Are you sure??? Did you try the Bayes formula? Where do you think is the trick? Can you capitalize on it?
all prior and conditional likelihoods are 50%, so the final conditional is 50% as well. it is like coin tossing. there is no probability different from random. there is no sequence ... it is repetition of identical trials. or i am really missing something here ... but i have no math background, so what do know? enlighten me. i would say: P(A) = 50% P(B) = 50% P(B/A) = 50% so P(A/B) is 50% as well. where is my flaw?
and: NO way of capitalizing on it. random in random out. no moderator know how entering on any level, no shift of any odds.
Exactly, when two losing systems are simultaneously in a drawdown - which they will be certainly at some point, since they're losing systems. That said, I imagine trading the equity curve can be a very profitable strategy (I've only briefly looked into the subject).
well, if your losing systems have streaks, meaning they show significant positive auto correlation, then you might be able to build a positive strategy out of two of which both are not be able to make it on their own as a stand alone. the point is that the P()s deviate from 50% once you have positive auto corr ...
A hint: Number 2 and 3 result in "0" loss/gain. So you can reduce the problem to this 3 outcomes: gain, loss, break even with 33% probability each. Similar to the "boy/girl problem" in the "twins" paradox http://en.wikipedia.org/wiki/Boy_or_Girl
"... If you knew that this stock has in fact gained $1 a share (moderator affect) what is the probability that this stock will gain another $1 a share?" that was your initial question. i stick to my answer of 1/2, since it is repetition independent trials. i see no way how this could go to 2/3. your assumption that the stock makes +1 in t1 does not cancel out just one path (-1 -1) but two (-1 -1 AND -1 +1). and if you use the bayesian formula you cannot treat the two paths with the end result of zero as one. otherwise the conditional P gets it wrong, which must assume that in t1 a +1 happened.
and the comparison to the twins does not hold water, since you cancel out not only Girl Girl but Girl Boy either. your assumption is that the first child is a Boy. and your question is: is the next child a boy. and the clear answer (here order matters) is: 1/2. IMHO.
my point is: since your assumption cancels out 50% of the paths you CANNOT throw cases 2 and 3 in one basket.
i do not see how the result 0 can have 1/3 probability with two paths from four leading there. in my eyes 0 has 50% chance, the others 25%. you have four outcomes, not three. since it is about conditional probs you cannot aggregate 2 and 3.