PhP Pair-Completion

[algojack]

The first three tests would be

•   threshold for a SP to repeat, with spin-by-spin % distribution (SU= 29or30 spins)

I'll take them one at a time. What do you mean by a spin-by-spin-distribution?  A file with 10M cycles showing every spin would be far too big. What do you mean by "Threshold"? Is it the maximum number of spins before a split repeats?

[Ross]

You're probably right. But where's the harm?

No harm at all!

[TRD]

I'll take them one at a time. What do you mean by a spin-by-spin-distribution?  A file with 10M cycles showing every spin would be far too big. What do you mean by "Threshold"? Is it the maximum number of spins before a split repeats?

What I mean is like redhot's post, just spins -- in essence this output format.

Code Select Expand

rouletteforum.cc/index.php?topic=26049.msg228070#msg228070

Total cycles
spin1  ......   Cycles     .....      % relevant to Total Cycles
spin2  ......   Cycles     .....      % relevant to Total Cycles 
spin3  ......   Cycles     .....      % relevant to Total Cycles
   ...
spin16 ......   Cycles     .....      % relevant to Total Cycles
spin17 ......   Cycles     .....      % relevant to Total Cycles
spin18 ......   Cycles     .....      % relevant to Total Cycles



Threshold.
We know that practically DS & ST can hit all unique/different in a row.
For SU, Straight-Up, we know r]that practically the first repeat treshold is 29th or 30th, test done over 10.000.000 spins -- so there's a window of 7-8 spins till 37 total numbers.




TEST1:   What's the window for SP, splits two-number position, -- if any?
                   Bet any appeared SP, add another on each spin,
                   at repeat restart betting the last outcome as the first bet.
                   Populus: 10.000.000 spins, even better cycles if not too much of a ask

[algojack]

This is the table for Splits.

Code Select Expand


                                      NO OF CYCLES     AS %
------------------------------------------------------------
First Repeat Occurred @ Spin no.  2      555753       5.558%
First Repeat Occurred @ Spin no.  3     1049581      10.496%
First Repeat Occurred @ Spin no.  4     1399462      13.995%
First Repeat Occurred @ Spin no.  5     1554231      15.542%
First Repeat Occurred @ Spin no.  6     1512373      15.124%
First Repeat Occurred @ Spin no.  7     1310145      13.101%
First Repeat Occurred @ Spin no.  8     1016764      10.168%
First Repeat Occurred @ Spin no.  9      712423       7.124%
First Repeat Occurred @ Spin no. 10      444302       4.443%
First Repeat Occurred @ Spin no. 11      247508       2.475%
First Repeat Occurred @ Spin no. 12      120768       1.208%
First Repeat Occurred @ Spin no. 13       50952       0.510%
First Repeat Occurred @ Spin no. 14       18600       0.186%
First Repeat Occurred @ Spin no. 15        5566       0.056%
First Repeat Occurred @ Spin no. 16        1310       0.013%
First Repeat Occurred @ Spin no. 17         239       0.002%
First Repeat Occurred @ Spin no. 18          21       0.000%
First Repeat Occurred @ Spin no. 19           2       0.000%


These figures are a little too high because for simplicity I didn't include the zero. So the first row % should be closer to 2/37=5.41%, whereas they are actually closer to 1/18=5.56%. So as you can see, all splits arrived in the first 18 spins in only 2 of 10M cycles.

I'll code the system next (bet first split and then add each one as they appear).

[TRD]

@alogojack; great, thanks - that ticks one unknown off the list.
= there no open window on SPs to the first repeat related to the pool of all splits (18/18),
   similar to DS (6/6) & (ST 12/12) all can show in a row; & unlike the SU district where the
   practical window is (29-30/36; 10mil spin test).
   
   Still, the test shows that expanding the bet by an additional SP position on each & every spin, it only makes sense to bet till including perhaps spin 12, but later no more.
  Why - to keep the max cycle drawdown low, & rather recover the outstanding debt with consequent cycles till in nominal profit â†' restart next individual game in the session.

-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18= 9*19= -171
-1-2-3-4-5-6-7-8-9-10-11-(12)= 6*13= -66 (-78)
Big difference, ≈3x between (171) & (-66) = keeping drawdowns recoverable.

With first 5 spins being positive.

â†'


This relative to the Php Pair completion (version1), betting the pair-repeat or pair-counterpart only on each outcome appearance.

Of course the above ain't totally precise, just the first of the parameters to regard, still narrowing the extremes we play within a bit further.
Why - for when both pairs numbers are out .. the whole pair is bet, so on that particular spin the same number of SP positions as a bet are repeated.



although,
1) since all my system's betting is continuous - it would have been great if you have used the previous' cycle repeat =end of cycle SP as the first position used â†' thus the 'spin1' parameter would also show as relevant
2) include 0, for this is integral part of the game & approximations derived from neglect do not work in our favor, especially when designing a system
3) when opting for actual split positions I would determine that upon the numbers already out in recent spins, current or previous cycle - top & bottom columns outcomes have  (except numbers 36 & 34) three possible positioning options & middle column each 4 options (except number 35); effectively if number 9 is the previous cycle & in current cycle 8 shows SP8-9 would be placed if placing the repeat counterpart, or it pair-counterpart first
4) the SP repeat bell curve shows peaks at spins4,5 (5,6 in the chart), so it may make sense to bet for the repeat-counterpart first & then complete it with
eg. Number 8 shows =splitNⁿ5 = split8-10
      (x+y)=19 â†' splitN°5 + splitN°14  as pair conterparts



Can you first re-run the test two times, each enriched by the next of 1) & 2) points; disregard 3) & 4) for now. No betting amounts, still only distribution percentages to get really precise results.

[algojack]

2) include 0, for this is integral part of the game & approximations derived from neglect do not work in our favor, especially when designing a system

Ok. Are you also betting on the zero when it appears, or only the splits?

[quos]

eg. Number 8 shows =splitNⁿ5 = split8-10
      (x+y)=19 â†' splitN°5 + splitN°14  as pair conterparts

Hi TRD!!
what is split nº5 and split nº 14?

Thanks

[TRD]

Although you, playing SPs only, can get a bit more creative â†' the splits being arbitrarily labeled differently in each of the cycles ie. as the board fills up, when opting for a specific SP placement option (vertical & horizontal) & the formation derived from that as the board fills up, based on previous cycle outcomes coupling them with the already shown (to repeat, as a refinement that may exclude unnecessarily betting the cold numbers, pushing them towards the cycles' end =full board )

the most simple categorization per the screenshot, construct pair's complementing SP counterparts according to   (x+y)=19, zero excluded



Otherwise, if you are essentially playing the game on SU Pairs, betting both counterparts simultaneously; but rather than playing SUs from the beginning opting for a parachute-style attack that closes many games much quicker in nominal profit  &  moves from 2x DS (spin1, 1st pair), then  4Q (spin2, 2 pairs), 6ST (spin3, 3 pairs), 8SP (spin4, 4 pairs), 10SU (spin 5, 5 pairs)   --   I suggest that you, irregardless of the position type, couple the pairs according to (x+y)=36 for the coupling to be consistent all the way through. After all the goal is to have 2 numbers shown within a threshold of numbers shown producing a complementing pair, thus a hit. Also, the attack varries ad-hoc cycle-by-cycle, since with  fewer positions all the pairs shown can be already covered
eg.

spin1 #16
   16+20=36
   no need for 2DS, one DS covers the pair (-1)

spin 2 #16,19
    16+20=36, 19+17=36
    no need for 4Q; with 2DS both pairs are already covered, resulting in an exposition
    that allows for another DS-based spin pushing the need for Qs a spin onwards ..
    as per the nominal profit requirement (-2, total-3)

spin3 #16,19,25
    .. 22+14=36
    no need for 6ST; 2Q does the job (-2, total -5)

spin4 #16,19,22,25
     25+11=2=36
     no need for 8SP, nor 6ST;
     you wanna break even & continue with 2DS theonwards anew ? --
     bet 3Q (+6) = Q11-14, Q16-20, Q22-25; or Q7-11 instead if 7,8,10 is a recent show ..
     (-3, total -8)

spin5 #16,19,22,25,13
     5ST does the job covering all 5 pairs;
     ST10-12, ST13-17, ST16-18, ST19-21, ST22-24, ST25-27 (-6, total -14)

if spin5 hit (+6, total -2)
    no need to continue with SU,SP,ST, or Q -- 2DS does the job (+3, total +1)
    take the last two outcomes & start the pair-completion anew

spin6 #1
    1+35=36
     DS1-6, DS31-36 (-2, -4)

spin7 #1,13
     1+35=36, 13+23=36
     4Q = Q1-4, Q11-14 (since of 11&13 show), Q22-25, Q32-36
     
spin8 #33 Q32-36 hit (+5,total +1); game done â†' restart