# VB and Expected Value

Disclaimer : I have some basis in maths and probabilities but I am not an expert so if you find errors, please tell me.

I woke up this morning and I was wondering “How often do I have to make a correct prediction for VB to have a positive expected value ?

Then I realized the answer was not so obvious and that it would depend on different parameters.
To calculate an expected value, you need to know all the possible events, their probability, and their value (how much they will pay or cost you).
So let’s start from the beginning.

Here are all the different cases I can see:

1- I miss the good revolution and I know it -> I don’t bet
2- I think I have the good revolution but I am wrong -> random outcome
3- I think I have the good revolution and I am right :
3a- the ball doesn’t hit the dominant diamond -> random outcome
3b- the ball hits the dominant diamond :
3b1- the scatter is ok and I hit the zone I bet (Hourra!) -> I win
3b2- the scatter is not ok and I don’t hit the zone I bet -> I lose

let’s determine for each their value (let’s assume I play a n numbers zone, each number with 1 chip) :
case 1:
I just don’t play this one, so we will just ignore this case (unless you are so undisciplined or game addicted that you can’t help playing it anyway … shame on you). So its value is 0.

case 2:
here I get a random outcome
I have n chances to get 36 chips and 37-n chances to lose n chips.
the value for this event is :
Vrandom = (n/37)(36-n) + [(37-n)/37]-n
Vrandom = [36n -nn - 37n + nn]/37
Vrandom = -n/37

case 3a:
Here I get a random outcome too, so the value of this case is Vrandom = -n/37

case 3b1:
Here I hit my betting zone, the value of this event is the number of chips I will win, it is Vhit = 36-n

case 3b2:
Here I miss my betting zone, the value of this event is equal to the loss of my chips, it is Vmiss = -n

ok for the values of each events, let’s now considerate their probability, but before doing that we will introduce some variables :
‘r’ will be the probability that I can identify the good revolution
‘t’ will be the probability that the ball hits the dominant diamond(DD), it is the tilt percentage
‘a’ will be the probability that the ball hits the betting zone (when I identified the good revolution and that the ball hits the DD)
Ok, so, now we can continue.

case 1: its value is 0 so what ever its probability will be, it will have no incidence. let’s keep ignoring it

case 2: this case happends when I miss the good revolution, so it’s probability is (1-r)

case 3a: this case happends when I have the good revolution but the ball misses the DD, so it’s probability is r.(1-t)

case 3b1: this case happends when I have the good revolution, that the ball hit the DD but the ball misses the betting zone, so it’s probability is r.t.(1-a)

case 3b2: this case happends when I have the good revolution, that the ball hit the DD and the ball hits the betting zone, so it’s probability is r.t.a

So now let’s write the expected value of a spin with all these parameters :
EV = (1-r).Vrandom + r.(1-t).Vrandom + r.t.(1-a).Vmiss + r.t.a.Vhit
EV = vRandom.(1-r+r-r.t) + r.t[(1-a).Vmiss + a.Vhit]
EV = vRandom.(1-r.t) + r.t[(1-a).Vmiss + a.Vhit]
EV = vRandom -r.t.vRandom + r.t[(1-a).Vmiss + a.Vhit]
EV = vRandom + r.t[(1-a).Vmiss + a.Vhit - Vrandom]

let’s replace the values:
EV = -n/37 + r.t[(1-a).-n + a.(36-n) + n/37 ]
EV = -n/37 + r.t[-n + a.n + 36a -a.n + n/37 ]
EV = -n/37 + r.t[36a - n + n/37]
EV = -n/37 + r.t[(1332a - 37n + n)/37]
EV = [r.t(1332a - 36n) - n] / 37

And you will tell me “and now what ?”.
Yes true, we now have a nice equation which give us the expected value depending on 4 variables. Let’s try to give values to these variables.

n : that will be easy, let’s say we bet a 5 numbers zone, so n=5
r : this is a bit harder, it is the percentage of times we can predict the right revolution. I did a bit more than 50% at my first try on a wheel I didn’t know, so let’s assume that somebody trained on a wheel he knows can achieve a 75% of success (if I am wrong here, please somebody tell me and give me a more realistic value). So r=0.75
t: that is the tilt of the wheel, let’s take a wheel where the ball hits the diamond half of the time. so t=0.5

and now we just have one parameter we don’t know, it is ‘a’ and that is fine because it is the answer of the question I was wondering this morning.
Let’s calculate the ‘a’ minimum we need to have a positive expected value.

EV > 0
[r.t(1332a - 36n) - n] / 37 > 0
r.t(1332a - 36n) - n > 0
r.t(1332a - 36n) > n
1332a - 36n > n/(r.t)
1332a > n/(r.t) + 36n
a > (n/(r.t) + 36n)/1332
a > (n/1332).[1/(r.t) +36]

so let’s replace with our numeric values:
a > (5/1332).[1/(0.7*0.5) +36]
a > 0.145

so, in these conditions(values of n, r and t ) it is enough to hit the winning zone more than 14.5% of the times that the ball hit the DD and that we had the good revolution.

Let’s now do some other numeric applications:

on normal conditions (without VB, just luck), just with betting 5 numbers you need to hit more than :
x*(36-5)>5100
x>5
100/31
x>16.13%

now with VB :
slightly tilted wheel, average revolution detection, 5 chips bet
t=0.3, r=0.6, n=5
the minimum needed accuracy is 15,6%

slightly tilted wheel, good revolution detection, 5 chips bet
t=0.3, r=0.8, n=5
the minimum needed accuracy is 15.1%

average tilted wheel, average revolution detection, 5 chips bet
t=0.5, r=0.6, n=5
the minimum needed accuracy is 14.8%

average tilted wheel, good revolution detection, 5 chips bet
t=0.5, r=0.8, n=5
the minimum needed accuracy is 14.5%

heavily tilted wheel, average revolution detection, 5 chips bet
t=0.8, r=0.6, n=5
the minimum needed accuracy is 14.3%

heavily tilted wheel, good revolution detection, 5 chips bet
t=0.8, r=0.8, n=5
the minimum needed accuracy is 14.1%

So now, what can we say about all this ?
First in the formulas for EV and minimum needed accuracy, we can see that the tilt of the wheel and the spotting of the good revolution are as important as each other. A 10% less in spotting the ball revolution can be compensated by a 10% plus in the tilt of the wheel.
So both should be considered with the same interest.

Second, we can see that the difference in needed accuracy on a slightly tilted wheel when spotting the good revolution on an average basis or on a heavily tilted wheel when spotting the good revolution on an very good basis is 1.5%
1.5% doesn’t seem big but here we talk about percentages around 14/15% so it is actually a difference of around 10%
So it means that a good wheel and good revolution detection can give you up to 10% more winnings.

There are probably other conclusions to get from this, but now I am tired, so I let you guys tell me if you see something else Good post,

We can say that everything is measurable, and can be calculated.
But sometimes it is so complicated.

Let’s say if the ball hits DD within our desired amount of rotations to the end.
It still doesn’t have to arrive on every spin in same time to DD, it can deviate.
How much it depends on how early in time we have prediction, then that change in time is directly related to rotor speed which together creates errors in pockets.

If we missed rotation it doesn’t have to mean that we will get negative result from that spin. One reason is if the scatter creates 2 pick points. So if we deviate in between 2 rotations one can predict early and we like ball to jump 11 pockets if we have prediction in next rotation then we would like that the ball stops instantly.

Also if we miss DD.

For example if we targeting 5 rev to DD hit.
We still may get reasonable results if the ball makes 4.25 or 5.75 revolutions.
With slower rotor results will be close to predicted as if the ball made 5 revs.
Than again how will scatter develop from there (4.25 or 5.75) it depends if common drop zone is created by damage on ball track or by real tilt on the table.

Everything depends on method we apply on tilted wheel.
Myrulet VB2 is very specific. It actually slightly increasing prediction if the ball is faster even within same ball revolution. In some occasions it is good in some it is bad and again it depends on rotor speed and wheel / ball construction. Under some particular conditions and strong tilt no matter where the ball was dropping it may producing advantage in total higher then what we could expect, because some parameters overlap and 2 errors may cancel each other.

[quote=“Forester, post:2, topic:416”]Let’s say if the ball hits DD within our desired amount of rotations to the end.
It still doesn’t have to arrive on every spin in same time to DD, it can deviate.
How much it depends on how early in time we have prediction, then that change in time is directly related to rotor speed which together creates errors in pockets.[/quote]

Do you mean that when spotting the good revolution, and the ball has 5 left revolutions to do, the ball will not always do these 5 left revolutions in the same time ?
Do you have any idea of how much that time can vary ? I would say a few 1/10s of seconds, it should not be more than 3 or 4 pokets difference between the quickest and the slowest. I am not sure it is very significant compared to the scatter.

Ho I didn’t know that, I though an analyse of the scatter would give a Gauss shaped curve. Do you have any image or datas showing that ?

[quote=“Forester, post:2, topic:416”]For example if we targeting 5 rev to DD hit.
We still may get reasonable results if the ball makes 4.25 or 5.75 revolutions.
With slower rotor results will be close to predicted as if the ball made 5 revs.
Than again how will scatter develop from there (4.25 or 5.75) it depends if common drop zone is created by damage on ball track or by real tilt on the table.[/quote]

I put a random value to this event, but yes , if we actually have 2 picks as you say, it is true that we maybe still have like 50% chances to have the same expectation as when hitting the diamond, and 50% chances to get a random result. But I am not sure it will change a lot the results. I could try to had this in the equation if I have some time [quote=“Forester, post:2, topic:416”]Everything depends on method we apply on tilted wheel.
Myrulet VB2 is very specific. It actually slightly increasing prediction if the ball is faster even within same ball revolution. In some occasions it is good in some it is bad and again it depends on rotor speed and wheel / ball construction. Under some particular conditions and strong tilt no matter where the ball was dropping it may producing advantage in total higher then what we could expect, because some parameters overlap and 2 errors may cancel each other.[/quote]

I was wondering, (but this has nothing to do with my post :p) how do you use devices when you are at the table, for nobody to notice ?
I assume that when you clock the ball it is just by clicking, but you need to somehow see the number that the device gives you for betting Do you mean that when spotting the good revolution, and the ball has 5 left revolutions to do, the ball will not always do these 5 left revolutions in the same time ? Do you have any idea of how much that time can vary ?
It can be even one sec.
Ho I didn't know that, I though an analyse of the scatter would give a Gauss shaped curve. Do you have any image or datas showing that ?
[url=http://rouletteplace.com/index.php/topic,706.0.html]http://rouletteplace.com/index.php/topic,706.0.html[/url] sometimes is more obvious
put a random value to this event, but yes , if we actually have 2 picks as you say, it is true that we maybe still have like 50% chances to have the same expectation as when hitting the diamond, and 50% chances to get a random result. But I am not sure it will change a lot the results. I could try to had this in the equation if I have some time

Yes it is about 50% chance.
5=0K
5,75 is close
4.25 is close

I was wondering, (but this has nothing to do with my post :p) how do you use devices when you are at the table, for nobody to notice ? I assume that when you clock the ball it is just by clicking, but you need to somehow see the number that the device gives you for betting.

VB2 doesn’t have to use any device, it is just another method of visual prediction.
FFA talks, FFZ zaps players skin

the graph of the scatter is very interesting
Maybe when targetting a zone, it is better to bet 2 separate zones rather than a big one

Hi bentaye.

Via a via is an intressting effect.

The bi-modal behavior can come in two, three and even four peaks.

The high-probability areas does not always have - for an exapmpel - an 100% oppisite effect regarding via a via.
It can also come with the shape of an boomerang effect.

Here is an simpel chart. Take care LS

Hi Lucky

can you explain  via …via

i dont thinkso

i dont think so

First you take an good look at my avatar then have sweet dreams Cheers LS

I don’t get what the chart is showing