Brain-training tasks related to trading in one way or another. Theorist, game theory, etc. - page 2

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Let the probability of A be p, the probability of B be q = 1-p.


m.o. the outcome of an odd bet:

MOnechA = p*1p + q*(-1)rupee = (2p-1)rupee.

Obviously, if we bet on B instead of A, then MOneachB = 2q-1 = 1-2p = - MOneachA.


m.o. of the outcome of an even bet:

p*2*MonechA + (1p)*4*MonechB =

= p*2*MonechA - (1p)*4*MonechA =

=MONECHA*(p*2 - (1-r)*4) =

= (2p-1)(6p - 4)


Remaining to add and divide in half:

1/2*(2p-1 + (6p-4)(2p-1)) =

= (2p-1)/2*(1+6p-4)) =

= (2p-1)/2*3*(2p-1)) =

= 3/2*(2p-1)^2 >= 0, h, etc.

 
Mathemat:

Let the probability of A be p, the probability of B be q = 1-p.


In other cases we get a profit:

MOnechA = p*1p + q*(-1)rupee = (2p-1)rupee.

Obviously, if we bet instead of A on B, then MOneachB = 2q-1 = 1-2p = - MOneachA.


m.o. of the outcome of an even bet:

p*2*MonechA + (1p)*4*MonechB =

= p*2*MonechA - (1p)*4*MonechA =

=MONECHA*(p*2 - (1-r)*4) =

= (2p-1)(6p - 4)


Remaining to add and divide in half:

1/2*(2p-1 + (6p-4)(2p-1)) =

= (2p-1)/2*(1+6p-4)) =

= (2p-1)/2*3*(2p-1)) =

= 3/2*(2p-1)^2 >= 0, h, etc.



That's a bit too complicated.

Let's calculate in a simpler way, namely by series of events:

Series AA wins +3.

Series AB wins -1

Series BA win -5

Series BB win +3

Let the probability of event A = p

Then the series AA will fall with probability p^2

Series AB and Series BA with probability p * (1 - p) = p - p^2

Series BB with probability (1 - 2)^2 = 1 - 2*p + p^2

Total expected payoffs: 3 * p^2 + 3 * (1 - 2*p + p^2) = 3 * (1 - 2 * p + 2 * p^2)

Total expected payoff: (-5 - 1) * (p - p^2) = -6 * (p - p^2)

Let's construct an inequality that needs to be proved:

0 <= 3 * (1 - 2 *p + 2 * p^2) - 6 * (p - p^2)

6 * (p - p^2) <= 3 * (1 - 2 *p + 2 * p^2)

2 * (p - p^2) <= 1 - 2 * (p - p^2)

4 * (p - p^2) <= 1

p - p^2 <= 1 / 4


All that remains is to prove that p - p^2 at any value of p from 0 to 1 cannot be more than 1/4. This is already uncomplicated. Since at extremes of p = 0 and p = 1, p - p^2 = 0. And at p = 0.5 we have an extremum, p - p^2 = 1/4 = 0.25

Consequently, we deal with the system of rates that has no negative expected payoff. I.e. with the worst outcome we still have some profits. In other cases we gain profit.


Looking at the series considering wins and losses, we can conclude that the betting system is trending, as series AA and BB give profits, while series AB and BA give losses.

 
Reshetov:

And no one said that the betting system is risk-free. It is win-win according to MO, i.e. at p(A) != 0.5 the profit will tend to rise. But the variance can produce drawdowns.

For information: I forgot to turn off the script from yesterday... it is holding around 1500-2000 RUR for a few hours. The number of cycles I'm afraid to imagine.
 
sever30:

For information: I forgot to turn off the script from yesterday... as a few hours around 1500-2000rub. held. The number of cycles I'm afraid to imagine.


It is better to rewrite the algorithm in some language that compiles to machine code, such as C or Java and in integer expression. Then hundreds of millions of runs will be executed in a few seconds. Here's an example in Java:

  private void test() {
    Random rand = new java.util.Random();
    int deposit = 0; // Начальный депозит
    for (int i = 0; i < 100000000; i++) {
      int number = 0;
      for (int j = 0; j < 2; j++) {
        number = number * 2;
       // Если сравнение с числом не равным 49,
       // то, вероятность не равна 0.5
       // и депозит будет расти
        if (rand.nextInt(100) > 49) {
          number++;
        }
      }
      if (number == 0) {
        deposit +=3;
      }
      if (number == 1) {
        deposit--;
      }
      if (number == 2) {
        deposit -= 5;
      }
      if (number == 3) {
        deposit +=3;
      }
    }
    System.out.println(deposit);
  }

And here are the results for p(A) = 0.5

58264
-4496
7560
41640
62312
-23208
-11952
32124

I.e. despite the fact that the PRGP is multiplicative with a fairly uniform distribution, nevertheless the number of profitable tests slightly exceeds the number of unprofitable ones due to variance.

And here are the tests where the comparison is with number 50, i.e. p(A) = 0.51

143484
133556
101844
152840
76956
90296

For p(A) = 0.49, i.e. comparing with number 48

100740
147924
80708
115648
128136
101544


The results are about the same, since MO for p(A) = x equals MO for p(A) = 1 - x
 

OK, we've dealt with the special case. Now the second problem, namely the generalised formulation:


Betting systems with non-negative expectation


Let there be two mutually exclusive events A and B with corresponding probabilities: p(A) = 1 - p(B).

Rules of the game: if a player bets on an event and this event falls out, his winnings are equal to the bet. If the event does not fall, his loss equals his bet.

Our player bets using the following system:

The first or any other odd bet is always on event A. All odd bets are always equal in size, e.g. 1 ruble.

The second or any other odd bet:

- If the previous odd bet is won, the next even bet is increased by x times where x is greater than the odd bet, and placed on event A
- If the previous odd bet is lost, the next even bet increases y = f(x) times, and is placed on event B

Problem: Find a function for y = f(x) such that expectation for any p(A) in acceptable range from p(A) = 0 to p(A) = 1 is non-negative and the condition that expectation for p(A) = x is equal to expectation for p(A) = 1 - x is satisfied.
 
Reshetov:

p - p^2 <= 1 / 4


All that remains is to prove that p - p^2 for any value of p between 0 and 1 cannot be more than 1/4. This is already uncomplicated. Since at extremes of p = 0 and p = 1, p - p^2 = 0. And at p = 0.5 we have an extremum, p - p^2 = 1/4 = 0.25

Consequently, we deal with the system of rates that has no negative expected payoff. I.e. with the worst outcome we still have some profits. In other cases we gain profit.


Looking at the series, taking into account wins and losses, you can conclude that the betting system is a trend betting system, because series AA and BB give profits, while series AB and BA give losses.

It is not clear what is the trick here. At p = 0.5 we have expectation 0. And when it is different from 0.5, we have a constant trend, on it we will win with any betting system, with or without martingale. If we determine the trend correctly of course :)
 
Reshetov:

Looking at the series with wins and losses, we can conclude that the betting system is trending, as the AA and BB series are profitable, while the AB and BA series are losing.

If events A and B are random with probability 0.5 and independent, no money management will make the system profitable. Its equity will be a random stray. And since a player by definition cannot have infinite capital, sooner or later he is bound to lose everything he has.
 
timbo:
If events A and B are random with probability 0.5 and independent, no money management will make the system profitable. Its equity will be a random stray. And since a player, by definition, cannot have infinite capital, sooner or later he is bound to lose everything he has.


Your statement is knowingly wrong. Learn the math - it's handy.


The correct way is this:

If events A and B are random with probability 0.5 and independent, then no money manager will make a betting system in a game of beagle or similar with expectation not equal to 0. His equity will be a random stray. And as the player by definition cannot have infinite equity, sooner or later he will either use up everything he has with probability 0.5 or win the equity equal to the initial capital, i.e. double the initial capital with the same probability 0.5 for the approximate time of x^2 bets placed.

Correspondingly MO = x * 0.5 - x * 0.5 = 0;

where: x is the amount of initial capital / bet size

 
Reshetov:


Your statement is knowingly wrong. Learn the math - it's the best.


That is correct:

If events A and B are random with probability 0.5 and independent, no money management will make a system with expectation not equal to 0. Its equity will be a random stray. And as the player cannot have infinite equity by definition, sooner or later he will either use up everything he has with probability 0.5 or win an equity equal to the initial one, i.e. double the initial equity with the same probability 0.5.

Accordingly, MO = 1 * 0.5 - 1 * 0.5 = 0.

Reshetov - you're a pathological threesome. This is classic random walk theory. A mathematical expectation of 0 does not save you from being drained. A player can make a lot, much more than his initial capital, but if the game goes on indefinitely, he is bound to lose it all.
 
timbo:
Reshetov - you are a pathological threesome. This is classic random walk theory. A mathematical expectation of 0 does not save you from being dumped. A player can earn a lot, much more than the initial capital, but if the game continues indefinitely, he is sure to lose it all.

Even a stake minus for your selves would be too high a theorist grade.


Nerdiness in the form of infinitely long game does not apply. Our life is limited in time.

Besides there is a proof of losing with limited capital for eagle player only when the probability of winning is less than 0.5 and only when the game is played against a player with infinite capital. In other cases, the player with finite capital may lose or double, triple, quadruple, etc.


Learn the basics - it's tame.
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