144 Number Roulette

[falkor2k15]

In the past I created 24 Number Roulette over 2 spins (stitched) using quads and lines (4 x 6 = 24), but this time I've created 144 Number Roulette using streets and streets (12 x 12 = 144)!

As you know Roulette is not about 37 numbers, but about proportionate risk/reward and trying to avoid the table limits. Unfortunately, by adding an extra spin for every outcome we double the house edge, so European Roulette becomes like an American wheel.

The reason I want to test this is for the following reasons:
1) Is the maximum cycle length over 1 million spins closer to CL1 or CL144? With 37 numbers it's closer to CL37 (CL25 to be exact or thereabouts)
2) Do all 12 streets get hit on each side of either 2 spins before either of the 144 pigeons repeat because that would affect our ability to bet certain pigeons? (also see next point below)
3) Can we "bet on behalf" of the streets using Halves, Dozens and Lines and win before the main repeat happens each and every time?



Any comments?

[Firefox]

It's an interesting concept. You can create any number roulette.

A common option is 74 number.

Bet an even chance. If it loses STOP -1
If it wins parlay the 2 onto a number.
If it loses STOP -1
If it wins STOP +71

So you either lose 1 or win 71. Looks attractive like most positive progressions as you use the "house money" to double your bet.

But ultimately losing, as you only make the big play half the time and subject bets to the edge from two spins.

[falkor2k15]

The problem with 70 odd number roulette is that if both even chances have appeared then it's not possible to express a bet that covers multiple pigeons. This problem doesn't occur with 36 x 36 number roulette obviously, and I'm hoping it won't be a problem with 144 number roulette based on 12 holes x 12 holes.

Something else just occurred to me that I haven't tested before. We can turn this (144 Number Roulette) back into a single spin game! We simply carry over the last street as forming the 1st street of the next outcome. Therefore, we only ever make each bet over a single spin - but the bet encompasses both sets of 12 + 12 streets akin to betting twice using only one spin. For example, if Street 4 had hit the most on both sides then that would be the street we should bet as representative of both spins that make up the one outcome. Hard to explain. But I won't mess with this simulation to test that particular concept, as I have another sim based on quads made up using High and Low:
LL = Quad 1
LH = Quad 2
HL = Quad 3
HH = Quad 4

So if we have Quad 1 and Quad 2:
12...
=
LL LH H...
Then we have more Lows that have appeared compared to Highs, so we should bet Low next or perhaps wait till we get something like:
LL LH HL L
and then bet Low!

Anyhow, that won't take too long to confirm as a small digression from the main test - would be nice to eliminate.

[Firefox]

I'm familiar with the principle, but don't believe it increases your edge as at some stage you have to make a bet, to win anything.

Whatever you call that bet, be it something to complete a certainty based on a cycle or that an AP that must form, or something else, the possibility is always that there are two or multiple ways to complete that item.  And the pay off associated with those ways does not reflect the risk of making those bets.

Therefore you will lose on average, and chasing those losses to achieve a high chance of success will be caught by house limits well before the chance of losing is negligible.

This has all been simulated previously  for multiple different betting patterns with far greater computer power than we have access to.

[falkor2k15]

Sure - I don't hold much hope. Theoretically, it will always be a break even game made up independent bets. However, playing above 37 numbers is not something I've ever seen discussed, and it was hinted by Priyanka's teacher, reddwarf, that this could be the key to winning. So far all the concepts they mentioned haven't escaped the break even game, but this one has not been properly tested and could be described as playing the game in a different dimension so to speak. If anything, I'm happy just to get further clarity on why we can't win, eliminating further concepts as well as understanding the table limits better and why another 1K can't give us the leeway to recovery our losses and find a positive progression that results in edge. So only by practical experimentation can we truly prove something wrong. Again, I'm always looking for new perspectives to an old problem, and who knows: these tests could have some surprises in store. So far the cycle lengths haven't gone beyond the 40s, so are way closer to CL1 than CL144. Also, playing this kind of game over 2 spins per single outcome might shed more light on the Birthday Paradox; something may reveal itself that is not apparent when testing under regular conditions.

[Firefox]

Nothing wrong with testing and good luck in your quest!

[falkor2k15]

I only tested 500K spins in the end - but the pattern is clear: all cycles based on 144 pigeons ended before spin 50 (Average is CL13 compared to CL6 for 37 numbers):

[falkor2k15]

There isn't many cycles that use up all 12 streets on either side of the dual spin outcome:

NCL22 8 10
NCL12 10 9
NCL27 11 12
NCL22 11 9
NCL5 4 3
NCL15 7 10
NCL19 10 7
NCL21 10 9
NCL25 10 11
NCL12 7 6
NCL19 8 10
NCL4 4 4
NCL15 7 9
NCL20 10 8
NCL10 6 7
NCL11 8 7
NCL30 12 12
NCL11 9 7
NCL20 11 11
NCL18 10 8
NCL9 6 7
NCL26 9 11
NCL26 11 11
NCL7 5 5
NCL9 7 7
NCL13 7 7
NCL12 7 9
NCL6 5 5
NCL22 11 11
NCL10 8 6
NCL13 8 10
NCL18 10 10
NCL27 12 11
NCL9 5 6
NCL18 9 8
NCL12 8 7
NCL19 9 10
NCL10 7 9
NCL19 9 10
NCL8 7 4
NCL14 8 8
NCL17 10 8
NCL24 12 11
NCL23 10 12
NCL16 8 9
NCL14 9 8
NCL30 11 11
NCL20 10 10
NCL17 11 11
NCL2 2 2
NCL13 6 8
NCL4 3 3
NCL7 6 5
NCL21 10 10
NCL13 9 7
NCL7 3 6
NCL7 4 6
NCL13 8 9
NCL15 9 8
NCL25 9 11
NCL19 10 8
NCL9 6 6
NCL5 4 4
NCL12 8 10
NCL14 9 8
NCL20 11 10
NCL10 5 7
NCL9 7 7
NCL10 7 5
NCL3 3 2
NCL5 4 3
NCL20 10 11
NCL20 12 9
NCL27 11 11
NCL23 12 11
NCL19 10 8
NCL21 10 10
NCL25 11 11
NCL24 10 10
NCL10 6 7
NCL8 6 6
NCL29 11 12
NCL15 9 9
NCL8 6 6
NCL12 8 7
NCL10 9 8
NCL15 8 7
NCL23 10 11
NCL12 9 6
NCL15 11 10
NCL14 8 7
NCL19 8 11
NCL6 4 3
NCL7 5 5
NCL13 7 9
NCL15 9 8
NCL7 4 4
NCL24 11 10
NCL19 9 10
NCL19 9 11
NCL12 7 6
NCL14 8 11
NCL19 10 8
NCL31 12 11
NCL5 4 3
NCL16 8 9
NCL7 7 7
NCL5 4 4
NCL8 6 6
NCL12 6 7
NCL21 11 9
NCL6 3 4
NCL13 8 8
NCL9 6 6
NCL18 10 8
NCL20 10 9
NCL23 11 11
NCL17 8 9
NCL14 7 8
NCL12 9 7
NCL19 10 10
NCL10 7 6
NCL6 5 4
NCL20 9 11
NCL31 12 11
NCL9 6 6
NCL10 6 6
NCL14 8 8
NCL13 7 6
NCL9 5 6
NCL28 12 11
NCL9 6 6
NCL17 9 9
NCL9 5 6
NCL9 6 5
NCL20 9 9
NCL15 9 10
NCL14 8 7
NCL5 3 4
NCL21 10 9
NCL8 6 5
NCL3 3 3
NCL3 3 3
NCL11 7 8
NCL20 11 10
NCL4 2 3
NCL16 11 9
NCL15 8 9
NCL21 9 11
NCL12 7 7
NCL14 8 8
NCL3 2 2
NCL7 4 5
NCL8 5 5
NCL20 8 9
NCL11 6 7
NCL25 12 11
NCL13 8 7
NCL7 4 6
NCL18 8 10
NCL7 4 4
NCL9 7 6
NCL9 7 8
NCL13 9 7
NCL11 7 6
NCL2 1 1
NCL22 10 11
NCL25 10 10
NCL21 11 10
NCL11 7 7
NCL19 9 10
NCL13 7 8
NCL20 11 8
NCL16 9 9
NCL16 9 7
NCL9 5 6
NCL1 1 1
NCL9 5 6
NCL19 10 11
NCL6 5 4
NCL6 3 4
NCL22 11 10
NCL21 7 12
NCL23 10 11
NCL21 11 11
NCL18 9 10
NCL10 8 8
NCL20 10 10
NCL22 9 10
NCL15 9 10
NCL23 10 10
NCL20 10 10
NCL18 11 9
NCL6 5 5
NCL23 11 10
NCL29 10 12
NCL13 9 7
NCL20 11 10
NCL11 7 8
NCL8 5 5
NCL18 8 9
NCL18 9 8
NCL19 10 9
NCL9 5 5
NCL23 10 10
NCL6 3 3
NCL5 4 4
NCL3 2 2
NCL6 5 5
NCL7 6 4
NCL4 4 4
NCL13 7 7
NCL17 10 9
NCL19 12 10
NCL19 10 10
NCL10 7 6
NCL19 10 9
NCL13 8 10
NCL5 5 5
NCL21 10 11
NCL4 2 3
NCL13 6 9
NCL19 9 10
NCL25 11 11
NCL21 11 12
NCL23 11 11
NCL6 5 4
NCL13 6 9
NCL16 10 9
NCL10 5 7
NCL3 3 3
NCL8 5 7
NCL17 10 8
NCL3 3 3
NCL8 7 6
NCL9 6 6
NCL15 8 8
NCL9 7 6
NCL21 10 12
NCL8 7 5
NCL12 8 7
NCL12 7 8
NCL16 9 10
NCL12 7 7
NCL12 9 7
NCL5 5 5
NCL4 3 3
NCL16 10 9
NCL13 8 7
NCL12 8 7
NCL4 3 2
NCL15 9 7
NCL28 12 12
NCL26 10 10
NCL13 8 6
NCL20 10 10
NCL24 11 10
NCL6 4 5
NCL26 12 11
NCL13 9 8
NCL15 9 9
NCL14 8 8
NCL33 12 12
NCL11 8 8
NCL13 8 8
NCL9 8 7
NCL13 9 8
NCL3 2 2
NCL16 8 9
NCL10 7 6
NCL9 7 5
NCL15 10 7
NCL8 5 6
NCL4 3 3
NCL7 5 6
NCL15 10 8
NCL16 10 9
NCL22 9 10
NCL11 6 7
NCL23 10 9
NCL5 5 5
NCL9 7 6
NCL23 11 11
NCL10 8 8
NCL14 9 8
NCL11 8 6
NCL30 12 10
NCL17 10 7
NCL18 10 11
NCL1 1 1
NCL6 5 4
NCL20 11 10
NCL21 12 10
NCL6 4 4
NCL7 5 5
NCL18 10 8
NCL4 3 3
NCL5 3 4
NCL1 1 1
NCL10 8 9
NCL16 10 9
NCL5 3 4
NCL14 8 9
NCL9 6 7
NCL29 11 12
NCL16 9 8
NCL14 9 9
NCL15 9 7
NCL7 6 6
NCL16 10 9
NCL13 7 8
NCL21 10 11
NCL26 11 10
NCL17 11 8
NCL14 9 8
NCL14 9 10
NCL8 6 6
NCL19 11 10
NCL13 6 11
NCL18 11 10
NCL11 6 8
NCL28 11 11
NCL16 9 11
NCL22 9 11
NCL6 5 4
NCL19 9 10
NCL24 9 12
NCL10 7 8
NCL6 5 5
NCL8 5 4
NCL13 7 9
NCL6 5 5
NCL16 8 10
NCL2 1 1
NCL4 3 2
NCL28 12 11
NCL17 9 11
NCL12 6 8
NCL21 10 11
NCL13 10 9
NCL24 11 11
NCL17 9 9

The shortest cycle was CL19!  We'll just have to see if betting on behalf wins out before the deadlock in the next (practical) test...

[Maui13]

I'll be honest - I HAVE NO CLUE .... what the system is or how to even play it.

Sorry - I'm not the sharpest tool in the shed.

[Firefox]

I'll be honest - I HAVE NO CLUE .... what the system is or how to even play it.

Sorry - I'm not the sharpest tool in the shed.

Well, there are 12 streets and if you bet on street 1 followed by any of streets 1 to12 you have 12 different bets. Street 2 followed by streets 1 to 12 gives you another 12 and so on, all the way up to street 12, so you can effectively bet on 144 virtual numbers. He's looking at the most common distances between numbers repeating and seeing if there may be a betting advantage.

[Herby]

144 Number Roulette, 10.000.000 Cycles

Cyclelengths:
{{2, 69193}, {3, 137906}, {4, 203662}, {5, 267035}, {6, 324173}, {7, 374291}, {8, 418669}, {9, 455543},
{10, 484646}, {11, 505041}, {12,515746}, {13, 520039}, {14, 516634}, {15, 505082}, {16, 489603}, {17, 468236}, {18, 442049}, {19, 413297}, {20, 381228}, {21, 348156}, {22, 314464}, {23, 280417}, {24, 249395}, {25, 219893}, {26, 190275}, {27, 163621}, {28, 139195}, {29, 116935}, {30, 97576}, {31, 81176}, {32, 66222}, {33, 53623},
{34, 42911}, {35, 34271}, {36, 26523}, {37, 20894}, {38, 16021}, {39, 12323}, {40, 9243}, {41, 6902}, {42, 5196}, {43, 3717}, {44, 2726}, {45, 1928}, {46, 1324}, {47, 938}, {48, 734}, {49, 445}, {50, 286}, {51, 192}, {52, 149},
{53, 76}, {54, 70}, {55, 42}, {56, 19}, {57, 16}, {58, 11}, {59, 12}, {60, 5}, {61, 2}, {63, 1}, {64, 2}};

2 times maximum cyclelength 64   

[Maui13]

Well, there are 12 streets and if you bet on street 1 followed by any of streets 1 to12 you have 12 different bets. Street 2 followed by streets 1 to 12 gives you another 12 and so on, all the way up to street 12, so you can effectively bet on 144 virtual numbers. He's looking at the most common distances between numbers repeating and seeing if there may be a betting advantage.

Thank you for the explanation 

[falkor2k15]

Well, I'm not even sure if I'll be playing 1 cycle let alone 10 million cycles!  Nevertheless, here's my thoughts on this so far:

Firstly, it feels like increasing the number of pigeons is counter-productive to how we envisage Roulette ought to be played, i.e. catch the repeat before reaching the end of the cycle and the table limits. Here we may not encounter cycle lengths in their 60s during our lifetime, but certainly in the 40s. And since we are paying at least 1 unit per spin then we aren't helping the cause with 144 pigeons instead of 36! Instead it feels like we should try to make a custom game of, say, only 12 pigeons ala Quadruplets (came close to almost avoiding CL12 entirely). Also, the payout for CL1 is ridiculously high @ +131 instead of allowing us to gain +1 at any one time throughout a game that should be expected to last longer without reaching table limits; 144 Number Roulette being contrary to our idealization.

However, the concept of catching a repeat before the end of a number cycle now feels even more synonymous with a negative progression, and here the negative progression game becomes ever more complicated to play and not viable whatsoever. Betting on behalf was already to difficult for practical play in the normal realm as demonstrated by Mist Trap, and here we would have many additional calculations based on the parlaying between the dual spins. 

So let's turn to the positive progression game, where things might actually look more promising and could get quite interesting! After spin 1 if we want to bet the 143 sleepers then it could cost us 143 units as expected:
Bet 1 unit on 11 streets - lose = -11
Bet 12 units on 11 streets = win = +12
=+1

However, if we won the first bet then we could take the +1 without playing the 2nd spin, and then the cost is only 11 units to cover 143 sleepers!

Now after the 2nd unique has shown we could be using up 3 streets or 4 streets (1-2 on each side of the dual spin outcome). If we are using up 4 streets then it becomes even cheaper to cover the sleepers:
Bet 1 unit on 10 streets - lose = -10
Bet 11 units on 10 streets = win + 11
=+1
So we only risked 120 units to cover 142 sleepers, and if we won the first bet then we only risked 10 units to cover the 142 sleepers.

Of course if we lose then we lose 120 units for the sake of +1, but isn't that comparable to 36 numbers or 6 lines where we lose -1 integer to the size of the number set? So if anything that could give us some advantage, and we could choose to end the game early if we had only 3 streets being used up after the 2nd spin. Suddenly, using up all 12 streets doesn't now seem like a threat anymore because we can quit anytime we are ahead.

The only way that the 36 number game can compare to the 144 number game is if our first few single hits fit snuggly a single street - nigh on impossible! If we did happen to strike numbers 1,2,3 on spins 1,2,3 then we could bet streets 2-12 to cover all the sleepers, but that's not going to happen. So this is where our custom 144 Number Roulette may offer us some advantage.

[falkor2k15]

So there's a very interesting perspective/hypothesis that's developed from this topic that was invisible before - still remains to be proven:

Helps Positive Progression

*More pigeons
*Individual outcomes made up of more spins

Helps Negative Progression

*Less pigeons
*Individual outcomes made up of less spins

The last line is the opposite of using 2 streets as 1 number and harks back to Priyanka's Funny Sequences topic and the Quads example above - based on looking back at the last 2 spins and playing the next spin knowing that it will also be the first spin of the next dual spin outcome, hence playing "2 spins as 1 spin" instead of "2 spins as 2 spins". This is because we carry over the last spin result to the next outcome like we carry over the defining element of a cycle to the next cycle.

LL = Quad 1
LH = Quad 2
HL = Quad 3
HH = Quad 4

2 spins as 2 spins as 1 outcome

Quad 1...
LL...

Quad 1 Quad 2...
LL LH...

Quad 1 or Quad 2 can repeat next outcome.

2 spins as 1 spin as 1 outcome

Quad 1...
LL L...

Quad 1 Quad 2...
LL LH H...

Above there's no chance that Quad 1 or 2 can repeat on the next spin, so we get a free virtual spin that could help us only in a negative progression game! 

[MoneyT101]

You’re  point of view is so off....

Plus trying to do the same thing you were doing with less numbers.  So you’re going to get the same results 

🤦‍â™,️😂😂