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.
Last edited by SiberianPIMP; 08-15-2018 at 10:52 AM.