Apple & Pear Cycles

[falkor2k15]

Here's my latest creation - based on a quote from Priyanka:

"line 1 - corresponding element in new set is 1.
Now on playing for repeat, you play line 1 and Pos 1 which is again line 1. Can you see that here instead of playing two
lines you are playing only one line as you are able to lose apples?"

Option 1: Dozen 1
Option 2: Dozen 2
Option 3: Dozen 3
Option 4: Order 1
Option 5: Order 2
Option 6: Order 3

One has to repeat and two pigeons can be the same dozen!

[ego]

I don't know if this helps, but I will mention it.
One dozen has 50% probability to hit once within two attempts.
Binomial probability calculation.

Cheers

[Blueprint]

I wonder who you STOLE that information from, Falkor?

[falkor2k15]

I've updated the rules for carrying over the defining element of the Option Cycle, so that I can come up with uniform stats:
*Always carry over the last 2 options (from previous spin of the previous cycle)
*Outer Options Length = 1,2 or 3
*s = single repeat; d = double repeat

[Blueprint]

I wonder who you STOLE that information from, Falkor?

REPEAT

[Joe]

I don't know if this helps, but I will mention it.
One dozen has 50% probability to hit once within two attempts.
Binomial probability calculation.

Not trying to be a smartass or anything, but that's a bit of an overestimate.



link:://onlinestatbook.com/2/calculators/binomial_dist.html

[ego]

Well, maybe I should say around 50% probability.

EC
Dozen 1 in 2 attempts
Line 1 in 4 attempts
Corner 1 in 6 attempts
Street 1 in 8 attempts
Split 1 in 12 attempts
Single 1 in 25 attempts

[falkor2k15]

Let's have a look at these stats then...

[falkor2k15]

Strange stats... swap dozen 1 for dozen 2 or 3...

[falkor2k15]

I tested with 9 options (overlapping) that result in more pigeons than pigeonholes as per Dyk's suggestion:
Option 1: Dozen 1
Option 2: Dozen 2
Option 3: Dozen 3
Option 4: Order 1
Option 5: Order 2
Option 6: Order 3
Option 7: CL1
Option 8: CL2
Option 9: CL3

I tried playing the outer cycle all different ways, but it just broke even. Therefore, this proves that Priyanka's idea about losing apples doesn't result in any profit either - it's still a collection of static independent bets. So it's becoming increasingly clear after 3 years that Non-Random/combinatorics cannot be applied to Random to make profit - though some really great ideas that certainly warranted such a long-term project, but unfortunately we have to accept that Random cannot be beaten.

[wiggy]

Falkor, just out of curiosity, did you ever look into Baccarat for VdW and Cycles etc...

For an example...

1. BBBB
2. BBBP
3. BBPB
4. BBPP
5. BPBB
6. BPBP
7. BPPB
8. BPPP
9. PPPP
10. PPPB
11. PPBP
12. PPBB
13. PBPP
14. PBPB
15. PBBP
16. PBBB

For cycles, suppose group 7 has just appeared = BPPB.....Now assuming cycle 7 doesn't appear again, you have 15 left.
You could arrange them into dozens as follows....
c8
c9
c10
c11
c12 = dozen 1

c13
c14
c15
c16
c1 = dozen 2

c2
c3
c4
c5
c6 = dozen 3

Voila! You now have dozens. (although granted the 0, in this case c7 carries a bigger edge than in roulette)

Baccarat is an interesting game because the B outcome is more likely than the P outcome, longer B events occur more frequently than longer P events, while shorter P events (especially 1s P) occur more frequently than shorter B events.

Also for anyone interested, Nicksmi wrote some interesting stuff regarding VdW/Non-Random over on the Betselection forum just recently.

link:s://betselection.cc/roulette-forum/use-mathstatistics-to-beat-roulettebaccarat-part-2/msg64190/?PHPSESSID=680sc45avj56u339qu5ulfcun7#new

I am sitting here thinking why be a slave to the roulette wheel when a lot of this stuff could be applied to other areas of gaming which may offer better opportunities.

cheers

[falkor2k15]

Hi Wiggy,

Could you please translate to roulette as I don't know Bac?

I've already tried VdW and cycles in many roulette contexts. I've made 3 options/custom dozens from the output of EC cycles and whatnot. I even tested your ideas regarding Penny's game/non-transitive betting - translating it to cycles:

EC Triplet   Average Waiting Time (spins)

111   14
222   14
121   10
212   10
112   8
122   8
211   8
221   8

EC Cycle   Average Waiting Time (cycles)

121   10
212   10
11   6
22   6
122   6
211   6

Unfortunately, it's not applicable in terms of edge in Roulette, a game which can be described through many mathematical truths/models in as far as number combinations go - but there's no sign of any exploit/edge/profit - maths doesn't help nor claim to help us there. Ramsey theory doesn't know we are playing Roulette with unfair payout odds and trying to overcome it.

[wiggy]

Falkor, with the cycles, I have had just as much success playing off the cuff as opposed to anything else.

I like to run the dozens and columns as two seperate streams and it really does seem easier to read than other types of random bet selection. I can't say why, but so far, for me at least, it just is and it's a pretty uncomplicated way to play.

What I am looking for is either 4s or 4d with an intermittent 3s or 3d.

So here is....

4
4
3
4
3
4
4
3
4

If you are playing for a 4, it doesn't matter if it's going to produce s or d because you know what the 3rd result is going to be either way.

1
3
2 (played for the 2)
3 = 4d

3
1
2 (played for the 2)
1 = 4d

2
3
1 (played for the 1)
1 = 4d

2
1
3 (played for the 3)
1 = 4d

1
2
3 (played for the 3)
2 = 4d

2
1
3 (played for the 3)
2 = 4s

You can't get much simpler than that.

My initial thoughts were to ignore the 2s (of course 2 always has to be an s) and concentrate on either the 3s or 3d and ignore any 4 (either s or d) altogether....but it's messy. When the 3's come, they are solid but we don't which way they are going to go (s or d) Runs of 2s can be done with a positive progression....runs of 4, I play like I have just shown.

[falkor2k15]

The problem is, Wiggy, that S and D are completely independent - as is each cycle - don't believe the stories that dependency occurs under the condition of a repeat.

S is a 62% chance bet at variable spins (1-3) with proportionate payouts (in the long term) resulting in break even.
S uses multiple bets per cycle that cannot be stitched and/or parlayed - negative progression
D is a 38% bet (opposite to S) again with proportionate break even payouts.
D uses multiple bets per cycle that CAN be stitched and/or parlayed - positive progression