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>5100/31
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