Occasionally I get PMs from members asking me about various stats and probabilities in roulette, so I thought it might be a good idea to post them here together with my answers. If you have any questions regarding the stats of the game, here's the place to post them! Please try to limit posts to questions, and try to make them clear, giving an example if possible.
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Q: What is the longest number of spins a particular dozen can go without a repeat? For example, 1,2,3,3,2,1,2,2,2,3,2,1,2,2,3,1,1 is a sequence in which dozen 1 has not repeated for 15 spins.
A: Theoretically, according to the mathematics, there isn't any limit to how long an event can 'sleep', but the probabilities obviously get smaller the longer the sleep. There is a handy formula you can use to answer this question and others like it:
N = z2 (1 - p) / p
where, N is the number of spins (what you're looking for), p is the probability of the event in question, and z is the z-score, which is a measure of the number of standard deviations from the mean (average). If you don't know what standard deviation is, then see the standard deviation in plain English (link:://rouletteforum.cc/math-reference/standard-deviation-in-plain-english/) thread in this section of the forum).
For a dozen, the probability of 2 in a row (the minimum requirement) is 12/37 ✕ 12/37 = 0.1052
Since we want to know the longest that a dozen can go without a repeat, the value for z should be at least 3.0 but preferably 4.0. A value of 3.0 for z means that the length of sleep will be less than or equal to the value found using the formula in 99.7% of cases, but it's better to take z = 4.0 if you're looking for a maximum. In all my computer testing the worst case scenarios are usually between 4.0 and 4.5.
So if you put the numbers into the formula you get:
N = 16 (1 - 0.1052) / 0.1052 = 136 spins.
A simulation over 1 million spins confirmed this, with typical results being around 140 spins.
Fantastic post Bayes
This has lit a light bulb in my head
:thumbsup:
Hello Bayes,
Glad to have found you again.
I had a question about the appearances of numbers.......
This is not an actually method but the only way I could clearly explain my stats question...
Scenario:
We are betting each number for two spins cycles, 0-37, 74 spins
First we start with zero, we spin 74 times
Then number 1, we spin 74 times
Then number 2, we spin 74 times
etc........
The question is, how many times do we expect the specific number being bet on to appear in each session? A session is 74 spins.
After 37 sessions (each number played) do you think that 70% of the time a number will only land once?
Sometimes a number might land 4 times or 2 times sometimes it won't.
What do you think the average will be for each number to land in the spin cycle?
Hope it is an interesting question that you have not pondered before :)
Quotehow many times do we expect the specific number being bet on to appear in each session?
I would estimate a number hitting max 4 or 5 times within 74 spins BUT you must also rememebr any number can sleep for over 600 spins so if you are running up the scale 0 to 36 chances are many of them will be missed out
Hi amk,
This is an easy one. Since each number will land on average once every 37 spins, then in 74 spins it will (again on average) hit twice.
By definition, 70% of the time (actually a little less, just over 68% of sessions) the standard deviation will be between +1.00 and -1.00. This means that within these boundaries, a number could hit only once in the 74 spins (corresponding to a z-score of -0.72) or 3 times (a z-score of +0.72). A number hitting less than once or more than 3 times will take the z-score outside +/- 1.0. Hitting twice is the perfect average, corresponding to a z-score of zero.
If you take a z-score of 3.00 which covers 99.7% of all sessions (in other words, it's quite rare for a number to hit more than this in 74 spins), the number of hits is 6 (2.87 to be exact). 7 hits takes it to 3.59, which can happen of course, but is fairly rare.
Thank you very much for your great explanation Bayes!!
You make almost make math fun :)
I can't get my head around the basic of mathematical concepts.
Math does give you a better understanding of the world around you. This is also a reason why I find roulette so intriguing. Incredible that after many hundreds or thousands of spins each number will appear the same amount of times. Amazing how everything averages out.
This brings me to another question. I believe your stand point is that regardless of how many times we play results will always average out. Lets say I play a method with a 6 step progression for 1500 spins per day (250 games). Over the course of a month the wins on step 1 through 6 will average out to that which can be expected when running a simulation on RouletteXtreme for 45000 spins. Is this correct? Even if we had played 45000 spins straight (7500 games) we would get the same results?
Thanks again Bayes for your thoughts and you to Superman!!!
Hi AMK
I just ran a test on the bot against European roulette (playtech) I played each number for 37 spins, will do a 74 spin one later, see attached text file for the results.
Was this all numbers at same time for 37 spins - or 0 - 37spins, then 1- 37 spins then 2 -37 spins etc ?
Quote0 - 37spins, then 1- 37 spins then 2 -37 spins etc
Correct
Thanks, interesting ...
I just tried covering all numbers, then removing the ones that hit - 3 quick sessions all won - can your bot do that ?
Quotecan your bot do that
Yes it can, it can do anything I program it to do, been there done that, it fails eventually, progression gets to steep
Thanks Superman........
ok thanks n regards
OK as promised, heres the 74 spin cycle version
Does anybody know the total profit of the "74 cycle" Superman made........?
hi amk
74 L Prog 2 Cash 1336 Peak 1397
im sure its 1336 units
Thats correct Andy, it had lost a bit after the last win which took it to Peak 1397
Thanks dchq_bam and Superman,
Question for something completely different.
I am very bad with math. What is the progression for a four wide pattern of dozens with covering the zero?
Usual progression 1,3,9,27 with covering zero .., .., .., .., ??
Quote from: superman on Dec 10, 11:02 AM 2011
Yes it can, it can do anything I program it to do, been there done that, it fails eventually, progression gets to steep
@Superman, which "gets too steep" were you referring to, the one Poundmaker discussed or the amk topic?
Quotewhich "gets too steep"
Was replying to Poundmaker, bet all number then keep removing the one that just hit, at some point, say, after you have spun 5 times and removed those 5, you are still covering 30 numbers, those 5 WILL hit 2 maybe 4 times which sends the progression sky high, if you dont use a progression then you lose too much also.
superman,
know this old but liked poundmakers idea, you say spun 5 and removed 5 will be covering 30 numbers is that right or should it be 32 to cover. Wont plus1/minus 1 be safe