Probability for PnL >= 0 for 1DTE

What is the probability for PnL >= 0 for a 1DTE/ATM_IV300 trade with the following PnL diagram?

  • about 66%

  • about 32%

  • about 88%

  • about 53%

  • it's much different from the above ones

Results are only viewable after voting.
I soon will present the results of my research on the problem of this topic in this thread.
Spoiler: all 3 probability results are correct at the same time: 88%, 66%, 53% :)
It's about the correct interpretation of what each exactly means & describes! As said the devil is in the details...
 
Quanto said:
Are you saying you know the correct answer to the question in this topic?
Which one is it?
:)
I wonder why you always generalize, instead of being concrete, ie. being exact.
To be honest, I doubt you know the answer to the question in the topic; your motives are nefarious: always trying to make me publicly down. Seems you are following an agenda or orders...
You think you are something better b/c you are a senior and veteran in the markets, and a sponsor here, and me "just" a user who needs to obey your sicko world-view of slavery...
Nope! I'm a free man and don't need to obey you or any such sicko in the World.
I'm following the site rules. If you think such sponsors like you should have the right to ban users then you can try to add it to the site rules, so that then everybody knows that one has to obey the sponsors, even sickos.
I hate group-thinking, and am not interested to join a sicko group to get any help.
So, forget about you speaking for a group; speak just for yourself as an individual.
Since you have no respect for me, then don't expect any respect from me from now on.

If you continue your observed negative behaviour against me, then I'll have to block you, since you steal my time with your such off-topic postings in my such important threads.
More...

Pit viper... lol
 
Quanto said:
I soon will present the results of my research on the problem of this topic in this thread.
Spoiler: all 3 probability results are correct at the same time: 88%, 66%, 53% :)
It's about the correct interpretation of what each exactly means & describes! As said the devil is in the details...
More...
Solution to the problem in the OP (p for CC payoff at expiration).
There are at least 3 probabilities, each having a different meaning:
Code: 
1) pWin_BEP              = 0.659665 (66%) -->  R3 = 1.938281
2) pWin_BEP+Pot          = 0.539563 (54%) -->  R3 = 1.171849
3) pWin_BEP+Pot+Weighted = 0.882536 (88%) -->  R3 = 7.513246
All 3 operate right of the Break/Even Sx point (BEP).

R3 ("Reward Risk Ratio") means reward expectancy for every $1 risked. Formula: R3 = p / (1 - p).
R3 replaces all the many different (and wrong) Expectancy formulas out there in the wild.

Pot is the money that was on the table (depends on the # of simulation runs, but we are interested only in the percentages).

The case #3 is the most advanced method, but also complicated to calculate (requires a numerical integration method).
It interprets the payoff curve as "probability-weighted": payoffs near the initial S have a higher p than those further away.
IMO, the R3 of it is the most realistic one. :)

Attached also pgm output, but for experts only. Uses long lines --> use "less -Sn filename.txt" when viewing in Linux terminal.
 

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