Phil,
Your approach of simplicity to this question is slightly skewed from a statistical point of view... allow me to shed some light.
The term "out of" corresponds to probability, which is defined as a number from 0 to 1 inclusive, usually expressed as a fraction, which is the ratio of the number of chances of a specific event to the total number of chances possible. In our common day language, it is more commonly spoken as a percent between 0% and 100%, inclusive.
The term Odds is invariably mistaken for the word probability. In fact, in the statistics world we call it the 'slang' word for probability. Odds are expressed as the number of chances for (or against) versus the number of chances against (or for).
So, in your die example, since there is a total of six outcomes... we would say there are 2 outcomes against you (ITM) and 4 outcomes for you (OTM). Based on your analogy of a single outcome of a die ...
The odds in favor of you being ITM, would be 2:4
The odds against you being ITM, would be 4:2
The odds in favor of you being OTM, would be 4:2
The odds against you being OTM, would be 2:4
The odds in favor of you not being ITM would be 4:2
The odds against you not being ITM would be 2:4
The odds in favor of you not being OTM, would be 2:4
The odds in against you not being OTM, would be 4:2
I hope this gives you some ammunition to use with your kids next time they ask you a question. Just answer them in eight equivalent statements...
As for Probability... "2 out of 6" as described in Phil's example, lends itself to probability, rather than odds. 2/6 would equate to approximately .33 or about 33%. The problem here being... the die example assumes a constant distribution of events or an equally likely outcome of each event, rather than the Expected probability of the event, Mu.
Unfortunately, the distribution of success and failure of options prices is not evenly distributed. Moreover, it's not normally distributed either....
So in response to "rd's" question earlier.... more risk or less risk... indepent events, dependent events, mutually exclusive or not, is really beyond the scope our this forum.
In short... the risk is dependent, and the probabilility of success is inclusive of that risk. What I mean to say... is if you define the risk to be what your maximium loss could incur... (whether you account for both sides of the condor or just one), that total in $$ is defineable, (spread - credit). The direction for loss is directional... one way. Where as having two credit spreads (IC) in place... also has defined risk (spread - credit), but is bidirectional and has two possible failing outcomes.
Since the distribution of the option prices is not normally distributed... integrating the current market skew may yeild a better approximation than simple independent probability.
I would suspect (guessing) the expected probability of success (defining success to be OTM at expiration) whether single tailed or double tailed with respect to IC's would differ less than less than 5% from each other.
This sounds like a weekend project...
Hope this helps to really confuse you... that's what us Math people do... we just don't tell anyone.... job security.
Murray
Quote from optioncoach:
I think the approach is that each side has a 10% chance of being in the money so at anytime your entire position has a 20% chance of having a short option in the money. I think someone best explained it as rolling a die where you lose if it is a 1 or a 6. Although only one can come up in one roll, the odds are really 2 out of 6 that you will have a loss on the entire roll.
Phil