Pigeon hole, position stream, vdw?

[RayManZ]

Hi guys,

The recent topics sparked my interest on these subjects and how to use them for roulette. There is so much information on this forum that i dont even know what im looking for anymore.

First of the pigeon hole principle. PHP.
I believe we are looking at the all wrong. We know that in roulette we have 37 numbers. (zero wheel) those are the holes. Now we need 38 spins (the pigeons) and we have proven the PHP.

With a big enough progresssion you could bet that a number has to repeat in 38 spins. But that isnt really an option i believe.

But that repeat just end the PHP. Why are we looking for the end of PHP? Alot of holes need to be filled before we get to the end. So why not some how make a bet around that? Why are we so focused on those repeats? We get alot more unhit numbers in 38 spins than repeat!

Second: position streams
We have been told to some how use this tool to know what to bet? We know for a fact that a repeat happens in the first 50% of our position stream. If we have a stream of 37 numbers a repeat happens like 92% of the time on a number in position 1 till 19.

But how the hell do we use this fact? To be honest i have no idea. I thought about creating new number with it. If we take lines and the position stream of lines we can create 36 numbers. But those number behave the same as 36 random numbers so that does work.

Another idea is maybe only use the position stream when a cycle end? That way we should have more number in the first half of the stream because of the fact given earlier...

Also if we somehow create a new number using the stream with a favor to the first 50% of the stream we could in theory know when to bet a certain number. Because that number has to be in the same spot in the stream as before. So we don't have to bet when that number isnt in the right position in the stream.

Third: vdw
I havent really gave this on much thought. Also it only works for 50/50% change. Maybe it could be usefull to figure out if we need to bet a repeat or a unhit. Not sure about this.

If this helped anyone any further in there process... Your welcome! Please let me know if you have found something i missed.

Please point me to posts or topic i need to re-read.

Keep sharing! Thanks!

[nottophammer]

Ray
Would you go to random org and download 185 numbers? Do this 1000 times.
Now you’re going to say fcuk that too much work.
Well, get Priyanka’s tester in KTF topic bottom page 80.
I’ve done that as well as collected 1000 air-ball games.
You’d get your answer to how many repeats hit; in my case 40 spins and 60 spins.
Even 10’330 live spins posted by the General; gave the repeat average for 40 and 60 spins.
Okay, I’ll save you doing the above. The answer is 1-3-5-7=16 repeats in 40 spins. At 60 spins you’ll have 29.5 of the starting 37 hit. Round up to 30; so, 30 non-hits and 30 repeats, in 60 spins.
But average is no good to some; but even Turbo says you’ll likely have a repeat with 9 spins.

[nottophammer]

Ray not bad 17 repeats, +1 for repeats. At 40 spins. Even the non-hits are only -3 at 60 spins.
That’s only 1 set

But to0 much work to see if the 1-3-5-7=16

[TRD]

Code Select Expand

Here's Luckyfella's post -- Vaddi Pairs

gamblingforums ... /threads/gamblers-fallacy-absurd-proof.22620/page-30#post-126551

One final matter as a gift to forum members.

Remember Vaddi?

The neighbor is wrong.

The neighbour with math significance is,

x+y = 36

0+36=36
1+35=36
2+34=36
3+33=36
4+32=36
5+31=36
6+30=36
7+29=36
8+28=36
9+27=36
10+26=36
11+25=36
12+24=36
13+23=36
14+22=36
15+21=36
16+20=36
17+19=36
18+18=36

36 is the "equivalent" of birthday.

Enjoy this gift from me.

Take care everyone.

[TRD]

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& another one relevant
gamblingforums ... /threads/when-does-it-lose.22770/page-43#post-129440

A running set of 7 pigeon holes based on the most recent spins has a greater than 50% chance of a match on a 38 pocket wheel. 7 of the most recent hits provides for a repeat within 8 spins (as the first spin can't be a repeat within a set of .

Any generalised birthday calculation will show this.

Oh look, here is one 8 spins and 38 pockets:
Screenshot_20211201-073528_Chrome.jpg


So based on the above birthday approximation we win with probability p = 0.5213770274 and lose with probability q = (1 - p) which is q = 0.4786229726.

It's quite simple math to verify at this point.

When we win we gain net 29 units, and when we lose we forfeit 7 units. Our expected value is (29 x p) - (7 x q) = 11.76 units which is +EV and slightly less than the authors paper due to a more pessimistic approximation to the birthday attack calculation.

You can verify the birthday attack math independently on Wikipedia en.wikipedia.org/wiki/Birthday_attack

I've laid it out on a platter and and the sky is falling.

Here is an extract of the paper and the authors proof:
.. see images on the linked post