Martingale: the maximum possible chain of continuous losses/profits - page 14

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sever30:

and anyway, it's complicated... I didn't need to read a free paper advertising forex in 2007... would have worked and lived my life in peace.

P.S. I had a couple of drinks:)

You've got some good ideas, you're just ignoring them.
 
Tantrik:
All profitable trading is random and temporary....(arbitrage options are not related to trading - so purely technique, speed)
We need to broaden our knowledge of arbitrage.
 
Tantrik:
There is no way and there never will be. All profitable trading is random and temporary.... (arbitrage options do not apply to trading - so purely technique, speed)

life is also temporary and random :)
 
Avals:

life is also temporary and random :)
Exactly. We are talking about a martingale that looks for a way out and always finds it, while here we are talking about entry points (which don't exist). If you are not satisfied with martingale, trade with the trend.
 
vasya_vasya:
There are sensible thoughts in your mind, you just ignore them.

So many thoughts, my head's pounding... Which one are you talking about?
 
about leaving martin alone
 
Mischek:
about leaving martin alone


:)))

What can I say, just smile stupidly.

 

I don't get it. I've been reading and reading, and I don't get it. What is the connection between martin money management and random entries/exits?

 
sever30:

For example, roulette, always bet on black, what is the possible maximum length of the series of losses / profits may fall in a series of bets, for example, 1 000 000?

There is a calculator from Meta Driver, but there are some restrictions when calculating the chains, or maybe my hands are wrong...

It turns out that for the maximum series there are about 13-15 continuous losses/profits ?

Created exactly 1,000,000 random numbers in matlab. ( randn(1,1000000) ). From this data using the following code:

% var - any positive and negative numbers.
% This function converts the array var to the form
% -1 2 -3 4 -8



function out=getSeries(var)
tic
iter=0;
iterp=0;
flag=0;
flag2=0;
index=0;
out=0;
% Locate negative and positive values
pozitive=find(var>=0);
negative=find(var<=0);

% Change found values to 0 - negative
% 1 - positive

var(pozitive)=1;
var(negative)=0;

for i=1:length(var)
if var(i)==0
iterp=iterp+1;flag=0;
if flag2==0
index=index+1;
flag2=1;
out(index)=-iter;
iter=0;
end
end
if var(i)==1
iter=iter+1;
flag2=0;
if flag==0
index=index+1;
flag=1;
out(index)=iterp;
iterp=0;
end
end
end

toc

end

This produces a sequence of series. The figure shows the distribution of these series over the entire sequence. Correspondingly, we get approximately 500,000 series per 100,000. The answer is in the extremes of the graph.

 
sever30:

For example, roulette, always bet on black, what is the possible maximum length of the series of losses/profits that could fall in a series of bets, e.g. 1,000,000?

And drunken hedgehog understands that if we do N bets, the maximum possible series of losses on the duration is equal to N, because to lose more than N times in a row well does not work. The probability of a maximal series of losses for N bets is (19/36)^N.
1...78910111213141516
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