Hello All,
Iv'e been playing this on SkyVegas in fun mode as well as using actual spins from a file of 1 millions spins I got from VLS Roulette forum.
It has been working well in practice.
I just wanted to get some input/feedback tips before I play it for real.
Zelx
Note from Moderator.
Welcome to the forum Zelx. [smiley]afb/tup.gif[/smiley]
I have had a look at your PDF, and it seems ok, and nothing any of us should worry about.
We have to be careful, especially with first time posters.
Would be good if you pop over to the Newcomers (http://betselection.cc/newcomers-lounge/) section (Home page), and say hello there too. :nod:
Cheers
Chrisbis
Hi Chrisbis, thanks for the welcome. Will drop into the newcomers lounge now.
You get a feel for these things.
The exact same can be done with streets.
It would make no difference.
A street can sleep for 80?....can't remember exactly
But you have me thinking.
What about generating 12 random 3's each session
Rand.org.....sequences...1-36.....12 columns.c
That's an interesting idea Turner - generate 12 random 3s each session. I am always looking for ways to exploit the most unlikely event to happen.
Will give this a try.
Quote from: Turner on August 01, 2013, 01:57:03 PM
You get a feel for these things.
The exact same can be done with streets.
It would make no difference.
A street can sleep for 80?....can't remember exactly
Hi Turner,
A professor of math one said, that one out of six can skive for at least 100 times.
If he was right, one out of twelve might refuse to turn up much longer.
There is no need to remember the exact number of spins, however.
There is no absolute limit here.
If anyone rules out that one number or any Group of numbers can only skive X number of spins, he or she is wrong.
It is only a question of TIME, before Lady Luck or the Laws of probability proves it.
It may happen tonight or any other night. It is definitely inevitable [smiley]aes/devil.png[/smiley].
If it did not happen in one million spins, just take another look or look at one more million spins.
Dane
Dane.....there's no can't or can...just not seen it or seen it.
I have seen a street sleep 110 spins....so I know they can from observation. I can't say it's a limit but i can say it could sleep 110.
In theory....if someone asked me if my mother can break dance....the answer is no. The real answer is I have never seen her break dance.
No one can say random has limits....unless they have seen every run ever produced. Even then...you haven't seen tomorrows.
Quote from: Turner on August 01, 2013, 03:07:46 PM
Dane.....there's no can't or can...just not seen it or seen it.
I have seen a street sleep 110 spins....so I know they can from observation. I can't say it's a limit but i can say it could sleep 110.
In theory....if someone asked me if my mother can break dance....the answer is no. The real answer is I have never seen her break dance.
No one can say random has limits....unless they have seen every run ever produced. Even then...you haven't seen tomorrows.
So you are not going to produce a video with your mother´s break dance?
Inspired by Prianka´s "speedy" videos I can tell you that I am able to shoot faster than John Wayne.
The reason is simple: John Wayne is dead :nod: .
But now I grab my cheap electronic calculator to find the probability, that one chosen street at one roulette table does not turn up in 111 spins:
(34/37)^111 = 0.00008390123. It is rather close to 1/33/360. So if we observed 111 spins 360 nights a year, we should expect it to happen approx. every 33Th year.
Dane
Quote from: Dane on August 01, 2013, 04:41:03 PM
So you are not going to produce a video with your mother´s break dance?
Inspired by Prianka´s "speedy" videos I can tell you that I am able to shoot faster than John Wayne.
The reason is simple: John Wayne is dead :nod: .
But now I grab my cheap electronic calculator to find the probability, that one chosen street at one roulette table does not turn up in 111 spins:
(34/37)^111 = 0.00008390123. It is rather close to 1/33/360. So if we observed 111 spins 360 nights a year, we should expect it to happen approx. every 33Th year.
Dane
Dane, the electronic calculator is right. But you might want to relook your formula.
Sam taught me an important lesson. When a dozen hits it takes twelve buddies. When a line hits it takes 6 buddies. When a street hits it takes 3 buddies. So you might want to use 12/13 instead of 34/37. Hope am not making a mistake here.
I am right here, in saying, a street is 3 numbers out of 37, the only difference, is a street result* has to land in
1 of 13 possible locations, if including the Zero.
So, i guess its 13-1(sleeper)=12/13
*result meaning where the ball will land relative to the carpet/felt/table layout.
Does math include guessing? And should I really relook my formula ???
I don´t have a math degree.
But I am sure that some educated math member can put things straight and enlighten us. Thanks!
Dane
I did degree level maths in my HND electronics and electrocal engineering. I suck at maths and probability. I've tried reading Bayes and it sounds like the teacher in charlie brown.....a slighty muted trumpet played by a 3 year old
I always was taught in maths, to make a educated guess first, then 'prove' the guess right or wrong with analysis.
Maybe Bayes can swing an eye over here!
@ Dane, Let me try explaining in simple language my understanding of things without confusing on heavy mathematical principles.
You flip a coin. There are two possible options Head or tail. The chance of a head coming is 1/2.
Now lets get a bit complex. You have 3 bags. 1 bag having 2 red balls. 1 bag having 3 red balls. 1 bag having 5 black balls. Assuming that you are picking a whole bag, the chance of picking a red ball is 2/3. However you throw all these balls on to a box and pick one, then the chance of picking a red ball is 5/10 or 1/2. Hope you get the difference.
Assume that the balls are straight-ups. The bags are streets. Then you will be able to get the corollary on why it is 12/13 and not 34/37. I can get into some complex terminologies, but I think it is better understandable and better practiced, if we take a simplistic common sense based approach.
On doing the formula for sleepers, you are trying to find a formula for identifying a sleeper street for lets say 100 spins. Now if you are using 34/37 then what you are assuming is 3 numbers out of 37 is sleeping for 100 spins. Lets take a simple example. Lets assume that the sleepers for 100 spins are 5 numbers 1, 7, 22, 23 and 24. Now there is only 1 sleeping street but there are 5 sleeping numbers. So that immediately breaks away the fundamental reason of chosing 34/37. Hope you are getting what I am saying.
I am a newbie when it comes to roulette. But when it comes to statistics, I can safely say that I have better than average knowledge :cheer:
Priyanka:
In my example I chose ONE street and ONE street only on beforehand. And I found the probability that this specific street did not turn up.
Someone (Bayes?) might help us.
Cheers
Dane
@Dane
There are 12 streets to choose from, + Zero as a landing place.
The ball can only land on one street, and that one street does have three numbers in it, I agree, but the
question U asked was the probability of a Street sleeping, not the probability of three numbers sleeping.
If you take away the 36/37 numbers, and just say we have a game with only 12 Streets in it + Zero, that is 13 opportunities
for the ball to land.
That gives us 1 to sleep, which is 13-1=12/13
In other words, One street less than the Total Streets divided by ALL the possible landing places (12 streets + Zero)
Do you see it now?
12/13 = 0.9230769
whereas
34/37 = 0.9189189
Its only slightly different. (then put it to the Power of your chosen spins)
Maths can be fun....agreed.....but i had 20 runs of 10k and every street, line, ec had the same max sleep.....street being near to 100. Maths dosnt denote where in that cycle you are as you place yer chip.
Bit like being told to stand on a trapdoor covering a pit of wooden spikes. You know it opens once a day at the same precise time...but no one seems to know when it last opened....do you get on it?
Depressingly...every bet selection is a group somewhere in the scheme of thing....even 1, 15, 36
Even if one is just off on a 345 spin sleep...collectively...it could sleep for 110....and you just chose it at random
It couldn't get clearer than this.
Quote from: Turner on August 02, 2013, 12:54:11 PM
Maths can be fun....agreed.....but i had 20 runs of 10k and every street, line, ec had the same max sleep.....street being near to 100. Maths dosnt denote where in that cycle you are as you place yer chip.
Bit like being told to stand on a trapdoor covering a pit of wooden spikes. You know it opens once a day at the same precise time...but no one seems to know when it last opened....do you get on it?
Depressingly...every bet selection is a group somewhere in the scheme of thing....even 1, 15, 36
Even if one is just off on a 345 spin sleep...collectively...it could sleep for 110....and you just chose it at random
I've just had an idea....and its in my post....thought of it by reading my own post...lol
@Turner
Go on.........................>>!
Spit it out.
[smiley]monkey/phew.gif[/smiley]
I don't like too before testing. But it would intetest John Legend.
I can't accuse someone of not contributing then keep something to myself...but im in the house alone tonight....ill come back tomorrow.
Its very doable too...and will prove something or put it to bed
Also....i need to generate and run a few million spins in rx.
Thanks to Priyanka and Chrisbis for trying to explain. I´ll have to reread your posts.
I once found a "formula for sleepers" playing with my cheap electronic calculator. My procedure is rather simple. But please remember that I do not have a math degree. An example:
How many of the 37 numbers are probably missing after just 20 spins? First let your calculator calculate (36/37) ^20. THEN MULTIPLY THE RESULT BY 37!
Dane
Well, although no 'practically viable', the direct logical answer to your question of:-
"How many number out of 37 could be missing after 20 spins" = 36
and that is because all 20 spins could have been the same single number. [smiley]aes/surprized.png[/smiley]
Quote from: Chrisbis on August 03, 2013, 10:04:14 AM
Well, although no 'practically viable', the direct logical answer to your question of:-
"How many number out of 37 could be missing after 20 spins" = 36
and that is because all 20 spins could have been the same single number. [smiley]aes/surprized.png[/smiley]
I asked: "How many of the 37 numbers are probably missing....."
PROBABLY my method shows just that. At least I hope so. What happens in most cases? And what COULD happen?
Two different questions! My cheap calculator can´t find the probability of (1/37) ^19. In real life you ought to call the media (or balance the wheel) if something like that happened in this millenium!
Dane
Too true!! lol
The answer will depend on several variables, one of which will be:-
"what is the average rate of repeater number showing within 20 spins"
That will have a significant bearing of how someone calculates the probability out for your given situation.
Out of the 20 spins, one could assume there would be at least 1 repeat of a number, going all the way upto the insane circumstance, of all 20 spins being the same.
Obviously the easy calculation, would be to say the 20 spins are all unique numbers, and therefore you will have 17 missing numbers from the intended 37
Quote from: Dane on August 02, 2013, 12:02:27 PM
Priyanka:
In my example I chose ONE street and ONE street only on beforehand. And I found the probability that this specific street did not turn up.
Someone (Bayes?) might help us.
Cheers
Dane
Just noticed this thread. The chance of a predetermined street showing up is 3/37, the chance of it NOT hitting is 1 - 3/37 = 34/37.
If you want to know the probability that the street will sleep for X spins you need to multiply 34/37 to itself X times. :thumbsup:
Thanx for clearing that up Bayes! :)
Yes, thanks to Bayes! Actually I am better piano player [smiley]aes/headphones.png[/smiley]than calculator; but the divine structure in the music of Johann Sebastian Bach might have helped me! Now we have all deserved[smiley]aes/coffee.png[/smiley][smiley]aes/coffee.png[/smiley] I think!
Dane
Thanks Bayes for clearing it up. Just relooking at the example, realized that there was an error. Apologies if I confused anyone, but I think it validated my position on the learning that I mentioned earlier in the post.