Optimal strategy under statistical uncertainty - unsteady markets - page 8
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There's something you've done wrong here, Jura. Wins at equal stakes (say 1) just equal p and q, but not p^2 and q^2.
It's OK, you have to take the previous roll into account when calculating here.
We have 4 events
In total we win p*p + q*q and lose 2*p*q. If the probabilities are equal, then we have the usual symmetric eagle.
By the way, it shows very clearly the winning strategy in case of non-symmetric:
There's something you've done wrong here, Jura. The winnings at equal stakes (say 1) are simply p and q, but not p^2 and q^2.
Oh, how glad I would have been if I'd actually taken it and fudged it! Because then the probability of winning would be p + q = 1
But I should have done it the way some nerds advise (let's not point the finger).
Oh, how glad I would have been if I'd actually done the trick! Because then the probability of winning would have been p + q = 1
But I should have done it, as some geeks advise (let's not point fingers).
Eight pages of idle gossip, and the problem has not been solved. Meanwhile, the solution exists. And it is actively used by those who know, in any game. But it is unlikely that those who know it will publish it here in plain text, it is expensive, and they do not attend such forums. >>Yes, it's Markov, just amazingly brilliant and simple solution of Progression Development Matrix, which gives a positive result at the end of the series.
We are talking about a naïve forecast here. For example in the presentation www.swlearning.com/economics/mcguigan/mcguigan10e/ppt/ch05.ppt more about it and how to improve it. Actually it's used to assess the quality of a prediction model, here I've already written about coeff. Theil: 'Types of standard deviations. stddev is there, is there anything else?'. Anyone interested can just google 'theil coefficient' ... Too bad it's not in Metatrader's tester as an optimization criterion.
Eight pages of idle chatter, and the problem has not been solved. Meanwhile, the solution exists, in fact it is actively used, those who know, in any game. But those who know it is unlikely to publish it here in plain text, it is too expensive, and they do not attend such forums. >>Yes, it's Markov, just amazingly brilliant and simple solution of Progression Development Matrix, which gives a positive result at the end of the series.
I understand that you are one of those who knows, but how did you appear on this forum? And who is "those who knows"? Your post turned out to be flood too.
But we should have done a little more work, as some nerds advise (let's not point the finger).
Then, you are a nerd squared. The position "I know more than some nerds, but I am not a nerd" does not work.
Andrey, you wrote on the first page:
Ставить на более частую сторону. В любом случае стратегия должна учитывать историю. В данном случае -- простая адаптация под нее.
It seems that later you changed your strategy and started betting on what fell on the previous roll.
OK, let's say the probability of an eagle is p. The bet is always the same and equals 1. Then the 4 events are as follows:
You got an eagle, bet on an eagle. It's also an eagle. Win equals 1. Probability of the full event is pp.
Heads, we bet on heads. Tails falls. The winning is -1. The probability of the complete event is pq.
It is heads up, bet on tails. Tails rolls. The winning is 1. Probability of the full event qq.
Tails rolls, bet on tails. Heads roll. The payoff is -1. The probability of the complete event qp.
The expectation: pp*1 + pq*(-1) + qq*1 + qp*(-1) = (p-q)^2 > 0.
At p=0.55, m.o. equals 0.01, i.e. one hundredth of a bet.
Profit factor equals ( pp + qq ) / ( 2pq ) = 0.505 / 0.495 ~ 1.02.
Not much, of course. Right, Andrew?
P.S. By the way, the stakes can be adjusted to improve the result. Well, let's assume that the sum of bets on different sides equals 2, and we need to find their sizes, so that m.o. becomes a maximum. Well this is an elementary task. Answer: the bet on the more probable side must be 2, on the less probable side - 0. I.e. when the less probable side falls out, we miss a move.
In this case the m.o. is equal to 2p*( p - q ) = 0.11. It is much better. The profit factor is equal to p/q = 1.22.
But of course this can only be done if we already know which side is better. If we don't know, the universal answer is the first strategy, i.e. with equal bets on what has fallen out before. Especially since in the first strategy we didn't specifically stipulate whether p is greater than 0.5 or not, i.e. we didn't reveal the statistical advantage of one of the sides.
P.P.S. And if you take into account not the last shot, but, say, the last three? The full space of events is 16 pieces. You may also experiment with rates, choosing some more complicated criterion, say, minimizing drawdowns...
But of course this can only be done if we already know which side is better. If we don't know, the universal answer is the first strategy, i.e. with equal bets on what has fallen out before. The more so, in the first strategy we didn't stipulate whether the p is greater than 0.5 or not, i.e. we didn't show stat advantage of one of the sides.
Well, there the question is about the betting system. First we divide the capital into 2 equal parts (in half): first part will be for betting on heads, second on tails. Include a fixed share and do not even need to consider what fell out before that. The part, which bets on the "right" side, will grow faster than the other part shrinks. The MO of the individual drawing in the money will be constantly growing. Probability of ruin=0 if bets are not discrete (unlike the proposed solution) :)
an example where this strategy worked ;) and in general, how do the conditions of this problem relate to the real market? :)
Everything has already been said above on the thread.
It's all been said above on the thread.
Is this about "experimenting with ready-made TC code"? :)
Where in the market is the level of stationarity that allows such a small statistical advantage to be played out? All calculations and assumptions are based on purely abstract stationarity and the definition of probability as the frequency of an event when tested under the same conditions in the limit of infinity . Probability theory is abstract and inapplicable to most real processes, there are other disciplines for them with other conclusions and criteria ;) The problem is purely botanical - in Reshetov style :)