Conditional probabilities when trading options and forecasting

#1
Dear Statistics-Gurus,

I am trying to figure out how to "combine" two dependent probabilities, namely the probability to forecast the market direction wrong P(wrongFC)=35% and the probability of an option to be in-the-money ITM (only if the option is ITM I am losing money). The option can only be ITM when the forecast is wrong. In the options world, the probability of an option to be ITM depends on the strike used, and for this example we pick a 1 standard-deviation-out option, which has a probability of P(ITM)=32% (and P(OTM)=68% - out-of-the money).

So we know:
Probabilities of forecast: P(wrongFC)=35% (and accordingly P(correctFC)=65%)
Probabilities of option used: P(ITM)=32% (and accordingly P(OTM)=68%)

In addition, the following conditions apply (I am using the "given that" bar notation):
1) When my forecast is correct (the price has moved above initial price) then I know that my option will be OTM all the time:
P(OTM|correctFC)=100%

2) When my forecast is wrong (price has moved below initial price) then there are two possible outcomes for the option, ITM or OTM. Now I would like to calculate these conditional probabilities (the probabilities within the red circle in the chart):
P(OTM|wrongFC)=?%
P(ITM|wrongFC)=?%

3) In the end, what I am really interested in is the "combined" probability Pc(OTM)=?%, which will tell me the probability to be profitable after "combining" the two probabilities (or equally the "combined" probability Pc(ITM)=?% to determine how often I will lose).

The typical non-statistician answer to this question is (and that's the answer you get from all those very smart traders :shakehead): You have both probabilities on your side, as you are a) right in 65/100 cases with your forecast and b) you have a 68/100 probability of success because of you use a 1-standard-deviation-OTM option. Let's see if those smart traders are right (I hope they are :eek: ).

I've spend quite some time going back and forth on this problem and by using graphical alterations and I even came up with an answer, but I am not sure if I got this right. Here my preliminary result: Pc(ITM)=22.4% and Pc(OTM)=77.6%

Any help would be greatly appreciated. Many thanks.

spjuliman