Roulette Ask Me Anything About Betting the EC's (Even Chances)

ok Last question:

Timewise, how long took it let's say from Newbie till your Winning method ?

 

If Spike was attending a college class regarding algebra, and the professor wrote a lengthy formula on the blackboard and asked Spike to conclude the answer, here's what he would say: Are you out of your mind, don't you understand as I do that the writing on the blackboard does not really exist, it's all in your mind. And therefore, it is impossible to have answers to things written on blackboards.I have to wait until that formula is playing my game, then I'll give you the answer.

 

Oops, I keep saying profile picture when I should say Avatar.

 

Yes, I would be happy to play online. How would that be accomplished?

 

that one pattern he mentioned wins half of the times 50-50 but flat betting way evantually it still stays - like with others
no matter you wait 13-2 13-3 winning virtually or not
you will have atleast 20 units drawdown with only that one trend

 

Dave,

Here are the waiting times for doublets and triplets.

Doublets

HT and TH = 4
HH and TT = 6

Triplets

HHT, HTT, THH and TTH = 8
HTH and THT = 10
HHH and TTT = 14

 

I never heard the word 'edumacated' used before, so I looked up its meaning. Your usage of it is very fitting.

 

Okay, so can that knowlwdge create a betting strategy? When you use =4 and so on, what does that mean? I know it meand wait time, but wait time in so many outcomes or what? Can you demonstrate this in a long series of outcomes?

 

Yep sure, of course HH or TT can mean anything ranging from RED RED, BANKER BANKER or even WIN WIN in a W/L registry.

So taking this W/L registry for example which someone posted the other day.

L W LLLL W LL W LL W LL W L W L W LL W LLL WW

The waiting time for the WIN WIN here was 28 spins. (way over the average of 6) some strong negative fluctuation.

The waiting times for LOSS LOSS here was 4,2,3,3,3,7,3. Add all them up = 25/7 = 3.5. much less than the average of 6.

The wait times for WIN LOSS is 3,5,3,3,3,2,2,2,3. Add them all up = 26/9 = 2.8. less then the average of 4.

The wait times for LOSS WIN is 2,5,3,3,3,2,2,3,4. Add them all up = 27/=9 = 3. less than the average of 4.

I suppose this is a bit of an extreme example but it explains things in a basic fashion without getting too fancy and following all three EC will quite often produce some skewed variance in one form or another where you can take advantage.

 

Empirical. Why?

& the populus is only +4000 spins -- for that exact version, taken further steps after thay, it has been improved, so it ain't what statistician would say a big sample. But enough to draw some sonclusions.

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Chart for Markov Chains expectation is done, although a bit simplified as its as it is complex enough already, since it can be stretching over so many spins .. including so many steps & variations it can unfold -- .. & that's only the major structure skeletron .. there are even variations upon that in play .. contingent on the variance type = what is currently happening, what is unfolding, that's not accounted for in the chart itself = basically, taking advantage of the temporary cohesion .. until the indication of it fizzling out, defaulting to the skeletron itself; plus not including some recent upgrades;

its pending .. to be done, by someone, since I do not possess enough skills in this field, yet.

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18x EC ain't thaat rare though .. maybe two in a row, perhaps with a zero in-between is a bit more .. how should I say .. not that expected. But as we all know, that ain't no guarantee to go by, meaning dismiss or disregard it per se .. as variance doesn't go by averages, & can spit out serving you on a platter such extreme sequences the first time played.

It has to be accounted for & addressed approprietly, should it come, if at all, for whenever that might happen...
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But although that, 18x EC streak (repeat or chop), gives me a notion to go by, as in ..

Two-step Marti [base unit 1→2] = 0.001613 = so 3nines of certainty
Three-step Marti [base unit 1→2→4] = 0.001613² = 0.000002601769 = 5nines of certainty
For-step Marti [base unit [1→2→4→8] = 0.001613³ = 0.000000004196653394 = 8nines of certainty

it still doesn't address this specific thing properly, the form as it is in itself.

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Therefore, I am gonna print some data for context, for more potential input:

Running the progression instance once, reaching the exposition point with of (-150) with no-hit at all -- going over EC all the way to Quads, Corners payouts, would take min 49 spins, out of which 45 would be 2Q with 8ⁿ played/spin.

That in itself is already at 99.99827% certainty, so 4nines; adding 3EC→DZ+CL→DS→DS or 3EC→2DS+DZ in front, how much would that amount cumulatively?

Also, that in itself is already 360ⁿ played;
adding the attack would be with either 3EC (-1) 71ⁿ in the first five spins
or 3EC (-3) 95ⁿ played in the first four spins,
so in total over 49 no-hit spins either 410ⁿ or 438ⁿ played.

Furthermore, with each hit along the way, feeding the progression to extend further in a number of spins, the probability only heightens up or increases, due to extra bets being place along the way (← is this correct?); & usually, the extremest type of game = terminated event is in the region of ≈200+ spin .. at the point of reaching (-150).

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If that is correct, drawing the parallels with 18x EC probability streak intel, there's still something that does not get addressed, or put into perspective, just yet:

due to the average session length in spins being ≈300, potentially extending towards & including 400 spins
+
the length of a terminated game being in the region of 200 spins
+
a session ranging anywhere from 25-70 games, or on average 50 games = events

.. when applying Marti to the base unit, or bu in short (the next instance of the fundamental progression or event is run) = then the expected number of spins relative to the number of terminated games would range ..

• for a two-step bu Marti [1→2] to be successful & 140 games (taking the top of the event range) -- the next terminated even = back to back, must not happen the next 25-70 games .. or 160-400 spins; or 320-800 spins in total only one such event happens
• for a three-step bu Marti [1→2→4], presuming that two terminated events have happened in the interval of 320-800 spins, must not happen again in the next 160--400; or in total a back-to-back-to back terminated event must not happen within 480-1200 spins

Now, looking at it from the mathematical perspective -- based on the presumption that the probability raises with each hit realized; this is something that does not happen in the 18x EC streak .. & besides the empirical probability is already lower than the mathematical including 2Q bets only & excluding the first four attack bets.

So, what would be the real probability to go by for the two points above .. ie. the terminated events within that specific interval of spins?


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Suppose I wanted to use this as a betteing strategy. I would not use the pattern WW because it has the longest waiting time. So then, using the other 2 patterns WW and LL, would I place a bet on L when the last outcome is a L, and on a W when the last outcome is a W? Maybe a dumb question because I am not familiar with this therory. Thanks.

 

Actually, looking at your doublets HT and TH having a shorter waiting time than TT and HH. I am already testing a strategy with that in mind. I simply am betting on a pattern I call Alternating. When I see XXXO, I bet X. When I see OOOX, I bet O. This is doing real well. I explaian this is more detail on page 113, 17 posts down.

 

This is the easiest way to play it IMO.

I will run the same results from above in horizontal fashion.

L
W here is the LW doublet
L
L here is the LL doublet
L
L
W
L here is the WL doublet

The WW doublet has not appeared as yet and so if it continues to go AWOL, you will have either LW, WL or LL which means at least one L is going to be in the doublet giving you two bites of the cherry. Suppose you use a negative 12 progression, let's take a look and see what would have happened.

L bet 1. win
W you don't need to bet here.

L bet 1. win
L

W bet 1. lose
L bet 2. win

L bet 1. win
W

L bet 1. win
W

L bet 1. win
W

L bet 1. win
L

W bet 1. lose
L bet 2. win

L bet 1. win
L

W bet 1. lose
W bet 2. lose


+6 units using a relatively safe 12 neg progression.

Somebody might not want to use the doublets in groups of two spins, you could use it on a rolling spin by spin basis as well which would have resulted in +11 using a 12 neg progression.

 

Thanks, I will take a look.

 

Thinking about it, that's a bit like Nathan's anti streak method. You have had a streak of 3 and although nothing is due, streaks don't go on for ever.
Taking into account that you have just seen the first chop, it's not the worst thing to do.

 

When you say Nathan's anti-streak method, you surely are not talking about Nathan Detroit are you?

 

The one and only! (There is only one)
One of his preferred methods is to wait for three of the same and then bet for the opposite.
We all have a rough idea of what's what despite all the banter. Take into account that 1's = 50%, 2's = 25% and 3's = 12.5% and he says he hates the maths. LMAO!

 

Well, isn't Nathan the one who said there is no strategy for EC, but yet this is his preferd methods? You cannot get ahead with this strategy flat betting. It has to be Martingaled.

 

I don't need a class in algebra or any kind of math. The method I use, guessing, is impossible to analyze with math. This is what drives the mathboyz insane.