a trading problem for mathematicians

[abattia]

Quote from tropicalknight:

This seems over-complicated and I do not see the logic in multiplying the P(win)'s.

The new number of trades = N x P(A gives signal) x P(B gives signal)

I don't see how this is disputable.

Hmm... I think you’re right.

So, thanks!

Let me see if I can redeem myself reformulating one last time:

New win rate = (new number of winning trades) / (new number of all trades)


New number of winning trades
= N x P(A gives signal) x P(B gives signal | A gave signal) x P(A wins) x P(B wins | A won)
= 0.6 x N x P(A gives signal) x P(B gives signal | A gave signal)

(NB this is because P(B wins | A won) = 1, see * below)


New number of all trades
= N x P(A gives signal) x P(B gives signal | A gave signal)


So, New win rate
= (new number of winning trades) / (new number of all trades)
= 0.6


Which I find confusing (if it’s correct).
a) So what’s the benefit of getting confirmation from another signal, if you already have one signal telling you to enter a trade?
b) What happens in the case that you have two signals that each lead to different win rates, say 50% and 60%. Depending on which one you treat as A or B above, you will get a different final result, either 0.5 or 0.6.

= = = = = = =
* According to the OP, both signals are for the same instrument, both are for entering a long only trade, both are for an entry at the open of a bar, and both resulting trades (if traded separately rather than together) would exit at the same close. Therefore, on occasions when both signals fire off together, the only possible outcomes are either that both win, or that both lose.

Thus,
P(B wins | A won) = 1

=======

Over and …

… ah, forget it, no-one believes me anyway when I say that any more..

 

[asiaprop]

gosh I think some of you people are seriously too dumb or simply lack reading capabilities to discuss a topic intelligently.

Long time ago it upset me to waste time with people who were really not on par to comprehend even simple topics. Now I smile because I know that such people partly feed me.

Back to the topic: Could you PLEASE stop talking about p/l, this is not the topic of this discussion, you are missing the point completely.

Quote from Rehoboth:

Well then I dont think you understand what you are asking without giving a p/l ratio per strat.

Who is to say you make 10 bucks per win but lose 10k per loss. Still a win rate of 60% but a giant sucky strat.

 

[asiaprop]

absolutely second that!!!

Quote from u21c3f6:

One should really be sure they understand something before they call somone else a bozo.

With all due respect, you are not understanding the problem presented in this thread.

Joe.

 

[asiaprop]

the probability is a measure that describes the likelihood of an event occurring, while you are talking about the realization of an event. Two completely different things.

This website really is the perfect realization of the 95% losers vs 5% who take from the majority. I am out of here, this is really getting too dumb.

Quote from SnakeEYE:

There is no such thing as probability in trading and there is no such thing as probability diapason.The 'probability' is always 100%.For 2 different events you always have 100% 'probability' outcome at the same time.

Are you that dumb to get it

When you are flying,for e.g., there are 2 different outcomes which are both 100% ,either you crash, or you`ll reach your destination.

 

[Texas]

Quote from jhiro:

No, you bozo's each have half of the story.. back a horse in at 60%, if it wins what is the chance of your very next bet being a winner if you back another 60% chance?

This means you would only get the two signals lining up together 36% of the time

So why would you trade a 60% system in for one that still only has a 60% strike rate but generates the signals almost half as often?

What ya gotta ask is why is this debated for so long isn't it clear right

 

[Alex123]

This is both hilarious and sad. The question was answered correctly in the 6th post on the 1st page of this thread by pbj and it's now on page 22…

 

[SnakeEYE]

Quote from asiaprop:

the probability is a measure that describes the likelihood of an event occurring, while you are talking about the realization of an event. Two completely different things.

Why even begin with probability?

 

[trend2009]

After more thoughts, I believe the answer is 69%. A few posts earlier correctly gave the answer. Here is my reasoning, which I wish to make it easier for people to understand how 69% is derived.

1) suppose there is a large pool of red and blue balls. red indicates for winning ball, blue indicates for losing ball.
2) system A is applied to the pool and takes out 100 balls. Now we have 60 red balls, and 40 blue balls.
3) system B is then applied to this 100 balls taken out by system A. System B tries to pick out all red balls. since system B has 60% correctness, thus:

3a) for the 60 red balls, system B can correctly pick out 60*0.6=36, the remaining 60*0.4=24 will be incorrectly discarded by system B.

3b) for the 40 blue balls, system B can correctly identify 40*0.6=24 as blue balls, thus this 24 blue balls will be discarded by system B. the remaining 40*0.4=16 blue balls will be incorrectly identified by system B as red balls, thus picked out by system B.

in the end, system B picks out 36 red balls, and 16 blue balls.

the red balls, ie, winning rate=36/(36+16)=69%.

Quote from swingtrader123:

Here is my logic: A game of flipping coins. You have two coins to flip. If either coin turns up heads, you win. You flip the first coin, and its a head, you win. The second coin flip does not matter. Your first win has a 50% probability of winning. Second try, your first coin flip is a tail. You flip the second coin. Now what is the probability of a head on the 2nd coin? It's 50%. So, you still have a 50% chance of winning. Mathematically, for two coins, you have 2 chances to win, but with 4 possible outcomes. So, the probability is 2/4 or 50%.

The key here is that only one system must be correct. So, the system with the highest probability is the probability of success. The second system's probability does not contribute or remove any probability to the first.

 

[trend2009]

the purpose of this thread is to demonstrate the importance of signal confirmation from uncorrelated sources. For two methods of 60% correctness, the combination has a winning rate of 70%. for two methods of 70% correctness, the winning rate of combination is as high as 84%.

then does holy grail exist if you have 10% uncorrelated signals? the problem here is that with more confirmation, you have fewer signals to trade, thus you lose the large pool to make your holy grail meaningful.

Quote from trend2009:

After more thoughts, I believe the answer is 69%. A few posts earlier correctly gave the answer. Here is my reasoning, which I wish to make it easier for people to understand how 69% is derived.

1) suppose there is a large pool of red and blue balls. red indicates for winning ball, blue indicates for losing ball.
2) system A is applied to the pool and takes out 100 balls. Now we have 60 red balls, and 40 blue balls.
3) system B is then applied to this 100 balls taken out by system A. System B tries to pick out all red balls. since system B has 60% correctness, thus:

3a) for the 60 red balls, system B can correctly pick out 60*0.6=36, the remaining 60*0.4=24 will be incorrectly discarded by system B.

3b) for the 40 blue balls, system B can correctly identify 40*0.6=24 as blue balls, thus this 24 blue balls will be discarded by system B. the remaining 40*0.4=16 blue balls will be incorrectly identified by system B as red balls, thus picked out by system B.

in the end, system B picks out 36 red balls, and 16 blue balls.

the red balls, ie, winning rate=36/(36+16)=69%.

 

[fishing]

OMG please read asiaprop's posts...

i dont know what's so hard to understand..