycle starts with 7 numbers of the 49 hitting (6 + B). These would be the numbers that hit on April 13th. So, the "unhit" numbers would be the 42 numbers that did not hit on April 13th.

Good Luck!