masterj
Joined: Dec 19, 2018
• Posts: 16
January 26th, 2020 at 8:45:46 AM permalink
Hello all,

the expected frequency of won bets on 1 number of a European Roulette Wheel is p=1/37.
In the long run, you should get this result. (e.g. you win 1000 out of 37000 spins.

What are the deviations after x spins? It would be great if someone could tell me how I can calculate this?

Thank you very much,
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4292
January 26th, 2020 at 11:19:30 AM permalink
Quote: masterj

Hello all,

the expected frequency of won bets on 1 number of a European Roulette Wheel is p=1/37.
In the long run, you should get this result. (e.g. you win 1000 out of 37000 spins.

What are the deviations after x spins? It would be great if someone could tell me how I can calculate this?

Thank you very much,

Well, statistics isn't my strong suit, but I think standard deviation is calculated like this:

SD = square root of variance

Oh, you want more?
In this case, variance = the sum of p(k) (v(k) - m)2, where k is each of the n possible results; p(k) is the probability that result k happens, v(k) is the value of result k, and m is the mean result.
In this case, for one spin, there are two possible results; 1, with probability 1/37, and 0, with probability 36/37.
This means p(1) = 1/37, v(1) = 1, p(2) = 36/37, v(2) = 0, and m = 1/37.
The variance = 1/37 x (1 - 1/37)2 + 36/37 x (0 - 1/37)2 = (362 + 36) / 373
= (36 x 37) / 373 = (6 / 37)2
and the standard deviation is the square root of this, or 6/37.

For x spins, there are (x + 1) possibile results - 0 hits, 1 hit, 2 hits, ..., (x - 1) hits, and x hits.
The probability of n hits is (x)C(n) (1/37)x (36/37)n-x, where (x)C(n) is the number of combinations of x items taken n at a time (also COMBIN(x,n) or C(x,n)); the value of n hits is n, and the mean in x spins is x/37.

If I am calculating this right, after X spins, the mean number of hits is X / 37, and the standard deviation is 6 sqrt(X) / 37.
Note that, while the standard deviation of 37,000 spins is about 31.2, the standard deviation of 74,000 spins is not twice that, but about 1.4 times that.
masterj
Joined: Dec 19, 2018
• Posts: 16
January 26th, 2020 at 11:45:21 AM permalink
Wow, that was quick! Thx man!

I have got some questions to your calculations,

and the standard deviation is the square root of this, or 6/37.
Question: does this mean that 6/37 is that you can run over / under expected value of +5 on average?

Note that, while the standard deviation of 37,000 spins is about 31.2, the standard deviation of 74,000 spins is not twice that, but about 1.4 times that.
Question: does this mean that after 37000 spins, there is a chance that 1 number runs over / under the expected value of 31.2 spins?

Thank you very much again!
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4292
January 26th, 2020 at 12:01:21 PM permalink
Quote: masterj

Wow, that was quick! Thx man!

I have got some questions to your calculations,

and the standard deviation is the square root of this, or 6/37.
Question: does this mean that 6/37 is that you can run over / under expected value of +5 on average?

Note that, while the standard deviation of 37,000 spins is about 31.2, the standard deviation of 74,000 spins is not twice that, but about 1.4 times that.
Question: does this mean that after 37000 spins, there is a chance that 1 number runs over / under the expected value of 31.2 spins?

Thank you very much again!

6/37 is the standard deviation over one spin. For N spins, the standard deviation for the number of hits of a particular number is 6 sqrt(N) / 37.

There is about a 68% chance that a set of results will be within one standard deviation either way (i.e. anywhere from one SD below the mean to one SD above), and about a 95% chance that the results will be within two SDs either way. The exact value for N SDs is 200 sqrt(2 PI) times the integral from 0 to positive infinity of the function 1 / (e to the power of N2); however, there is no known "easy " way to calculate that - you have to use approximation methods.

For example, over 10,000 spins, the expected number of times 1 will come up is 10,000 / 37 = 270.27, and the standard deviation is 600 / 37 = 16.22, so 65% of the time, 1 should come up between 254.05 and 286.49 times, and 98% of the time, 1 should come up between 237.83 and 302.71 times.

Pardon me for asking, but why the interest in this particular situation?

Given 37,000 spins, it's almost certain that at least one number will go over the expected number of 1000 spins - in fact, the only way that no numbers have more than 1000 hits (and, for that matter, none have fewer than 1000) is if every number comes up 1000 times, which, despite being "expected," is almost impossible.
masterj
Joined: Dec 19, 2018
• Posts: 16
January 27th, 2020 at 12:58:49 AM permalink
For example, over 10,000 spins, the expected number of times 1 will come up is 10,000 / 37 = 270.27, and the standard deviation is 600 / 37 = 16.22, so 65% of the time, 1 should come up between 254.05 and 286.49 times, and 98% of the time, 1 should come up between 237.83 and 302.71 times.

Pardon me for asking, but why the interest in this particular situation?

if you choose 1 out of 37 numbers, most of them are within one SD. if you can make a bigger profit with these numbers, then you loose on numbers within 2 SD, then it might be possible to make some money.
charliepatrick
Joined: Jun 17, 2011
• Posts: 2114
January 27th, 2020 at 1:25:27 AM permalink
For a two way result where the chances of winning are p and losing are q = (1-p) then the average is Np and the SD = SQRT (Npq). However for small N you can also work out the probabilities exactly by Pr(0) = 36/37 ^ N etc.
weezrDASvegas
Joined: Feb 2, 2018
• Posts: 69
Thanks for this post from:
January 27th, 2020 at 2:33:52 AM permalink
Quote: masterj

Hello all,

the expected frequency of won bets on 1 number of a European Roulette Wheel is p=1/37.
In the long run, you should get this result. (e.g. you win 1000 out of 37000 spins.

What are the deviations after x spins? It would be great if someone could tell me how I can calculate this?

Thank you very much,

The roulette numbers will never come out equally. Its like the lotto numbers. they come out close to one another only in percentages after lots of draws. The standard deviation separates them in absolute terms. it is called the normal probability rule. One app does good calculations (search SuperFormula).

For 1000 spins you got

The standard deviation for an event of probability
p = .02702703
in 1000 binomial experiments is:
BSD = 5.13

The expected (theoretical) number of successes is: 27

Based on the Normal Probability Rule:

* 68.2% of the successes will fall within 1 Standard Deviation
from 27 - i.e., between 22 - 32
* 95.4% of the successes will fall within 2 Standard Deviations
from 27 - i.e., between 17 - 37
* 99.7% of the successes will fall within 3 Standard Deviations
from 27 - i.e., between 12 - 42
I submit to you that no roulette number will show fewer than 12 hits and no more than 42 wins. That happens 99.7% of the times.

I saw results for about 8000 real spins from a German casino (I think the only who publishes roulette resulst).

The standard deviation for an event of probability
p = .02702703
in 7990 binomial experiments is:
BSD = 14.5

The expected (theoretical) number of successes is: 216

Based on the Normal Probability Rule:

* 68.2% of the successes will fall within 1 Standard Deviation
from 216 - i.e., between 202 - 230
* 95.4% of the successes will fall within 2 Standard Deviations
from 216 - i.e., between 188 - 244
* 99.7% of the successes will fall within 3 Standard Deviations
from 216 - i.e., between 174 - 258

The coldest number had 192 hits and the hottest won 239 wins – inside the calculated limits. It is convincing that you can win at roulette with particular numbers * you might find the bias *
Roulette Numbers Ranked by Frequency
weezrDASvegas
Joined: Feb 2, 2018
• Posts: 69
Thanks for this post from:
January 27th, 2020 at 2:43:06 AM permalink
Quote: weezrDASvegas

The roulette numbers will never come out equally. Its like the lotto numbers. they come out close to one another only in percentages after lots of draws. The standard deviation separates them in absolute terms. it is called the normal probability rule. One app does good calculations (search SuperFormula).

For 1000 spins you got

The standard deviation for an event of probability
p = .02702703
in 1000 binomial experiments is:
BSD = 5.13

The expected (theoretical) number of successes is: 27

Based on the Normal Probability Rule:

* 68.2% of the successes will fall within 1 Standard Deviation
from 27 - i.e., between 22 - 32
* 95.4% of the successes will fall within 2 Standard Deviations
from 27 - i.e., between 17 - 37
* 99.7% of the successes will fall within 3 Standard Deviations
from 27 - i.e., between 12 - 42
I submit to you that no roulette number will show fewer than 12 hits and no more than 42 wins. That happens 99.7% of the times.

I saw results for about 8000 real spins from a German casino (I think the only who publishes roulette resulst).

The standard deviation for an event of probability
p = .02702703
in 7990 binomial experiments is:
BSD = 14.5

The expected (theoretical) number of successes is: 216

Based on the Normal Probability Rule:

* 68.2% of the successes will fall within 1 Standard Deviation
from 216 - i.e., between 202 - 230
* 95.4% of the successes will fall within 2 Standard Deviations
from 216 - i.e., between 188 - 244
* 99.7% of the successes will fall within 3 Standard Deviations
from 216 - i.e., between 174 - 258

The coldest number had 192 hits and the hottest won 239 wins – inside the calculated limits. It is convincing that you can win at roulette with particular numbers * you might find the bias *
Roulette Numbers Ranked by Frequency

If you average the coldest and hottest numbers you get the expected number of successes (216)

192+239=431 average 215.5~216
masterj
Joined: Dec 19, 2018
• Posts: 16
January 27th, 2020 at 6:41:43 AM permalink
Hello weezrDASvegas,

The standard deviation for an event of probability
p = .02702703
in 7990 binomial experiments is:
BSD = 14.5

The expected (theoretical) number of successes is: 216

Based on the Normal Probability Rule:

* 68.2% of the successes will fall within 1 Standard Deviation
from 216 - i.e., between 202 - 230
* 95.4% of the successes will fall within 2 Standard Deviations
from 216 - i.e., between 188 - 244
* 99.7% of the successes will fall within 3 Standard Deviations
from 216 - i.e., between 174 - 258

do these deviations get smaller or bigger the more spins you play?
weezrDASvegas
Joined: Feb 2, 2018
• Posts: 69
Thanks for this post from:
January 27th, 2020 at 8:02:10 AM permalink
Quote: masterj

Hello weezrDASvegas,

The standard deviation for an event of probability
p = .02702703
in 7990 binomial experiments is:
BSD = 14.5

The expected (theoretical) number of successes is: 216

Based on the Normal Probability Rule:

* 68.2% of the successes will fall within 1 Standard Deviation
from 216 - i.e., between 202 - 230
* 95.4% of the successes will fall within 2 Standard Deviations
from 216 - i.e., between 188 - 244
* 99.7% of the successes will fall within 3 Standard Deviations
from 216 - i.e., between 174 - 258

do these deviations get smaller or bigger the more spins you play?

standard deviation gets BIGGER and B I G G E R the more spins you play. the gaps between roulette numbers get bigger although percentages get closer.

100 spins - SD = 1.6
1000 spins - SD = 5.13
7990 spins - SD = 14.5
37000 spins - SD = 31.2

SuperFormula.exe calculates

The standard deviation for an event of probability
p = .02702703
in 37000 binomial experiments is:
BSD = 31.19

The expected (theoretical) number of successes is: 1000

Based on the Normal Probability Rule:

* 68.2% of the successes will fall within 1 Standard Deviation
from 1000 - i.e., between 969 - 1031
* 95.4% of the successes will fall within 2 Standard Deviations
from 1000 - i.e., between 938 - 1062
* 99.7% of the successes will fall within 3 Standard Deviations
from 1000 - i.e., between 907 - 1093

You can expect no fewer than 907 wins but no more than 1093 hits. hopefully you'll be on the righthand side
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