So that's a minimum of 10 cycles to get a repeat! Going by, say, dozen repeats, we can calculate the cost to recover our losses:
DOZEN REPEATS
Dozens per betUnitsWinLose
0
112-1
221-5
131-8
291-26
1142-40
2411-122
1622-184
21851-554
12782-832
28331-2498
112502-3748
237491-11246
156242-16870
2168711-50612
1253072-75919
2759201-227759
11138801-341639
23416401-1024919
So to play Priyanka's Dozen Cycles we would need an astonishing BR of 1,024,919 units - not to mention a big favour from the House to increase their table limits to 341,640 x 2 = 683,180 units for the final 2 dozen bet!?
But guess what? Let's say you find yourself in the following scenario:
CL1 = 0, CL2 = 0, CL3 = 1
O1 = 0, O2 = 1, O3 = 0
Defining Element = 1,2 or 3 (doesn't matter which)
One of them (CL, Order or DE) MUST repeat on the very next Cycle! In fact, most times the repeat will be on the Defining Element, followed by Order 2. Least likely to repeat out of those 3 would be CL3 - but one of them constants, nevertheless, MUST repeat!
The maximum for any repeat to occur is 4 cycles (outer CL3) - but we expect a repeat to happen mostly on the 2nd cycle, i.e. the first cycle we bet on - influenced mainly by the defining triple combi carried over from the previous set/outer cycle:
CL100 o100 d100>>>CL110 o110 d101>>>CL120 o120 d111
CL010 o010 d010>>>CL020 o110 d020
CL010 o100 d010>>>CL020 o110 d011
CL010 o010 d001>>>CL110 o110 d002
CL100 o100 d001>>>CL110 o200 d002
CL010 o100 d001>>>CL020 o110 d011
CL010 o010 d010>>>CL110 o110 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL110 o200 d020
CL010 o100 d010>>>CL110 o200 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL110 o200 d020
CL010 o100 d010>>>CL020 o110 d110
CL010 o010 d100>>>CL020 o020 d110
CL010 o010 d010>>>CL020 o020 d110
CL010 o010 d100>>>CL110 o110 d200
CL100 o100 d100>>>CL200 o200 d200
CL100 o100 d100>>>CL110 o110 d110>>>CL210 o210 d120
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL101 o101 d011>>>CL111 o111 d111>>>CL121 o121 d112
The defining combi has more chance of being defined the "same" as previous cycle - but contrary to singular DE repeats and anticipating DE to be the same over the course of any 3 "inner" cycle lengths (we cannot say which length the DE repeat will likely fall on) we now expect the defining element (representing the triple combi) to be no more than "outer" CL1 only.
Continue reading...
DOZEN REPEATS
Dozens per betUnitsWinLose
0
112-1
221-5
131-8
291-26
1142-40
2411-122
1622-184
21851-554
12782-832
28331-2498
112502-3748
237491-11246
156242-16870
2168711-50612
1253072-75919
2759201-227759
11138801-341639
23416401-1024919
So to play Priyanka's Dozen Cycles we would need an astonishing BR of 1,024,919 units - not to mention a big favour from the House to increase their table limits to 341,640 x 2 = 683,180 units for the final 2 dozen bet!?
But guess what? Let's say you find yourself in the following scenario:
CL1 = 0, CL2 = 0, CL3 = 1
O1 = 0, O2 = 1, O3 = 0
Defining Element = 1,2 or 3 (doesn't matter which)
One of them (CL, Order or DE) MUST repeat on the very next Cycle! In fact, most times the repeat will be on the Defining Element, followed by Order 2. Least likely to repeat out of those 3 would be CL3 - but one of them constants, nevertheless, MUST repeat!
The maximum for any repeat to occur is 4 cycles (outer CL3) - but we expect a repeat to happen mostly on the 2nd cycle, i.e. the first cycle we bet on - influenced mainly by the defining triple combi carried over from the previous set/outer cycle:
CL100 o100 d100>>>CL110 o110 d101>>>CL120 o120 d111
CL010 o010 d010>>>CL020 o110 d020
CL010 o100 d010>>>CL020 o110 d011
CL010 o010 d001>>>CL110 o110 d002
CL100 o100 d001>>>CL110 o200 d002
CL010 o100 d001>>>CL020 o110 d011
CL010 o010 d010>>>CL110 o110 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL110 o200 d020
CL010 o100 d010>>>CL110 o200 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL110 o200 d020
CL010 o100 d010>>>CL020 o110 d110
CL010 o010 d100>>>CL020 o020 d110
CL010 o010 d010>>>CL020 o020 d110
CL010 o010 d100>>>CL110 o110 d200
CL100 o100 d100>>>CL200 o200 d200
CL100 o100 d100>>>CL110 o110 d110>>>CL210 o210 d120
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL200 o200 d020
CL100 o100 d010>>>CL101 o101 d011>>>CL111 o111 d111>>>CL121 o121 d112
The defining combi has more chance of being defined the "same" as previous cycle - but contrary to singular DE repeats and anticipating DE to be the same over the course of any 3 "inner" cycle lengths (we cannot say which length the DE repeat will likely fall on) we now expect the defining element (representing the triple combi) to be no more than "outer" CL1 only.
Continue reading...