Quote from dagnyt:
I'm sure you understand this, but it's worth mentioning: After you have two consecutive losing months, the probability of losing the next month is UNCHANGED.
Thus, don't feel overconfident that the 3rd month is going to work out well.
Edit: I see someone beat me with a similar response.
Mark
Yes I do know that the single option probability remains unchanged in a random system Mark - thanks. But also note the price volatility changes each and every option period. This complicates the probabilities even further. So I just use averages.
Observe that a very unlucky person in theory could lose 100% of the time on any option or equity he every bought or sold no matter the quality or risk of his "investment". We just do not see it that often in practise (or maybe such losers keep it to themselves and don't tell anyone and that explains why its not seen).
The probability I was expressing was the probability of losing 3 or more times in a row for a static or average win probability. That is simply calculated by taking the loss rate (100% - win rate) multiplied by itself for each attempt (e.g. 90% win IC win rate [if a real valid probability] would result in the following probability of 4 consecutive chances of losing in a row: .1 x .1 x .1 x .1 = .0001 = .01% probability) . Is it possible to have 4 or 1000 or more runs of losses with an IC win rate of 90%? Yes it is. Is it probable - NO, it is very improbable. Do we see it often? No, since few have the financial balls to do it and since after too many consecutive bad luck runs there is not enough demand for the products (over supply) or enough free market money available to make a market and net credit probably trends toward zero to compensate for over supply ...

TrendSailor
