Sensation! A profitable strategy for playing beagle has been found! - page 7
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No way! Of course I won't watch it, I don't know what kind of books you have... I don't know if it's worth watching. In terms of pure mathematics, you can win a game of eagle at martingale.
If the deposit is infinite.
No way! Of course I won't watch it, I don't know what kind of books you have... I don't know if it's worth watching.
Come on, they're great books, thick, with nice rough paper. They don't make them now, the secret of making them has been lost for centuries. No pictures, though.
In terms of pure mathematics, you can win a coin toss using the Martingale method.
Only in one case, in infinity, with infinite initial capital. I wish you to live exactly that long and have exactly that much capital.
If the deposit is infinite.
That's why I wrote that in terms of "pure mathematics"
And where do you have an even/integer check? You have grossly violated the conditions by not checking for parity.
By the way, I want to know what your program will output if all the conditions are satisfied. More precisely, I'm interested in the quality of the PRNG built into it.
but is the condition "more than zero" or "less than zero" not good enough? Is there any difference? It seems to me that the author has simply made a mistake when converting to bars, it's kind of obvious, you can't get such figures there. And the most reliable method "by eye" confirms it - there is no correspondence between them. I am not even talking about infinite number of process implementations.
1. Come on, great books, thick, with lovely rough paper. They don't make them now, the secret of making them has been lost for centuries. No pictures, though.
2. Only in one case, in perpetuity. I wish you to live that long.
1. Oh! I definitely won't read without pictures. 2. Better wish so much money;)
but is the condition "more than zero" or "less than zero" not good enough? Is there any difference? It seems to me that the aftar has just made a mistake when switching to bars, it's kind of obvious, you can't get such figures there. And the most reliable method "by eye" confirms it - there is no correspondence between them. Not to mention the endless number of implementations of the process.
No, you have to use even/non-even. That's exactly the error, otherwise it's OK.
In general, my colleague, you should learn MQEl and write programs in a proper language!
When I have a conversation with people who claim in all seriousness that they can win at orchestra, I have a firm grip on my pockets. You can expect anything from these people. So, I can't believe your bewilderment.
No, it's because it's rigorously proven mathematically. You can't win at orbit. It's also rigorously mathematically proven how to change the oracle to make winning possible. But you probably haven't read a book about it.
If you'd read carefully, you'd come across a phrase like this
...The most famous among the paradoxes of probability theory should be considered the St. Petersburg paradox, first stated in the "Memoir", which a famous mathematician Daniel Bernoulli presented to the St. Petersburg Academy. Suppose I flip a coin and agree to pay you a dollar if heads roll. If tails rolls, I flip a coin a second time and pay you two dollars if heads roll at the second flip. If it's tails again, I'll flip a third time and pay you four dollars if it's heads on the third flip. In short, I double the payout each time. I continue to flip the coin until you stop the game and offer to pay me back. How much should you pay me so that I agree to play this "one-way game" with you and you are not left at a loss? The answer is hard to believe: no matter how much you pay me per game, even if it's a million dollars, you can still more than recoup your expenses. In any given game, the probability of you winning one dollar is 1/2, the probability of you winning two dollars is 1/4, four dollars is 1/8, and so on. In the end, you can expect to win a sum of (1 x 1/2) + (2 x 1/4) + (4 x 1/8) ... This infinite series is divergent: its sum equals infinity. Consequently, no matter how much you pay me before each game, if you play a long enough match, you're bound to win. In making this determination, we are assuming that my capital is unlimited and that we can play any number of games. Of course, if you paid 1000 dollars for the right to play a single game, you would lose, but that chance is more than compensated by the chance, albeit small, to win an astronomical sum with a long series of eagles alone. If my capital, as it is in reality, is limited, then a reasonable fee for the right to play a game should also have an upper limit. The Petersburg paradox occurs in any game of chance with doubling stakes....
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the kind of approach I've used in what's called Sixlines... is when playing (conventionally) TP10 SL10 with 5 wins in a row (basically 50 pips in one direction) the winnings are not calculated 5 to 1... but 31 to 1
1. Boo! I definitely won't read it without the pictures. 2. Better wish so much money ;)
Wished!
If you had read carefully, you would have come across a phrase like this
... The most famous among the paradoxes of probability theory should be considered the St. Petersburg paradox, first stated in the "Memoir" which the famous mathematician Daniel Bernoulli presented to the St. Petersburg Academy. Suppose I flip a coin and agree to pay you a dollar if heads roll. If tails rolls, I flip a coin a second time and pay you two dollars if heads roll at the second flip. If it's tails again, I'll flip a third time and pay you four dollars if it's heads on the third flip. In short, I double the payout each time. I continue to flip the coin until you stop the game and offer to pay me back. How much should you pay me so that I agree to play this "one-way game" with you and you are not left at a loss? The answer is hard to believe: no matter how much you pay me per game, even if it's a million dollars, you can still more than recoup your expenses. In any given game, the probability of you winning one dollar is 1/2, the probability of you winning two dollars is 1/4, four dollars is 1/8, and so on. In the end, you can expect to win a sum of (1 x 1/2) + (2 x 1/4) + (4 x 1/8) ... This infinite series is divergent: its sum equals infinity. Consequently, no matter how much you pay me before each game, if you play a long enough match, you're bound to win. In making this determination, we are assuming that my capital is unlimited and that we can play any number of games. Of course, if you paid 1000 dollars for the right to play a single game, you would lose, but that chance is more than compensated by the chance, albeit small, to win an astronomical sum with a long series of eagles alone. If my capital, as it is in reality, is limited, then a reasonable fee for the right to play a game should also have an upper limit. The Petersburg paradox arises in any game of chance with doubling stakes....And?! What conclusion do you draw from all this? Isn't it already "winnable"? if you have less than infinite money.
And?! What conclusion do you draw from all this? Isn't it already "winnable"? if you have less than infinite money.
fuck... :-) my personal experience... it's possible to win - 3 years of practically daily trading... and about infinity of money.... in my case the minimum lot is 0.1 of a $6000 deposit...
but if you don't have 10K$ or more to trade with, of course... You have only to read books and flounder in forums... :-) like Michuil by the way... ....
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here are those "advisors" from the articles you cited ---
What is Martingale?
What is Martingale and does it make sense to use it?
i'll just "tweak" them a bit - and i'll post the versions that are profitable as a result....