How can i calculate the probability of sucess between 2 strategies? - page 2

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Mrluck07:

Correct me if i'm wrong,

A If you have 2 strategies with 70% wins, the chance of lose trades in the same sequence, on both strategies is only 9%. So that means 1 in 10, right?

B If the chance is only 1 in 10, that means 9 in 10 times, there will be lose trades, but only in 1 strategy, and the other one on the same sequence will win, is that correct?

C So you have a 90% win strategy, because you just need to 1 strategy to lose money, and trade the next one, because you know the chance of lose that trade will be only 1 in 10.

You are wrong.
C is not correct.
The probability of 2 loss is calculated before you take the trades, and then it is right.
But in C you change that, and you say that you already did a trade and you know that it lost.
So  now we have just one deal left to calculate the chance for and the chance for that is 1/3 as any other deal.

 
Marco vd Heijden:

Some people think trading is high speed money.

But then they discover that a snail moves faster then the trend.

You can combine anything you want, even with a 90% win strategy, a snail can only go so fast.

so what does it mean? i did not understand, can you be more clear?
 
Mrluck07:

If someone has 2 strategies with 70% win rate each one, and they have very different time entries , what's the probability of having 2 loss trades at the same sequence of trades?

example

sequence of trades:

 X  I   I   I  X  I   I  X  I   I     strategy 1

I   X  X   I   I  I  I   X  I   I   strategy 2

What's the probability of something like  X X loss in both strategies at the same sequence,to happen?

thanks for any help

I can say simple strategy always works even big players using a simple way without complex. 

But when anyone tries to catch more in short time they starts and experiments tons ideas 

So just walk on simple way 

 
Amir Yacoby:
You are wrong.
C is not correct.
The probability of 2 loss is calculated before you take the trades, and then it is right.
But in C you change that, and you say that you already did a trade and you know that it lost.
So  now we have just one deal left to calculate the chance for and the chance for that is 1/3 as any other deal.

great, i see what you mean
 

I came up with a similar strategy. At least I think what you are looking for. But I trade with the same strategy that I know works for me. But I have it working in only SELL mode in strategy 1 and only BUY mode in strategy 2. So far it's been doing a great job, but the EA I am using was already working great in the regular way by changing directions. But with my new strategy one of them is making wins while the other one is trying to solve the market going wrong.

So I came up with something very simple, but also very great. Been running it like this for around a month and so far every day have been profitable. Not one single day with a loss. But my system also involves a grid and a special type of martingale, but in my eyes it's not as risky as the usual strategies, since I am running it in two directions. And I have open trades all the time.

 
More clean and more simple strategy is the best for me. I don't need a lot of indicators attached on my chart.
 
Mrluck07:

If someone has 2 strategies with 70% win rate each one, and they have very different time entries , what's the probability of having 2 loss trades at the same sequence of trades?

example

sequence of trades:

 X  I   I   I  X  I   I  X  I   I     strategy 1

I   X  X   I   I  I  I   X  I   I   strategy 2

What's the probability of something like  X X loss in both strategies at the same sequence,to happen?

thanks for any help

Just some noob probability calculation. Lets assume there are 10n trades in each case. Out of which 7n are wins and 3n are losses. Total no. of various possibilities are 10n C 3n. in each case. If there is to be no overlap of loosing trades then number of such possibilities in second case is 5n C 3n. So total probability of having two losses in same sequence = 1-((5n C 3n)/(10n C 3n)) .


After some more simplification this equals. 1 - {(2n+1)(2n+2)...(5n)}/{(7n+1)(7n+2)...(10n)}

if n =5   then the probability of overlapping losses = 1- (6x7x ... x24x25)/(36x37x...x49x50)

if n =10   then the probability of overlapping losses = 1- (21x22x ...x50)/(71x72 x ....x100)


if n -> infty, it can easily seen that this probability tends to 1. So for large number of trades. The probability of losses on same sequence is almost sure.

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