What is the probability for PnL >= 0 for a 1DTE/ATM_IV300 trade with the following PnL diagram? a) about 66% b) about 32% c) about 88% d) about 53% e) it's much different from the above ones What formula do you use for computing the said probability? Info: the trade resembles a CoveredCall (S=10) or its synthetic, ShortPut, as both have a similar looking outcome (curve) at expiration (the orange line on the PnL chart below). Wikipedia: "The payoff from selling a covered call is identical to selling a short naked put. Both variants are a short implied volatility strategy." https://optioncreator.com/st24lqk or https://optioncreator.com/stzqa0r
My calc gives an unbelievably high 88%, but I'm not sure, maybe buggy, hence asking you the crowd for help...
I just performed a Monte Carlo simulation, and the winner is.... 66 % ! Problem solved! Case closed. PM me if you have questions regarding the MC simulation (I used standard C++ w/o any external library, that's C++11 or higher std).
Hmm. after thinking some more on the problem, I get the impression that p should be higher than the said 66%. Maybe the MCsim was not correct yet. The question is open again. If someone gets a different value than 66% then lets talk about it...
Underlying is not necessary since we know the outcome at expiration (ie. the posted PnL chart). The B/E point is 9.3742. We need the p for greater equal that stock price.
you know the BE. What if 1 DTE the underlying is $10 below BE? What if 1 DTE the underlying is $10 above BE? Don't you think that matters to answer your question?
I'm not sure your MC simulation takes into account the expectancy of the stocks going down well below break even. I'm not a fan of these trades and feel the risk reward is very low.
At close on 1DTE the stock price is the shown initial $10 and IV is 300. For the close on the next day (ie. expiration), p for Sx >= BEP 9.3742 is sought. If one does the "usual calculation" then one gets the said 66%, but the usual calc is maybe insufficient. Need to find the correct solution to this math problem... The devil is in the detail, as usual.