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[Chrisbis]

@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)

[Turner]

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

[Priyanka]

It couldn't get clearer than this.

[Turner]

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

[Chrisbis]

@Turner


Go on.........................>>!
Spit it out.

[smiley]monkey/phew.gif[/smiley]

[Turner]

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.

[Dane]

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

[Chrisbis]

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]

[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]

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

[Chrisbis]

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

[Bayes]

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. 

[Chrisbis]

Thanx for clearing that up Bayes! 

[Dane]

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

[Priyanka]

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.