Can you explain me 2 different approaches of standard deviation in roulette?

Hello,
I hope this isn't stuppid question but I need a little more explanation,

When we calculate SD of bets (ie. SD of straight-up bet), this is obviously not the same thing as when we calculate SD of weel frequency related to straight-up outcomes.

For example if one would calculate SD of straight-up then result will be 5,837837....
In the same way if one would calculate SD of straight-up bet according to expected weel frequency the result will be 2,655637... which is not the same as above.

I know how to calculate both of them, all I need is a little better matematical explanation.

Thank you.

The standard deviation (sigma) is the square root of the variance .

If you have a random variable noted X, then

D^2(X)= M[(X - M(X))^2]= M(X^2) - [M(X)]^2 -> this is the formula for the variance.
The standard deviation is the square root.

M(X) stands for the average.

edward,
tank you very much,
As I said, I know how to calculate SD and variance and what sigma is, that was not my question.

My question was difference between two approaches of SD in roulette.
1 Standard deviation of bet and
2 Standard deviation of well frequency for some particaular bet.

Kind regards.

Quote: codekiddy

My question was difference between two approaches of SD in roulette.
1 Standard deviation of bet and
2 Standard deviation of well frequency for some particaular bet.

Kind regards.

This is so funny. In school last year, I could not get over the difference between the 2 also.
My professor explained it this way. Now I understand the difference.

Let's use 18/37 as the example for one spin
#2 method is the BSD (binomial standard deviation) N*P*Q is the variance. easy stuff. Now take the square root for the SD.
sd = 0.499817352
#1 method returns 0.999634703
(For a wager that pays x:1 with a probability of winning p, the standard deviation (per unit bet and per 'root decision') are:
std = (x+1)*Sqrt[p*(1-p)])

You should see the factor between the two values is 2.
Because BSD treats each outcome (Success and Failure) as 1 or 0
But for the bet, we need to be able to treat S and F as 1 and -1
So, for the sd of the bet we must double the BSD value when p = 18/37

when p = 1/37 the factor is now 36. Do you see why? (35+1)
when p = 12/37 the factor should be 3. Do you agree?
Now you have some homework to complete

In summary, one can use the BSD formula for the sd of the bet, but you need to know the *factor* to multiply by.
It is easy once you do it a few times. ;)
Sally

Hehe :D
Now that was very well explained and easy to follow!

I see now, Binomial SD doesn't care about payout so the result is low, per one spin this is SD =(1PQ) ^ 0,5
While, when calculating for the bet, result per one spin is: SD = Payout * (1PQ) ^ 0,5

Thank you very much Sally for this valuable explanation.
Much appreciated!


Have a nice day :)

Quote: codekiddy

I see now, Binomial SD doesn't care about payout so the result is low, per one spin this is SD =(1PQ) ^ 0,5
While, when calculating for the bet, result per one spin is: SD = Payout * (1PQ) ^ 0,5
Thank you very much Sally for this valuable explanation.
Much appreciated!
Have a nice day :)

Here are the 2 formulas together.
There still is one piece of the puzzle missing. I thought you might ask about it.
Maybe you already know it. So then this is for all that follow.

BSD = (N*P*Q)^.5 where Q = 1-p
Betstdev = (x+1)*Sqrt[p*(1-p)*N]

BetSD = Bet * Payoff * (N*P*Q)^.5 (Payoff = (x+1) where x = the payoff of X to 1)

Most BetSD values are shown for 1 unit, 1trial so you can just take N^.5 and go from there.
So, the Bet * Payoff completes the formula.

Have fun with your math!
Sally

Hello again,

Yeah, I didn't think about that, cos what I wanted was general result for 1 chip in one or more spins.

Just to be sure, let's say if we calculate SD for 100 spins where we're flat beting 2 dollars on each of some 7 numbers combination then to solve this we do it so:

SD = 2$ * 36 * ((7/37) * (30/37) * 100) ^ 0,5 = 281,99$

Which means that we may expect a SD of 282$ around the mean in these 100 spins? where mean is (2$ * 7) * -0,027 *100 = -37,8$

I my equation and conclusion correct?

Thank you!

Correct.
You could have checked yourself and solved it for 1 trial then multiplied that result by 10.

2 handy formulas.
house edge = (x+1)*p - 1 
standard deviation = Bet*(x+1)*Sqrt(p*(1-p)*n)

Another way to use the formulas for the (x+1) is to figure the payoff as (x to 1) +1
This would make it easier to calculate the combined house edge and variance of different multiple bets, if you want to do that.
Remember we can add ev and variance, not he and sd.

stdev = Total Bet*(x+1)*[Sqrt(p*(1-p)*n)]
what is x?
since there are Only 2 possible outcomes (that is what Binomial means)
-$14 loss or $58 win
58/14 would be the 'to 1' value (you are paid $58 for every $14 total bet)
58/14 + 1 = 58/14 + 14/14 = 72/14
$14*(72/14)*Sqrt(p*(1-p)*n) = your answer.

Also remember all this is just an approximation to the binomial distribution.
You can easily use that function in Excel. (Number of trials below 1000 before more error starts to creep in)
=BINOMDIST()

with 19 wins and 81 losses you will have exactly a -$32 result
with 20 wins and 80 losses you will have exactly a $40 result
(see the $72 difference)
Now, probability of any win after 100 trials is 43.0852% using the binomial distribution.
Using the normal distribution we get
0.446630352 showing the error to be on the high side.

Once I get my Excel set up in my new computer I can show you how to use that function if you need some help.
You should be able to crunch the numbers yourself now.
Easy to make a table in Excel.

Glad you have had fun learning the math
Sally

The best way to calculate the standard deviation in roulette is by recording 5000 spins. Most numbers will hit a very even amount of times. The average number of hits for each number will give you your base line for example if the all the numbers seem to hit an average of say 39 and two or three numbers hit say 49 or 59 times your standard deviation for the most often hit numbers would be 10-20. Some numbers will hit quite a bit more than the other numbers.

I have 6 wheels I have recoded 5000+ spins. I know the identifying mark for each wheel and I know how much I should bet per spin based on my bankroll and the standard deviations for my biased numbers on a biased wheel.

I dont get to play them very often because they are European style with single zero but when I do it pays quite good. Its a euro wheel how could it not lol.

Quote: codekiddy

Hello again,

Yeah, I didn't think about that, cos what I wanted was general result for 1 chip in one or more spins.

Just to be sure, let's say if we calculate SD for 100 spins where we're flat beting 2 dollars on each of some 7 numbers combination then to solve this we do it so:

SD = 2$ * 36 * ((7/37) * (30/37) * 100) ^ 0,5 = 281,99$

Which means that we may expect a SD of 282$ around the mean in these 100 spins? where mean is (2$ * 7) * -0,027 *100 = -37,8$

I my equation and conclusion correct?

Thank you!

1 sd for 100 trials playing 7 numbers is= square root of(100x0,1891x0,8109)=3,91

The mean for 7 numbers at 100 trial is 100x7/37=18,91

You reach +1sd hitting 22 times in 100 trials betting 7 numbers, 2 sd at 26 and 3 with 30 hits.

To know the bad luck you go 18,91 hits to 7 in 100 trials

Multiply hit for the amount you wage and you have the results

I donnot understand the formula you used.

Mustangsally, I see you know big about SD and math. I have a couple of questions.

I´m investigating french singlezero roulette.

Some questions

1)In a 1000-spin-sample(random or not, we do noy know it yet), we have 37 pockets, the chance to find 1 number with 3 st dev is 1/10, to pick one of the 37 before the test and get 3st dev is 27/10000, is that correct?

2)the same chance we have for a section of 3 neighbor numbers, but, what is the chance to find 3 numbers that the sum of the 3 get 3 st dev? they do not need to reach 3st dev each of them. The chance to get 2 st dev is 47/1000, the chance to get 3 at the same sample is 47/3=16/1000?

3)supose you have a 1000 2000 3000 or 5000 sample(random or not), with numbers and sections having 1 2 3 st dev, is there a way to have a degree of certancy over a number of sector to get 3 or more st dev the next sample?

4)how do you calculate st dev after collectig more data. What is stronger? 4 st dev on 1000 trials or 3 st dev on 3000 trials? is there a table toknow it.

5)What is stronger on a 1000-spin-sample: 1 number with 3st dev, a section of 3 numbers, or a section of 10 numbers with 3 st dev?

6)what is the cause that after collecting knew data the st dev raise and the % of winning drop? is there a way to predict he bounderies of the raise or drop of the elements?

Weird questions, aren´t they?

I hope you´ll be able to answer them

Quote: ybot

1 sd for 100 trials playing 7 numbers is= square root of(100x0,1891x0,8109)=3,91
The mean for 7 numbers at 100 trial is 100x7/37=18,91
You reach +1sd hitting 22 times in 100 trials betting 7 numbers, 2 sd at 26 and 3 with 30 hits.
To know the bad luck you go 18,91 hits to 7 in 100 trials
Multiply hit for the amount you wage and you have the results
I donnot understand the formula you used.

Hi ybot, you don't understand my equation because you are talking about binomial SD and my formula is about SD of normal distribution, that's what this thread is about, and it's already explained, thanks to mustangsally :D

Quote: i0r0retardod

The best way to calculate the standard deviation in roulette is by recording 5000 spins. Most numbers will hit a very even amount of times.

It looks that you are talking about binomial distribution as well, which in your case probably applies to chi-sqare testings since you are talking about biased numbers in your second post.
anyway thank you for leting me know :D

Quote: mustangsally

2 handy formulas.
house edge = (x+1)*p - 1 
standard deviation = Bet*(x+1)*Sqrt(p*(1-p)*n)

Another way to use the formulas for the (x+1) is to figure the payoff as (x to 1) +1
This would make it easier to calculate the combined house edge and variance of different multiple bets, if you want to do that.
Remember we can add ev and variance, not he and sd.

stdev = Total Bet*(x+1)*[Sqrt(p*(1-p)*n)]
what is x?
since there are Only 2 possible outcomes (that is what Binomial means)
-$14 loss or $58 win
58/14 would be the 'to 1' value (you are paid $58 for every $14 total bet)
58/14 + 1 = 58/14 + 14/14 = 72/14
$14*(72/14)*Sqrt(p*(1-p)*n) = your answer.
Sally

Thank you for all your time Sally, now this is much more clear to me :)
also thanks for mentioning BINOMDIST formula! althought this is old version... now new versions are BINOM.DIST and BINOM.INV in new excel 2010.

Cheers mate!

codekiddy: we always apply binomial distribution/satndard deviation in roulette.

Quote: codekiddy

For example if one would calculate SD of straight-up then result will be 5,837837....
In the same way if one would calculate SD of straight-up bet according to expected weel frequency the result will be 2,655637... which is not the same as above.

I know how to calculate both of them, all I need is a little better matematical explanation.

Hey codekiddy,

In you opening post you showed to 2 different numbers for the Standard Deviation of a straight up bet.

5.837837
And
2.655637

I know how you got the first number of 5.837837.

36 * Sqrt( (1/37) * (36/37) ) = 5.837837

But I can't figure out how you calculated the 2.655637.

I know this post is really old so you might not be around anymore. If anyone else knows what that 2.655637 means and how it is calculated. Much appreciated.

Thanks!

No idea how to get the second figure, I tried various prizes using this method and changing the numbers.
Contribution = Combinations * Deviation^2.
Std Dev = SQRT(Total Contribution / Combinations).

	Value	Combinations	Contribution
Prize	0	36	0
Prize	36	1	36
Average / Totals	0.972 973	37	36
	Deviation	Combinations	Contribution
Prizes	-0.972 973	36	34.080 351
Prizes	35.027 027	1	1226.892 622
Std / Totals	5.837 838	37	1260.972 973

Thanks for trying charliepatrick. I imagine it was just a miscalculation or maybe a typo. To me it was just one of those times where it kept bugging me all day that couldn't figure it out.