Hi. I will be very glad for every comment about this. Is it correct? Is it wrong? Why? Criticism also welcomed.
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EXPLOIT RANDOM THEORY

I will explain this with help of the next game:

Rules of the game Coin Flip game:
- Flipping a fair coin
- We do not have infinity amount of units to bet
- Our lowest vs highest bet is 20x
- We decide about our next pick(we can also decide not to bet):H or T, after last one falls
- If we do make correct prediction for next H/T we get +1 unit
- If we do not make correct prediction for next T/H we lose -1 unit
- Infinity number of flips

First some calculations:
Number of trials:4
Number of combinations:2 on 4=16
All possible combinations( H: Heads; T: Tails)
1.HHHH
2.THHH
3. HTHH
4. HHTH
5. HHHT
6. TTTT
7. HTTT
8. THTT
9. TTHT
10. TTTH
11. HHTT
12. HTHT
13. HTTH
14. THTH
15. TTHH
16. THHT

The rule of the game is to bet on each flip (or not to bet on each flip;we can also decide not to bet some flip if we want). My betting strategy is to bet in the way that I have in 4 flips covered all the outcomes that I want (a group bet out of one bet): these combinations are 11,12,13,14,15,16. I have covered 6 out of 16 possibilities.

Odds probability of this is 6/16=0,375=37,5%. But that does not mean that our exact probability odds are the same. With our system we do not have classic binomial calculations(for example at flip no.2 there is not yes or no question and so on).

We win our system if this combinations (from combinations list above): 11,12,13,14,15,16 falls, and we lose if other 10 combinations falls. If we want to make winning system we need to hit our combinations in more than 37,5% of time.

With our betting system we are basically picking combinations with same end results (same difference of mean), that are closer to mean. We chose which combinations we want to bet on. The combinations that we do not want, we leave it to »house«. Offcourse, we are picking the combinations with same end result (same difference from mean;2H2T), that are closer to mean, especially because the probability odds are the same for every single combination.

How do we bet (we are basically betting every of 6 mentioned bets from no.1 flip to no. 4 flip at same time; Table shows how all of the bets look together):


Table 1

Betting strategy explained (from Table 1) on example #1:

We allways start with betting (we allways start with 6 units;we look at the line: on every flip) 3 units on T and we also bet 3 units on H, if T(1. Flip; we have 6 units) falls, we than bet 2 units on T and we also bet 4 units on H. If H (2. Flip; we have 8 units) falls, we than bet 4 units on T and we also bet 4 units on H. If H(3. Flip;we have 8 units) falls, we than bet 8 on T. If T falls(4. flip) we have won 16 units. Offcourse, we can miss a bet at a flip when a line tells us that we need to bet same amount of units on H or T, but because this is example how betting works I did not make any simplicity.
We repeat this betting system to infinity. If we lose our betting system before 4. bet. We wait until the end of 4.th flip, and start betting on 5th flip. We allways bet on our next four sequence of 4 coin flips. For example if sequences are:HTHH,TTHT. The bold sequence is not winning or losing for us(we are not interested in the bold part). Only four sequences, than next four and so on.

Explanation with graphs:

-(Source of all graphs: https://www.di-mgt.com.au/binomial-calculator.html; all graphs are modified by me)
-every line that I wrote on the graph is aprox. estimate


Graph 1:

Our graph(binomial probability distribution) is not a linear, it does not have straight lines. Therefor, we can exploit it (show leaks of randomness). We can exploit it exactly with our betting system.


Graph 2:

Above line1(l1) we have results 2H2T, between lines l1 and l2 we have results 3T1H and 3H1T, and below line l2 we have 4T and 4H. We are sure that there will be 50% of T and 50% of H in a long run. What can we conclude from this(at least 2 different things):The group of combinations with same end result vs others group of combinations(devided with the number so result is 1) have same % chance. Or, we can also conclude that the combinations that are closer to mean are more likely to happened than the one that are far from the mean. Last sentence is correct (at least with our betting system).


Graph 3:

In this graph m:l1,l2 is median for l1 and l2. It crosses our graph line at 3T1H and 3H1T.
Line 3 and 4 are tangents of a graph line at a dot, when our m:l1,l2 crosses graph line. As soon as we did grouping of combinations(and covered all 4. coinflips), l3 and l4 comes to effect(and graph turns from curvy to linear;this is what happens when we do our betting system: when multi grouping we get average: in this case average straight line).
Why l3 and l4 comes in effect:Will explain that with(we can also make a graph for this example) a simple game. We have 6 apples. 6 apples are in 3 groups. In group no.1, there is 1 apple, in group no.2 there are 2 apples and in group no.3 there are 3 apples. The goal of a game is to profit the most apples at each game. The rules of the games are that we pick one of this group of apples as of buy-in for the game(we can decide which group exactly we want), than give these groups of apples in the basket and draw random group out(we do not know which one we will pick from the basket because we have put them in black bags). In our game, we offcourse pick group no.1 (with one apple) as buy-in, because our average at all groups is 2 apples (straight line) and because group no.1 is under the average. We will average profit the most if we take basket no.1 as a buy-in. And this is aprox. the same thing that we are doing with our ERT bet.


Graph 4:

If dot (that we are interested in) on our graph line, is higher(has bigger y) than red lines l3,l4 (depends on which site it is) at same x(x axis), than it is overrated (it will fall less time that average binomial calculations tells us), if dot is lower than red line(has smaller y) at same x (x axis), than it is underrated (it wil fall more times that binomial calculations tell us).
Our 2H2T(green circle) are avg. lower at same x (axis x) vs red line(diff to red line at y axis), than all other possible combinations, therefor is underrated with binomial calculations(our 2H2T will fall more time than the aprox. number tell us). Offcourse percent difference will be very small at our real percent combination vs normal binomial percent combination, but it is there.

Therefor our 2H2T, will win more time than odds probability number tell us(37,5%). We can conclude that 2H2T will win more than 37,5% of the time. As for if we go back to our game, because this game is made with only 0EV possibilities and our conclusion shows that we will hit our predictions in more than 37,5%(0EV odds probability) of time. That also mean that if we change rules of the game, so it cointains very small fee(for example:0,0000001%) on every flip we are still beating the game with the system that I described here.

This graph is the source of your confusion. The true graph of the binomial n=4 has no curvy bell shaped line!. You drew that line in yourself. The original, true graph consists only of the 5 bars above 0,1,2,3,4 heads, the height of the bars representing the respective probabilities. The only thing you accomplish by drawing in the curvy line is to mislead yourself into following a rabbit down a rabbit hole into an Alice in Wonderland of nonsense.

If you think your curvy line represents some kind of randomness from multiple trials of flipping 4 coins then you are simply wrong. There are curvy lines to represent that. For example, if you flip 4 coins a million times and compute the 5 ratios of 0-heads, 1-heads, 2-heads, 3-heads, 4-heads to the 1 million 4-sequences flipped, then those ratios are random variables and they each will have a curvy line looking like a tall, narrow camel hump centered on the probabilities 1/16, 4/16, 6/16, 4/16, 1/16 respectively. Each curvy line will then represent a continuous probability density function approximating the probability distribution for the respective random variable ratio.


If you don't understand that language you need to study more probability. You gave a very interesting link Parrondo's Paradox at the end of your first thread.

https://en.wikipedia.org/wiki/Parrondo%27s_paradox


That's a fascinating and live subject. If your interest in it stimulates you to study probability theory so that you can really understand it then that would be great. However, if all you get out of it is an excuse to continue chasing this rabbit down this rabbit hole of nonsense then it's a waste of your time. I will not contribute to your delinquency by entertaining this idea of yours further.



PairTheBoard

(0.5 * 1) + (0.5 * -1) = EV
0.5 - 0.5 = EV
0 = EV

You do not need to do anything more complicated than this.







Is this a game to see how many threads can be started and linked to the next one before a containment thread or a ban?

Im probably just stupid and dont understand.

But basically what you are saying is:

Once a "certain" sequence comes, the next throw isnt entirely "random" anymore?

@PairTheBoard
TNX for reply. I only draw line for better understanding. I could do the same without graph line also. After all we do know that 2H2T+4T+4H vs 3H1T+1H3T=50% vs 50%(l1 and l2 from graphs).

@Lego05
I really do not understand how is the thing that I am doing wrong. I mean, what???? What I will wrote under is also meant for some trolls...

You will ban me for what? Is it wrong because I am "wasting" my time and studying some things that I am interested in and posting here why I think something may not be exact(and because I think some things deserves new thread:so people do not go looking in the middle of the thread for what I am saying:making easier for people to read)? Or is it wrong because I am presenting maybe for somebody something interesting(taking my own time for that)? And not just trolling without anything done by myself(like some do:I do not mean you by this)? So what if I am wrong... Or is it wrong what I am doing because I am discussing something that for some reason some mod do not like, so that means ban? After all we are all guests here and we should feel appriciated here. Maybe somebody is interested also in this topic. After all, we people know everything what is there to be known in whole universe... I am here to discuss some things.. I never even slighty insult anybody. But some people did insult me.. I mean ok, not much of a deal..but...I do not feel like explaining some things if like this... Not how it works in my book.. No need to ban me. I will never login in my account again. Or maybe I will ban myself.

Maybe somebody will find the threads that I opened interesting(and will read them with joy) or maybe somebody will make "something huge" out of these conversations...I think this is also cool...Maybe I am correct with first post in this thread...

@Ronny Mahoni
Hi. No. I am saying that maybe when we do a "grouping", some things that have 37,5% vs 63,5% chance, turns to 37,505% vs 63,495% chance (if we pick our spot correctly:graph 4) at fliping a coin.

That's a little sensitive. And you didn't answer the question. (it was a rhetorical question though anyway)

If you were posting all of this disingenuously and your real intent was to see how long you could make it go and how many threads you could run through, then the thing you would probably be banned for would be trolling (if you were banned at all - the threads might just be locked). If you are not doing that, then you would not be banned for trolling. However, I would not ban you or lock the threads in any event.




And this:

is really all you need to do.

All you are doing is betting on whether a coin flip is heads or tails. You have a lot of more complicated looking stuff in front of that which is used to determine whether you bet on heads or tails. But at the end of the day when you actually make a wager, what you do is bet on a coin flip being either heads or tails. If the coin lands on the one you chose, then you win and if the coin lands on the one other than the one you chose, then you lose.

Or:


Assume:
Chance of H = 50%
Chance of T = 50%
(i.e. a fair coin)



.......Sequence...............Probability of Sequence Occurring

1.......HHHH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
2.......THHH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
3.......HTHH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
4.......HHTH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
5.......HHHT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
6.......TTTT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
7.......HTTT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
8.......THTT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
9.......TTHT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
10......TTTH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
11......HHTT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
12......HTHT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
13......HTTH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
14......THTH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
15......TTHH.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625
16......THHT.....................0.5 * 0.5 * 0.5 * 0.5 = 0.0625


Each sequence is equally likely (6.25%). And to check it and make sure we did not miss anything:

0.0625 * 16 = 1

So, yep, that adds to 100%, so did not miss any sequences.


So, there are 6 equally likely sequences counting sequence 11 through, and including sequence 16, so:

0.0625 * 6 = 0.375

So, yep, 37.5% chance that flipping a fair coin 4 times will result in one of the sequences set forth in sequence 11 through, and including, sequence 16.

Hey PIMP,

Your delusion can be modeled in game theory. The penny is your opponent, it is trying to outsmart you.

Every action the penny can take, is denoted in your betting tree diagram.

Every action available to a participant creates game space. For your penny, each action is either heads or tails, binomial.

So, this is how the penny outplays you. The penny has only symmetrical game spaces. There is no distribution of action by the penny that can create anything other.

Your .0005 bias does not exist, because no sequence of moves by the penny can create such game space.

Have a nice day.

-Rob

Siberian, in some way I do like to read your stuff but forget to try to grasp something from 50/50 independent propositions.

For example, we have tested about 15 millions of REAL roulette spins (even chances), trying to ascertain whether the EV can change within "short" intervals after a given strong deviation (sigma 3.5 or more) had taken place before betting, of course ignoring zero.
The results, obviously, were that no matter the point we started the betting EV remained the same forever and ever.

Things can change significantly when you consider one side having a p>50% working into "infinite" successions of finite dependent propositions.

In my view, this is the fundamental problem: We've already answered your question and you've been shown to be wrong in your knowledge.

Reposting the same wrong theory but adding in pictures doesn't make your theory any stronger or any better. And you're right at that nexus where you're losing the benefit of the doubt because you haven't shown any true engagement with the ideas. That also puts you right on the border of being seen as a troll and simply being seen as misinformed.

Getting hyper-defensive when people point this out only leads to you being seen as more trolly.

Again, it's *true* that this is possible, but that's a feature of the physical reality of coins and how you flip them and not part of the theoretical construct. The "grouping" is useless to tell you more information about this. It's just a way that you're obfuscating the problem to yourself, and making you feel clever when you're not.

Just go to a single coin toss. If you *KNEW* that a coin was biased, you could just bet on that bias an make an infinite amount of money. Otherwise, you can't. It's that simple.

No to the bolded. You drawing a curved line onto a discrete distribution that can only have fixed values, does not magically put continuous values in between the points. There is no curve on the original graph because it does not exist. The values for a binomial can ONLY be the ones on the points, there is no curve. You cannot flip half a heads, or .005 of a heads. You can only flip a whole heads or not a heads.

Hi. First I have to say that I am really thankful for all the constructive replies to my questions. Some of you guys are amazing(u have no idea how amazing;the opposite goes for trolls and some others). Anyway, I just wanted to say that I have learned how to make simulations. I have tested a lot of theories, in which i thought this could be done: many EV-=sumEV+. I did not find any leak... But than again this: If this is impossible: Many EV-=sumEV+.. How wormholes, blackholes exist? Isn´t that "Many EV-=sumEV+" is what is happening there?

No.