Hi guys. Please tell what you think about this. Is it correct? Is it wrong? Why? All comments are much appriciated. Criticism also welcomed. I posted same in Probability forum, but I am not getting any answers, so I hope you guys can maybe help me. Thanks.

COIN FLIP GAME

Rules of a game:

- Flipping a coin

- We do not have infinity amount of units to bet

- Our lowest vs highest bet is 20x

- We decide about our next pick:H or T, after last one falls

- If we do make correct prediction for next H/T we get +0,999 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

We bet all the combinations that has as a result 2H+2T, HHHH and TTTT. There are 8 combinations like this, mentioned above (combinations with numbers: 1,6,11,12,13,14,15,16).

How do we bet:


We repeat that to infinity. If we lose before 4th bet, we wait until 5th bet and repeat as mentioned.

More calculations:
N=4
u= 4*(0,5)=2; 0,5 because it is coin flip; u is mean
o2=4*(0,5)*(1,0-0,5)=1; 0,5 because it is coin flip
o=square root 1=1

Let us now take 0,5o and 1,5o.
0,5o=0,382924; rounded on 6th decimal; not in favour of this theory
1,5o=0,866386; rounded on 6th decimal; not in favour of this theory
1-0,866386=0,133614

We are interested in u+-0,5o. This in our case means from 1,5 to 2,5 (2 is mean).
We are also interested in +-1,5o. This in our case means from 0,5 to 3,5 (2 is mean).

If we would have odds 2 for every winning prediction we would need to hit our system in at least:
winning combinations/all combinations
8/16=0,5=50 % of the time.
The numbers from SD calculations tells us that we will hit this system in: 38,2924% + 13,3614 %=51,6538 % of time.
51,6538 % is more than 50 %.

Therefor, we made a +EV bet out of only 0EV bets.

Now, let us calculate EV in our case from the example (with odds 1,999).
In our case we have odds 1,999.

That means that every of our 8 combinations has got odds:
1,999 on 4=15,968; rounded on 3 decimal; not in favour of this system
Together our system has got odds: 15.968
We calculated before, that we will win our system in 51.6538 % of time.
15,968 x 0,516538=8,2480; rounded on 4 decimal;not in favour of our system;
8,2480 is more than 8.

In every 4 flips we make profit of 8,2480 - 8= 0,248 units.
That is profit per flip: 0,248/4=0,062 .
Profit per every flip =0,062/8=0,755%.
The more time we play this game the more units we have (if we are betting like I wrote).

Therefor, we made a +EV bet out of only -EV bets.

If above calculation is correct. This means that those 2 sentences are wrong:

- The EV of the sum of random variables is always equal to the sum of the EVs, whether the variables are independent or not (for example in the same game).

- Optimal strategy is allways the best strategy.

I don't understand your betting chart nor anything after that. You're betting on coin flips. Where do the 4 flip combinations come into it? Maybe explain an example.


PairTheBoard

Your notation is a bit strange and you're making a very bold claim. So you might be getting no response because you're not being taken seriously. (I'm not taking you seriously, either. So I just gave it a quick skim and it's far from clear you're saying anything meaningful.)

Try explaining your betting pattern and don't worry about computing the EV.

For example:

On the first flip, bet on ________.

If the first flip is a head, then on the next flip bet on _______.
Otherwise, bet on _________.

Hi, PairTheBoard.

I am not english native and not math expert so it is a bit hard for me to explain in correct math words. So please feel free to ask if anything else is unclear. Would be glad to answer.

The rule of the game is to bet on each flip(or not to bet on each flip;we can also miss betting 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 (kind of a group bet out of one bet): these combinations are 1,6,11,12,13,14,15,16). I repeat this bet to infinity. I have covered 8 out of 16 possibilities. Probability should be 50%, but is it really? In exact example that I am reffering to?

I have used SD calculations for probabilities...I think this is fine?

Maybe we calculate probability in some cases wrong (very close to real, but not exactly). As for maybe this (in my example) 8/16 is just average, and not exact number?

For example: SD graph is not linear for a reason.

If what I am thinking is correct this can also be used in St. Petersburg paradox and many other math problems.

Hi Aaron W.

Please read my next post to yours. Even if mentioned post was not reffering to you, I think it has what you are looking for.

Probably not. What are you actually doing with the standard deviations? It looks like you're just combining them somewhat randomly. For example:

What purpose is there in subtracting 1.5 standard deviations from the number 1? And it looks like you turned these numbers into percents:

There's no logic to these calculations.

No, it's not. If you can't tell me *exactly* what you're betting on each flip and each possibility, your system is no good. If you want to do this in reality, what are you actually betting on at each event?

Ok, I think now I understand what is unclear.



We are basically betting on all the combinations that we want from 1st bet to 4th bet. For example one of these 8 combinations is: TTTT. If we would just want to bet on this combination we would bet on first flip T(2*1), second flip we would bet on T again(we would have 2 times more money as pre bet 1: 2*2), on 3rd bet we put units on T again (2*4) and on 4th bet we put units on T again (2*8)= aprox. 16, if we win the bet. We basically have 8 combinations of bets like this. The table shows just more simplified, what we do at each bet, if H or T. From bet 1 to bet 4.

We win our bet (in our system) if those combinations come(otherwise we lose):
HHTT
HTHT
HTTH
THTH
TTHH
THHT
HHHH
TTTT

Same combination is in the same horizontal column.

I hope now is more understandable.

Also, I am sorry, I am a bit in a hurry right now. I will answer your other post after some time.

I think it is the right thing to calculate with SD calculations in my betting system.

Exact SD numbers can be seen here (in the middle of the page):

https://en.wikipedia.org/wiki/68%E2%...80%9399.7_rule

I will try to explain this as simlpy as I can. Hopefully will be understandable.I needed to know what is percent number of combinations HHHH and TTTT to happened. These combinations are not in the 1,5 standard deviation, therefor are out. If we know percent for standard deviation 1,5= 86,6386 %. We know our combinations are not there. So 100%(=1 in mentioned case)-86,6386=13,3614%.

I hope it is more clearer now.

I think I got it!

OP wants to know which is the probability of having two heads out of four. Since for N=4, the mean value is 2 (correct) and the standard deviation is 1 (still correct), he thinks that he can use a normal distribution:

P(H=2) = P(H>=1.5 AND H<=2.5) (i.e. a one half sigma interval)

and, if H is normally distributed (which is of course lol not), than you have P(H=2) = 0.3829249.

Then, he wants to know how often you have 0 or 4 Heads:

P(H=0 OR H=4) = P(H<=0.5 OR H>=3.5)

which is the complement of a 1.5 sigma interval. So the above probability is about ~ 0.13.

So, he tallies the two cases and arrives to the conclusion that:

P(H=2 OR H=0 OR H=4) is about 51%.

OP, I don't know where you learned to use a normal distribution in this case. It's so wrong in so many ways. The probability of having a given number of heads out of four tosses follows the binomial distribution. The sigma intervals you are referring to are valid only if the variable you are considering follows a normal distribution.

Just to state the obvious, any of the 16 possible sequences is equally likely. You have that:

P(H=0) = 1/16
P(H=4) = 1/16
P(H=2) = 6/16 = 0.375

The combined chance of the above is obviously 50%.

Thanks. This was the piece that was missing in order to understand what he was saying.

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Here's maybe a way to see why this approach is problematic:

P(H=2) = P(H>=1.5 AND H<=2.5) = P(H>=1.999 AND H<=2.001) = P(H>=1.001 AND H<=2.999) = P(H>=1.999 AND H<=2.999)

Clearly, there's something funny going on if you can use ALL of these ranges and get a whole bunch of different values following your technique.

A quote:
...."A stronger rule states that the normal approximation is appropriate only if everything within 3 standard deviations of its mean is within the range of possible values."....

Source: https://en.wikipedia.org/wiki/Binomial_distribution (approx. middle of a site)

I do not really know this stuff, but:

ONE

I would think that the normal distribution not being appropriate if everything within 3 standard deviations of its mean is not within the range of possible values does not mean that if everything within 3 standard deviations of its mean is within the range of possible values, then the normal distribution is necessarily appropriate.

Or put another way: For the normal distribution to be appropriate it is necessary that everything within 3 standard deviations of its mean be within the range of possible values, but that alone is not sufficient for the normal distribution to be appropriate.


TWO

Also, that section of the article actually says more fully (with the formulas omitted):



So I think what this is saying is that even when the binomial distribution is the correct thing to use, in some cases the normal distribution will yield a reasonable approximation of what would be yielded by the binomial distribution if n is large enough and p is not too close to 0 or 1, and the article provides some rules of thumb for how to determine when n is large enough and p is not too close to 0 or 1. The "stronger rule" you quote is one of those rules of thumb for when those conditions are satisfied such that the normal distribution may reasonably approximate the binomial distribution. So I think the calculations you did yielded an answer that is approximately the answer the binomial distribution would have yielded, which answer yielded by the binomial distribution would be the correct answer. The binomial distribution would have yielded the correct answer of 50% and by using the normal distribution you got a reasonable approximate of the correct answer (51.6538 is reasonably approximate to 50 (although obviously it would be a very important difference with respect to the bet you describe)).


If someone could confirm whether that is correct, that would be cool.

Look closer at that section, specifically the requirements for how large n needs to be in order for the approximation to be OK.

(hint: n>=10, and even then, it's still an approximation. I think you'll find that if you modified your sequence to 10 bets, you'd still find a small amount over 50% with your method, but that amount would be the amount that the approximation differs from reality)

Siberian:
Normal is by definition a continuous distribution. Binomial is by definition a discrete distribution with only two possible values (win/loss heads/tails etc).

The only reason to use the normal approximation for discrete distributions is because it's usually easier to calculate. It never gives the exactly correct answer no matter how large N is, but merely approaches the correct answer as N approaches infinity. But as you have seen by your own calculations, with small N you can have a significant error.

Appropriate for what purpose? It is appropriate to *approximate*. It is in a section called "Normal approximation."

That is, it will get you an answer that's kind of close to right. (And I guess in some sense you were. You were within a couple percent of the exact value.) This is much different from saying that the calculation you get is exactly right. So the magic 1.6% can be understood as an error due to approximation.

Finally some good stuff for OP to learn. That is what should be. Ask a curious question, get thoughtful responses. I knew OP would have a well thought out system, if he would share it he could learn.

Roulette solid, btw.

No. I think this is wrong in our case(in my betting system). Will explain more detailed now... If I am correct, this is huge thing...

@Lego05, RustyBrooks, NewOldGuy, Aaron W.

My quotation was meant in the way that what some of you are saying is not exactly correct. It is close, but not exactly. My betting system is not classic binomial(if it actually is). I google all over the internet and my system do not feet in any system I found.

https://en.wikipedia.org/wiki/Binomial_distribution
From this link you go to: Sums of binomials(it has some parts of it)...

Most importantly my system contains, for example in a flip no.3: past, present, future,...

And yes offcourse: our next coinflip is 50% vs 50% percent. But that does not mean that in our system everything is linear... I will try to explain this. And because it is maybe huge, I am opening new thread about my theory.

Your game uses coin flips (trials). A coin flips is exactly binomial, and your system is a series of binomial trials. Any sequence of coin flips will follow a binomial distribution. The term does not refer to your "system". You can apply any system of bets you like and it doesn't change the fact that coinflips follow a binomial distribution, by definition.

Also, your game has only one binomial variable. So the math for sums of binomial variables (your link above) doesn't come into play. Even if it did, that sum also follows a binomial distribution, not a normal one.

@NewOldGuy

OK, next time please read everything what I wrote...

I am on this forum to learn and maybe contribut something. And I am most certainly not in this forum that people will troll me (like some do). I believe in civilised conversation. And I also believe in the most constructive discussions possible. I am also not wasting my valueable time anymore trying to explain some things and than I am getting some "10 seconds posts" back. Not how it works in my book.

Also, yeah, everything is linear in binomial distribution. lol... You can just figure this out by yourself...

Nothing in my last post was less than civilized, and it absolutely was constructive. You just don't want to hear it, but instead you want to continue trying to invent perpetual motion and create your own math to justify your theory. I can stop pointing out your mistakes if that satisfies you.

Have you considered inviting John Gabriel to this thread?

Juk

It actually shows that what you're saying is not exactly correct.

You can't find anything that fits because you have not been able to meaningfully describe it.

I'm going to go back to my basic premise. If you can't tell me *EXACTLY* what I'm betting on in any given moment, then your system is worthless.

----

I don't even know if you know what you mean by linear. Do the exercise above. Or write a computer program to simulate your betting strategy and let us read through the code.

Or else there's a clear reason that you're not being taken seriously. And if you're right despite all of this, extend your strategy to play roulette and make millions of dollars and laugh at us all.

Yeah I still do not actually know how the betting sequence works. An example would be useful.

But here's what's happened so far
you: here's some math that shows something weird
us: the way you're doing this doesn't work and here's why
you: ok but my situation is actually not covered or explained by this math

If that's how this is going to go, we have a problem: this is a goal post shift.