Optimal strategy under statistical uncertainty - unsteady markets - page 6
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We know for a fact that it is a sandwich toss. The probability of one side falling out is p, the other q = 1 - p. Bernoulli's scheme.
I have this strong intuitive feeling that skipping deals in Bernoulli's scheme does not statistically change it in any way. It will still be the same Bernoulli scheme with the same probabilities. The reason is that deals are independent of history.
Expectation of a deal when the deal's reward is equal to its loss and the value of the deal is constant is not equal to zero anyway:
| p * M + ( 1 - p ) * (- M ) | = | ( 2 * p - 1 ) * M | # 0
So whether or not we know p > 0.5 or vice versa, it's still not a martingale. Varying the size of the bets... I don't know what it can do yet - but it's unlikely to change anything in terms of m.o. sign either.
2 PapaYozh:
No stat advantage of 11 over 9 in a series of only 20 trials is out of the question. It's just a very small deviation of frequency from probability - even if the coin is correct.
1.
If we have 0<p<1 and consequently 0<q<1, we can distinguish series in the sequence of events and bet within the series according to the rules:
1) we bet on each flip of a coin;
2) during the series stakes will be placed only on one outcome, the choice of a winning outcome (heads or tails) will be made before the series starts;
3) The size of the next bet in the series Vi = 2^i, where i is the number of unfavourable outcomes in the current series of deals.
The series finishes if you get a favourable outcome, the next event will be the beginning of the next series.
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2.
Of course, we can't speak about any representativeness of the sample of 20 elements. I just wanted to show that the rules
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- If the previous trade signal resulted in a loss, the next position should be opened against the previous trade signal interpretation
- If previous trade signal resulted in profit, next position should be opened against the previous trade signal interpretation
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cannot guarantee a positive payoff, even if there is a statistical advantage of one outcome over the other.
The probabilities for this betting system:
Let's take the probability of a wrong coin being struck by heads as p, and by tails as q
By the full probability theorem we have only two incompatible outcomes (two sides of the coin), hence: p + q = 1 <=> p = 1 - q
Since we are going to bet on the previous outcome, i.e. only on the side that fell in the previous coin flip, respectively, p part of the bet will be for heads and q part for tails.
Since the probability of winning with a bet on heads is p, and the odds of betting on heads are only p -y part of all bets, the winnings of bets on heads are p * p = p^2
As the probability of winning for betting on heads is q, and the odds of betting on tails are only q - a part of all bets, then the winnings of bets on tails are q * q = q^2
Total probability of winning in this betting system will be: p^2 + q^2 = 1 - 2 * p * q
probability of losing (an incompatible outcome relative to winning) in this betting system is: 1 - p^2 - q^2 = 2 * p * q
Expectation for this betting system:
Let's denote the size of the win per individual bet to the size of the stake as profit, The size of the loss is equal to the stake by an absolute value of stake. If stake = profit = 1, then expectation in this betting system is
MO = profit * (p^2 + q^2) - 2 * p * q * stake = p^2 - 2 * p * q + q^2 = (p - q)^2
Accordingly, the zero mathematical expectation in this case is possible only in one case, i.e., when p = q = 0.5, because we obtain MO = (0.5 - 0.5)^2 = 0^2 = 0
In all other cases, when p is not equal to q, expectation is positive, since everything in parentheses is squared. Therefore, it makes no difference whether it is greater or less than p or q.
This is a generalized case, for example, when winning size is not equal to losing size. Expectation is calculated by formula:
MO = profit * ((p - q)^2) - (stake - profit) * 2 * p * q = profit * ((p - q)^2) + (profit - stake) * 2 * p * q
1.
If we have 0<p<1 and consequently 0<q<1, then it is possible to allocate series in the sequence of events and to bet within series according to the rules:
2.
Of course, the representativeness of a sample of 20 elements is out of the question. I just wanted to show that the rules
The rules of the series cannot guarantee a positive payoff, even if there is a statistical superiority of one outcome over the other.
1. The initial condition is that only the previous toss-up is available for analysis. However, yes, you can take the last n, I think three is already enough :)
But again, let's not forget that in general, if the Shannon strategy works, we can restore the skew we want with a high confidence probability.
2. This is empty reasoning - of course they can.
The probabilities for a given betting system:
You can get the required probabilities differently, the result will be the same.
Let there be two coins, with probabilities of tails p1 and p2, respectively eagles q1 and q2.
Since the probability for simultaneous occurrence of two independent events is equal to the product of the probabilities of these events, we have the probability of falling two tails p1*p2, respectively, the probability of falling two eagles q1*q2.
Since the probability of occurrence of at least one of two incompatible events is equal to the sum of probabilities of these events, we have the probability of two tails or two eagles p1*p2+q1*q2.
Since p1=p2, it follows that p^2+q^2.
The hardest part is explaining to people how two independent coins came out of the same row. :)
The hardest part is explaining to people how two independent coins came out of one row. :)
Independence is a consequence of the fact that coins have "no memory", whether they are right or crooked. Therefore, if two coins are absolutely identical, it makes no difference whether we flip just one of them or alternate in any order of flipping both of them.
Independence is a consequence of the fact that coins have "no memory", whether they are right or wrong. Therefore, if two coins are absolutely identical, it makes no difference whether we flip just one of them or alternate both in any order of flipping.
A lot of people can't understand that.
A lot of people can't understand that.
I don't care what others understand or don't understand. It is more important to me that my balance curve is slowly growing on such primitive mathematics.
And everyone else's notions or misunderstandings are their own problems.
By conditions, it is necessary to create a profitable betting system, which does not allow to calculate statistically the advantage of one side of the coin, therefore, its algorithm should be built on the knowledge of only two parameters:
1. The number of the next toss-up.
2. The side of the coin which was struck on the previous flip.
This is a typical example of a Markov chain. The result of the flip does not depend on the previous flip, no matter how bent the coin is. It is impossible to talk about strategy in this context, because the task is to guess which side of the coin will fall in one single test - it's not a strategy.
Without statistics here can not do, and the statistics will be simple to the point of obscenity. Bets every time heads, if there was a profit means all cool - continue in the same vein, if the amount of money in your pocket began to decrease, then we must "change the strategy" and put all the time on tails.
You can start this chain of bets with the same thing that was in the first flip, theoretically, the probability of hitting the right probability is higher at once.
1. The initial condition is that only the previous toss-up is available for analysis. However, yes, we can take the last n, I think three is enough :)
But again, let's not forget that in general, if the Shannon strategy works, we can restore the skewness we need with high confidence.
2. This is idle reasoning -- of course it can.
1. What do history and the last n tossups have to do with it?
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п.1.
Choose a favourable outcome for the series (heads or tails).
Null i.
п.2.
Bet Vi = 2^i on the outcome chosen in item 1;
п.3.
If the outcome coincides with the one chosen for the series, the series is over, pass to step 1.
Otherwise i++, go to point 2.
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And no history.
2. You can call your line on point 2 as empty reasoning.