# Learn roulette visual prediction VB2 in 21 minutes - video

#21

Losing the liner decay of the ball on Huxley wheels when the wheel is spinning at 2.2 to 2.5 seconds per spin is a big deal. Here in the US the tracks are not as smooth as they are elsewhere. On some wheels, depending on the dealer and track the ball will sometimes shake itself from the track prematurely at the DD.

Over the last couple of days I’ve been making loads of video spins on my home wheel (Huxley Mark Six) and have been running them through the avidemux to view the spins in great detail while applying VB2.

#22

Problems with earlier prediction.

If the ball is crossing 50 pockets per second and reference time changes by 150ms that’s 10% different therefore and the error is 5 pockets., which is 10% of 50.

Greater problem is with rotor.

If we time the rotor within 1.5 sec and remaining time until the ball drops is 12 sec. For each additional pocket that rotor makes we need to add 8 to the prediction. It means if we make one pocket error when timing the rotor it is 8 pocket differences to the final result. Errors as that are highly possible when counting time.

When multiplying factor is 9 it is easy to play since area of ¼ of the wheel (9 pockets) can easy be projected to full wheel.

It would be much easier predicting 3 rotations to the end of spin as some videos on net. Remaining time 6 sec, reference time for rotor 2 sec , All you need is multiplying factor of 3 for each additional pocket that the rotor makes.

Rotor timing explained on video is simple once someone understands it. There is no need to see how many pockets the rotor made within reference time, but only to spot the difference it makes from spin to spin than with multiplying factor apply it to the prediction.

#23

I never played with such fast wheel as 2.2-2.5 sec most common wheel speed for me is 4-5 sec. I not change starting point for prediction i simply add pockets that i do because it is more common to start always at the same point. After i get prediction ( with corection for wheel speed ) i look if duering reference time ball done somewhere around 1.5 rot. Then i observe that predicted point and look at which place it is when ball drops (stops ) .

In most cases i achieve that most of the time that point is in between 1/4 or 1/3 of round . If dispersion in finishing position of rotor is greater then is dangerous to play, but that will happen rare. If wheel is fast around 2.5 sec that dispersion must be greater of course.

In the same time i observe where ball mostly stops. I mark all that on wheel map. After some amount of such observations i get some data base on which i can calculate if i can have advantage on that wheel or not and from that data i find the best starting point for prediction. To choose corect starting point is very important and aim is that some ball falling (stopping ) places will have double chanses to be shown by prediction. It means say ball done 6 rot and fall to 0, but if ball will do say 6.75 - outcome still will be 0. So longer travel of ball rotor compensate. I calculate all such possibles double chance areas for ball, and acordingly this calculation i chooose starting point for prediction and places where i must to bet.

#24

In usual, after rotor is clocked there is time to think and to adjust, once you define where to start VB when you get prediction its exact number where to play.

Snow is still boiling his brain to understand why it works?

#25

During the last few days, I have added over 200 spins of new video that are being run through the Avidemux software. I’ve applied the VB2 to them. The problem is still ball chatter distortion. Smooth spins = better prediciton. The traditional VB later in the spin is still by far more accurate because of the distortion.

Just to give people an idea as to how much ball chatter can distort spins, here are some examples of the total ball travel time,. next to total revs. The travel time is from the first point that the ball passes the DD and continues until the ball smacks the same DD at the end of the spin. (The information listed below are measurements showing ball distortion, not VB2 preditions.)

16 revs, total time 14.05 seconds
16 revs, total time 15.61 seconds
16 revs, total time 15.05 seconds
16 revs, total time 14.04 seconds

18 revs, total time 15.58 seconds
18 revs, total time 15.15 seconds
18 revs, total time 16.42 seconds

19 revs, total time 14.98 seconds
19 revs, total time 16.78 seconds

Notice how on one of the spins it took the ball only 14.98 seconds to travel 19 revs, but on another spin it took the ball 15.61 seconds to travel just 16 revs.

A note regarding ball chatter distortion: When the ball is snapped, the ball slides for several spins early in the spin before rolling smoother. Later in the spin there is usually another wave of distortion that creates a rapid drop off in the ball speed (some people call this second wave the knee point). How the dealer snaps or spins the ball in the beginning has a big effect on the ball distortion. This is why I listed some of the examples above.

Later, I will post some ball lap times in order to demonstrate how it can distort the linear predictability and see if I can also upload some of these video spins to Youtube.

-Snowman

#26

19 revs, total time 14.98 seconds
19 revs, total time 16.78 seconds

Can you get 19 rev times for this rotations ?

#27

That’s from the ball chatter distortion. When I snap the ball hard I get a heavy ball slide. If I change the way I release the ball just a little bit I can get the ball to slide less.

#28

Perhaps it may have some influence but the difference is normal phenomena on the tilted wheel.
One spin the ball of 2100ms hits dd and previous rotation was 1900ms

The other spin ball of 2200 may hit dd and previous rotation was 2000

If you add times its 4s and 4.2s.
Only in 2 rotations the difference is 200ms .
It keeps accumulating and it’s the biggest enemy for advantage play since slower ball actually crosses more pockets.

More here
Tilt, slower ball travels longer!

[URL=http://imageshack.us/photo/my-images/51/tilttravelslonger.png/][/URL]

On the chart 2 spins green and blue
One row rotation times the other total time until the ball drops form particular ball rotation

Only 8 rotations to the end of spin you can see marked in red color one sec difference.

First 0.660 ms rotations are same for both spin , maybe they are not same but I couldn’t measure them form video with good enough accuracy.

Also it looks as 0.84 ms ball time at green spin make one extra rotation.
This were 2 extreme spins, blue hits very top of dd, green just passes it makes additional rotation, slides on top od H diamond in front of DD but on the end still hits it.

New program the Acrobat will be useful here. ;D

#29
Just to give people an idea as to how much ball chatter can distort spins, here are some examples of the total ball travel time,. next to total revs. The travel time is from the first point that the ball passes the DD and continues until the ball smacks the same DD at the end of the spin. (The information listed below are measurements showing ball distortion, not VB2 preditions.)

16 revs, total time 14.05 seconds
16 revs, total time 15.61 seconds
16 revs, total time 15.05 seconds
16 revs, total time 14.04 seconds

18 revs, total time 15.58 seconds
18 revs, total time 15.15 seconds
18 revs, total time 16.42 seconds

19 revs, total time 14.98 seconds
19 revs, total time 16.78 seconds

Notice how on one of the spins it took the ball only 14.98 seconds to travel 19 revs, but on another spin it took the ball 15.61 seconds to travel just 16 revs.

A note regarding ball chatter distortion: When the ball is snapped, the ball slides for several spins early in the spin before rolling smoother. Later in the spin there is usually another wave of distortion that creates a rapid drop off in the ball speed (some people call this second wave the knee point). How the dealer snaps or spins the ball in the beginning has a big effect on the ball distortion. This is why I listed some of the examples above.

Cool - i will run some sampels my self regarding distortion - nice input.

Later, I will post some ball lap times in order to demonstrate how it can distort the linear predictability and see if I can also upload some of these video spins to Youtube.

That would be great - Cheers.

#30

In short, you must also pay close attention to how cleanly the ball is rotating on the track.

#31

So the question is - can we looking to few ball rounds, when ball speed is around 0.8-1.2 predict with good accuracy the distance which is left for ball to travel and remaining time. I think that we can do something at least that this prediction will be better than random. I played roulette in many countries in many casinos but i never see wheel with greater than 35% tilt, normally tilt which i noticed observing about 50-100 spins is around 30% and if it is greater then hits to the same diamond are very very diferents, that is the same like ball with hit to two diamonds. Why i think that it is posible to predict better than random ball position at the end and left time from some specifick ball speed - that is because each event have his own reason. If ball travel longer than normally but done less distance for that is some reason and i think that from observing ball speed in some points, observing how changes that speed, we can see something, what will happend in future.

#32

Hi Forester.
The video doesn’t work after closing MEGAUPLOAD/MEGAVIDEO.

#33

[quote=“rremoroulette, post:32, topic:761”]Hi Forester.
The video doesn’t work after closing MEGAUPLOAD/MEGAVIDEO.

#34

You make a good point and raise one interesting question.

So the question is - can we looking to few ball rounds, when ball speed is around 0.8-1.2 predict with good accuracy the distance which is left for ball to travel and remaining time.

As i state before so does not the force of the ball during the spindevelopment have any memory, so when or if we use a static reference point to estimate deceleration there will be a small error or a natural physic penhomen witch occur - that is due to that the ball will divide (naturaly) at 0.8 or 1.2 on different places on the ball track, spots, so some ms plus or minus will allways be the case using one static reference point.
But this does not mean it does not exist acc ways that give the exact same timeframe to estimate - witch give a error with one early or late but same seconds to end or hovering around plus/minus 30 ms.

#35

From; Roulette, dealer signature myth or not.

This is one of posts from closed forum, with some editing I will try to explain how well prediction based on dealer’s signature can be.

Red line at bottom of the picture represents remaining time until ball hits rotor.

Coloured triangle with blue green and yellow areas represents quality of wheel that we may play. With better quality wheel with of coloured area will reduce and we will have better accuracy.

First vertical red/black bar 1 located at 6 sec to the end, has 4 numbers in blue area which means that ball with same speed will 50% of time start failing within 4 numbers of accuracy.

20 % of time it may be 7 numbers,
20 % of time it may be 10 numbers
And 10 % it may be anything.

Simple meaning of this is that from 100 spins where the ball is 100% same 50% will hit at same spot within 4 numbers of tolerances, but remaining will be around or even completely out of order. Reasons for that are deformations on the wheel, slight tilt, the ball is never 100% rounded dints on the ball or ball track etc.

If we look bar 2, at 10 seconds to the end we can see that errors do increase. For example 50% of hits will be within 8 numbers. This is only simple representation; of course that increasing does not have to be linear with time but definitely is proportional.

Once when we get prediction there is no way that we can predict how ball will be affected by all deformations on the way to the end of spin and it is regardless what was the ACC at that moment.

It’s same as shooting arrow all the time with same force, on rainy and windy day.
Even the arrow starts with same force and same angle, first drop of rain effects it. Based on how firs rain drop effects it, next drop may have completely different effect. On top of that slight changes in wind, different at any single point of arrow traveling time would affect final result even more…
It will result that some arrows will finish 100 meters from us + or - 5 meters, but some +or -10 or even more. However if we observe same arrows with same speed where the target is only 50m distanced, differences in between final results will be smaller.

Looking above picture try to imagine how much error will increase with 15 sec. to the end. That would be one of shorter spins for dealers signature play. So even if dealer spins the ball exactly same every time we would not have much chance. On top of that consider back spinning of ball at first few revolutions which again is not constant. Finally consider change in rotor speed. If rotor changes from 3 sec per revolution to 2.8s it is not significant change for naked eye but it makes 1 number per second difference therefore result after 15 sec will be 15 numbers different. I think that this explains why even if dealer spins the ball every time 100% with same speed we cannot expect that ball will end up at same place.

If we play tilted wheel that may help but that is only helping in previous example if dealer spins the ball with force to make 15 or 15.5 revolutions. On tilted wheel result will not be 0.5 rotations different but it will be close to same or it will shift by how much the wheel changes until ball makes extra ball rotation.

It is because tilted will makes various ball speeds to fail at same point. Sounds good but there is additional problem, the ball that pass dominant diamond even if it has similar speed may continue for additional full rotation and in most cases end up on opposite side. So it does not help much. On tilted wheel because of all described problems that ball may face on the way to final destination even if dealer spins it 100% same, some balls will make 15 rotations some 14 some 16. On top of that do not forget that we still did not add ball bouncing which buy itself in most cases is 50% random.

Let’s say that someone is using an electronics timer to help him to define when the ball is at particular speed range. If he does it about 6 rotations to the end he still is facing problems as described with picture above.

Putting that aside let’s consider wheel as perfection and look players chances to gain an advantage.

On tilted wheel to find only right ball revolution may not be good enough and way to detect ball speed within particular time definitely will not find it every time.

Even if player finds it within his range, who says that all ball speeds will hit dominant diamond, for that chance is probably again 50 %.
On top of that, ball speeds that will hit dominant diamond will not travel with same speed, so let’s say if there are 6 revolutions to the end, time for that 6 revolutions will change and of course rotor will not make same distance in traveling.
It produces error of about 10-12 numbers. (on very slow wheel it may be less and part of that will be compensated buy scatter)

So player who is trying to do it first has chance of 50 % to get right revolution, if he does not he is about half wheel wrong. It would be ok if that is half wheel but he has 10-12 numbers shifting on each half. Then again within time window that he defined for targeting particular ball revolution not all ball speeds will hit same diamond. Add slight error if wheel changes speed, add randomness of ball bouncing and he really does not have much chance to have advantage. In some occasions it may be but it is very minimal. His chance may increase with very slow rotor, very strong tilt and of course if he starts making accurate adjustments for change of rotor speed. Skilled visual player gets only 10-20% advantage.

Knowing all this if we go back to 15 sec at start of spin if we consider that dealer will never spin with 100% same strength, if rotor is never 100% same, if ball while spinning can be differently effected, if ball bouncing is partly random, if even defining close ball speed at only 6 revolutions to the end, hardly can produce any advantage, we definitely can say that dealer signature will not work on today’s casino wheels.

Some people in casino trying to play it even ignoring that few spins with ball lending same distance from starting number were actually when ball was spinning in different direction.

I place my early bets mostly based on theory of dealer’s signature. Sometimes it can be very profitable, but it is only trend. In most cases I see that ball and wheel are going with different speed and I still get same result. It is luck. Trend comes and goes, we never know when it will come or when it will go, and therefore we really cannot take advantage of it.

#36

Reading some posts and also in conversations with other players i noticed that many not understand situations when player can have edge against casino when to have edge is imposible and how to count absolute edge ( as absolute edge i mean minimal edge which we can have if we all do corect without mistakes ) For simplycity assume that wheel have only 12 numbers and wheel and ball have only 12 fixed positions at the end.

So lets say ball can stand in 12 diferent positions and wheel can be in that moment also in 12 diferent positions. so this way we can have 12*12 = 144 diferent wariations. Despite theese 144 possible variations some of them are the same - so wheel position at 0 and ball at 3 is equal to wheel position at 11 and ball at 2 and equall at wheel position at 10 and ball at 1 and so on. So really we have only 12 diferent outcomes only that every outcome can happend in 12 ways. So when we talk about possible edge for player we must look only to diferent outcomes. Lets say wheel speed is 0 so we always know exact wheel position at the end, unfortunatelly it can not produce for us any edge if ball stoping position will be totally random ( how can be if wheel is absolute leveled ).

So in that case only way to get edge to predict something about ball final position. Now lets say we play VB2. VB2 really shows us finall wheel position and acuracy of this showing is in averidge 0.3-0.6 round . From where i take that ? Lets say we predict about 12 sec til endand we think that in that moment till ball drop is left 12 sec, but we cant be sure for that and really remaining time can varry between 13 to 11 sec.

Lets say wheel speed is 4 sec per rev, so 13/4=3.25 but 11/4=2.75. So 3.25-2.75=0.5 of round. Because if we predict that remaining time is 12 sec then 13 an 11 sec will be very rare cases maybe 1 from 30-50 spins and about 80% of spins will be inside of 0.25 of round. Again for simplyficity lets say that all spins will be inside that 0.25 of round distance so if we have 12 numbers then it will be only 3 possibles positions for wheel. So what we need in this case for getting edge ? We need that ball will stop in such amont of positions that if we will add that number to possibles number of wheel posibles positions and result will be less than 12.

In other words there must be some outcomes which not be able to happend. If w have tilted wheel and ball always will hit to the same diamond ( of course that is miracle situation but lets say it is ) and scater varry from +0.25 to +0.5 of round, then we will have only 3 possibles ball positions at the end ( if we have total only 12 possibles positions for ball ) So this way we have 3 positions for wheel and 3 positions what is totall 6 posibles outcomes. So this way we can cover theese 6 numbers from 12 and have hit in all spins we play. How big edge we will have in that case ? If payout is 11-1 we cover 6 to get 12 after every spin so our edge is 100% We risk 6 units to have for sure 12 units after spin.

As i said that is miracle unrealistick situation because we never have such conditions, but we can have such conditions with some expectation. Lets say for wheel to take that position at the end we have expectation 80% and for ball be in thic positions at the end expectation is 70% . Then 80%70% = 56% So totally we will be right in 56% and 56%100% = 56% and when that happend we will win the same amont what we will placed . So 656=336. In other cases we will lost theese 6 chips so 100-56=44 and 644=264 and 336-264=72 so in 100 spins placing 6 units we will be 72 units in plus and 72/600=12% edge. Usually edge can be biger because whern we missed our prediction we still can in some spins win simply by luck, but lets say that we cant. So how we can use that all mathematick ? There are one very simple way in which we can use that.

Lets say we observe 10 spins ( all numbers i give small because of simplificy really we must abserve much more spins ) and have wheel positions at the end as 1112222337 and ball positions 7889991010110 we can write all theese numbers to table there say in vertical will be writen wheel positions and horizontal will be writen ball positions. On intersections of columns and rows we will write reall outcomes. Then we have as 100 possibles outcomes but some of them will be more often others will be less often.

From that sample we will get such outcomes and its frequency :
0 1
1 3
2 7
3 12
4 19
5 22
6 17
7 9
8 3
9 2
10 3
11 2

from that we see that if we place bets on outcomes 3-4-5-6 then we will win in 70% cases or in other words if we cover 3-6 in theese 10 spins we will win in averidge 7 hits from 10 so 711=77 and 34=12 so 77-12=65 . So if we will place 4 chips in every of theese 10 spins after them wee will have on 65 chips more 65/40=162.5% edge.

Again i repeat that is speciall case you will never have such big edge in reall play and samples spins which we observe must be much more than 10 to do some conclusions, I only wanted to show one possible way based on previous data to find place where you must bet to have edge. And if you repeat all that with more spins few times and always will get similar betting position with more or less similar edge than you can do assumption that this wheel is good for play.

#37

Nicely explained.

Lets say we predict about 12 sec til endand we think that in that moment till ball drop is left 12 sec, but we cant be sure for that and really remaining time can varry between 13 to 11 sec. Lets say wheel speed is 4 sec per rev, so 13/4=3.25 but 11/4=2.75. So 3.25-2.75=0.5 of round. Because if we predict that remaining time is 12 sec then 13 an 11 sec will be very rare cases maybe 1 from 30-50 spins and about 80% of spins will be inside of 0.25 of round.

VB2 may be hard to understand since it’s created by brilliant mind ;D so let me explain something.

What you were explaining it relates to traditional VB. Perhaps someone is trying to guess X amount of rotations to the end of spin with remaining time 11-13 sec . You are trying to say most likely he will be at 12 sec to the end of spin compared to the 11 or 13, but for the VB2 it is totally irrelevant. The player even doesn’t have to know is he most of the time 11 or 12 or 13 seconds to the end or if that changes during the play. Why would he, since he is always getting same reference number as if he is all the time in same moment during the spin.

If as in your example the rotor is 4 sec/rotation when he plays he will read same number in 12 or one rotation earlier or later than that. If he uses 12 sec as a remaining time when it is most of the time 11 all his mistake is slightly different multiplying factor for rotor speed adjustment. If he observes rotor with 2 sec reference time than for 12 sec the multiplying factor will be 12/2=6 and for 11 it would be 5.5.

How much difference does it make?

4 sec rotor spins 37/4=9.2 pockets per second.
If the rotor changes to 5 sec it would be 37/5=7.4 the difference is 9.2 -7.4=1.8
Since the player observes the difference during 2 sec time, it would be spotted as 2 x 1.8 = 3.6 pockets.
If he uses multiplying factor 6 it’s 3.6 x 6 =21.6
Or for 5.5 it would be 3.6 =5.5 =19.8
The difference is 21.6 - 19.8 = 1.8 pockets which is a reasonably small error.

Someone who is following this tried to understand VB and to implement it may be puzzled. If so, than why I do not get so accurate predictions in casino when observing ball drop point?

There is many factors that can effect prediction on the wheel itself but I will explain where the VB2 has the weakest points.

Of course change and instability in used reference time is one factor but it more relates to the users skill.

VB2 can handle well if prediction is in wrong ball rotation or if rotor speed changes but not both of it.
The truth is during the game bot factors change. If the players prediction time deviates more he needs more stable rotor and vice versa.

The accuracy of prediction is directly related to change of both factors.
It’s as errors are X times Y. If for example max tolerated error is 16 player can have it if x=4 and y =4, if X rises to 8 then Y needs to go down to 2 or less for errors to be under 16. If X or Y are zero then and the error is zero.

For example the rotor is 4 sec we predict 13,12,11 sec to the end of spin or even with greater differences Vb2 predicts perfectly. Or we predict 12 sec to the end of spin and the rotor changes we still predict perfectly. But if rotor speed and the time of prediction changes the player start getting errors.

I’ll not go in to detail since it has been explained somewhere in Vb2 section.
But if the player deviates with prediction 11, 12, 13 that’s 3 seconds wide and if rotor changes in between 4 and 5.1 sec (change of 2 pockets per sec). The players prediction may deviate in between 3x2=6 pockets (change in time of prediction times rotor speed change).

Even prediction so early in time the player should have accuracy of let’s say 4 sec, and reasonable rotor may change up to 3 pockets per sec, then the players errors created by VB2 will float in between 12 pockets. If traditional VB player trying to estimate right ball revolution earlier in spin and deviates so much (4 sec) he will have up to 44 pockets errors with such rotor speeds. (No it’s not that he is only 7 pockets wrong since 44-37 is 7, since it can be anything in between 0 and 44. Therefore 44 means he will be floating withing 44 pockets with his predictions)

Now is coming part Bebediktus also explained, most of the time the rotor may be x speed and the player may predict more times around targeted time so even the errors are 12 pockets most of the time it may be less then 6. Higher errors are coming with faster rotor and more different rotor speeds. For example the dealer spins rotor equally in between 3 and 6 sec .

Key to success
Regardless the fact VB2 predicts in wider range of rotations, learn to predict in the closes possible moment during the spin. That will be reducing your errors and let you play wider rotor speeds. Also avoid significant rotor speed changes.

For example try to identify moment in spin when the ball has 11 rev to go then count them. If most of the time you can get it right with accuracy of + and - 2 rotations rotations you are ok to play. It’s 5 rotations range but since at that time during the spin the ball is faster it is most likely less than 4 sec difference in time.

#38

Here’s a question for you. Based on the following information, how would you apply VB2?

Wheel speed is 2.5 seconds. The ball revolution times at each spin are as follows:

0.43
0.43
0.47
0.50
0.50
0.57
0.57
0.60
0.63
0.70
0.77
0.76
1.08
1.27
1.70
1.94
2.10

(The above split times are very common to Huxley Mark Six roulette wheels spread all over Las Vegas, Nevada, and Atlantic City, New Jersey. I’ve left off the first four revs of the ball. )

In the above example, traditional VB seems like a much better option.
Thanks,

Snowman

#39

I assume 0.76 should be 0.86.

You are correct if you apply traditional VB you are better off.
If you spot third last ball rotation you have only 2 rotations = 4 seconds, to place bets if dealer lets you.
Not sure how good so late prediction can be?

Third last rotation because its 1.70s compared to previous one 1.27
there is a huge ball slow down of 1.7-1.27=0.43 s/rotaton

Perhaps 0.86 to 1.08 can be identified then you will have reasonable remaining time ~8 sec.

Predicting that rotation is harder and many times you’ll be one rotation wrong. On such rotor it is 15 pockets .Sometimes it’s not a problem if scatter overlaps. For example some of spins you are 15 pockets wrong the ball doesn’t jump so you win. The others you predict the ball jumps mostly ~15 pockets.
If you applying VB2 with 2.5 sec rotor you probably need 4 sec reference time and for that you may need a timer as FFZ/V has.
T=1000/(2.5 x 100) / ACC is about 100
=1000/250 = 4
If your VB time ending around 1.08 rotation the time may be a bit shorter because ACC around 1.08 is greater.

We could also calculate time this way
t=700/(2.5 x 70)
ACC 70 is taken where the difference in between rotations was ~70 ms and it was when the ball was 700 ms.
So you get
t=700/175=4

#40

I calculated all very precize and Snowman say me that he want to predict in moment when ball is about 0.6-0.7 in speed. the best reference time what i find is 2.06 sec. So this wayif we start in the begining of that round which is 0.6 sec in 2.06 sec ball will do 3.18 of rounds we read number . When we start in the begining of 0.63 round then ball will do 2.96 in theese 2.06 sec but starting point is shifted by 2.5/0.6=0.24. 2.96+0.24=3.2 3.2-3.18=0.02 diference of round that is less than one pocket. If we start in the begining of 0.7 round then ball in 2.06 sec will do 2.69 rounds but starting point is shifted by 0.24 + 2.5/2.63=0.25 and totally 0.49. 2.69+0.49=3.18 exactly the same number like we started in 0.6 round. So we can say that when we started in theese three rounds no matter in which after 2.06 sec we will se exact the same number.