Hmm... I think youâre 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 itâs correct). a) So whatâs 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..
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.
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.
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 22â¦
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%.
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.