How many consecutive 8 spin sessions do you think there can be max before we have a 8 spin session in which one number repeats?

What are your estimations?

My guess is no more then 6 or 7 sessions???

Example:

1,2,3,4,5,6,7,8    8 spins and not one number repeated   (1 session)

again

1,2,3,4,5,6,7,8    8 spins and not one number repeated   (2 sessions)

again ....  etc.....


Useful Stat:

Several million spin tests have shown that the average number of spins before the first repeat appears is after 8 spins.

After millions of spin tests it has been shown that there will always be a minimum of 4 repeaters in a 37 spin cycle.

This then implies that 1 dozen, which one we don't know, will always have 2 or more repeaters in it during a 37 spin cycle.

Infact, 1 dozen and 1 column, will always have 2 or more repeaters in it during the same 37 spin cycle.

I wanted to present this observation to see if any good methods can be developed around it?

Any ideas out there : )

Started the thread to understand what the possibilities are.



With a few friends we play roulette at home.




Can we play in a poker fashion?


Have not thought about this yet and don't want to rack my brain too much : )


Could we perhaps play through the internet against each other instead of the casino?


How would this look and is it in everybody's favor?


Any insights?

Start an affiliate program with a casino.

Then play on a friends account.

What ever you lose you will be returned +-25%



Could this be used to our advantage?

You will have to review Soggets "An Early Xmas Gift" method to understand this tweak, perhaps not.


http://betselection.cc/private-general-board/an-early-christmas-gift-(-hopefully-)/msg4814/#msg4814

This tweak also had one loss over Soggets 10,000 spins but would have won on the next spin.


This one loss was -1200, Soggets one loss was -3000





We mark down all numbers which come in twice and bet on them up to 4 numbers.


Then we continue tracking all numbers which come in twice as with Soggets method.


Once we have a number which has hit 3 times we erase the most recent from our 4 two number bets and put the number which has hit 3 times in its place.


We continue to track all new two numbers and three numbers.


If another number hits 3 times we add this to our 4 numbers and remove the second most recent two number bet and add the newest number which hit 3 times.


It would look like this



1       (2)  (3)  (4) (5)  (6)

2       2      2      2     3     3
2       2      2      3     3     3
2       2      3      3     3     3
2       3      3      3     3     4
2
2
2
3 ... first 3 (2)
2
2
3 ... second 3  (3)
3 ...  third 3  (4)
2
3 ... forth 3  (5)   we now have 4 numbers hit 3 times and we continue betting these until a number
2                          appears which has hit 4 times then remove the most recent 3# and replace it with a 4#
3
3
4 ... first 4  (6)
3 ... our first number to hit 3 times is landed (this is the coldest of all numbers which hit 3 times




My theory with this approach is that we are continually playing the coldest repeaters and adding the hottest repeaters. These we continue to move down the 4 numbers we bet.



ps
Thank you Sogget for all your hard work. Without you this method and its stats would not have been found!

=:)

Alice
You're nothing but a pack of cards!




Alice
Let me see: four times five is twelve, and four times six is thirteen, and four times seven is -- oh dear! I shall never get to twenty at that rate!




Alice
It would be so nice if something made sense for a change.

PB HORIZONTAL                                                                                                 




We track all EC's at the same time just as with PB.


Instead of checking each EC group off vertically we check them off horizontally as one group.


Each line represents 3 spins:


  H/L    O/E     R/B


  7        5        2
  7        2        4
  2        1        3
  1        8        3   ....  at this point group 6 is missing horizontally
  ...       ...       ...   ....  in the next 3 spins we bet against a group 6 appearing in either H/L  O/E  R/B
                                  we select the EC with the most unique groups/numbers
                                  it will be more unlikely for another unique number, 6, to appear next in this EC row
                                  we select to play O/E



Each EC group has its own listing but matches with the other 3 EC groups



      H/L      O/E     R/B


1   HHH    OOO    RRR


2   HHL     OOE    RRB


3   HLL      OEE    RBB

ETC



In the example above 6 was missing from all EC's horizontally, so we look on our group list and see



6   LLH      EEO    BBR




Now we select  EEO (#6)  not to appear in the next three spins in O/E. 


O/E had the most unique numbers.



I feel PB HORIZONTAL makes it a little more difficult for random and a game would only last 10 to 20 spins




AMK                                                                                                                     

Yesterday I was thinking about surpassing the table limit by asking a friend to play with me. I would give him a certain BR. When on a progression I reach the table limit my friend would start placing my second BR so that the progression actually continues.


For example, lets say the table limit on a certain bet is $200. Once I reach $200 my friend places the next betting amount to continue my progression. Say the next bet in the progression was $270. Now I would place $200 again and my friend using my BR places the remaining $70 etc




I have no got my head around all the aspects or even if there is a feasible table limit which can be used but thought it was interesting.

Hello everyone,


First post. I like the vibe of the forum.






PB 4 DL




PB 4 DL is based around JohnLegends Double Loss results for PATTERN BREAKER.  Strikerate 594/8  (74/1)


My assumption is that this win to loss rate will be seen even when we halve the number of patterns. PATTERN BREAKER revolves around 8 patterns. Will there be a proportional strikerate when we only use 4 patterns? Remember we are waiting for a loss, playing against a double loss.




On all EC's we can find 4 patterns:


HL         OE        RB
HH        OO        RR
LH         EO        BB
LL          EE        BR




We translate this into numbers 1 through 4


1        1        1
2        2        2
3        3        3
4        4        4




We start playing just as regular PATTERN BREAKER, but only with 4 patterns on each EC.


We wait for a loss on one of the EC's


R/B


3
3
1
3
2
4....  pattern 4 (last unhit pattern) appears directly after pattern 2


Now we observe the other two EC's which have not seen a loss yet.


First EC to qualify we bet against the last unhit pattern to appear directly.


H/L


2
1
2
4
..   here we bet against pattern 3 (LH) appearing directly. A three unit bet: 1,2






I am inclined to think that we can expect a proportional Double Loss strikerate as with PATTERN BREAKER.


I hope to start playing this live soon.